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Exercise 10.2

1. Compute the magnitude of the following vectors:

Solun:- Given {"code":"\\begin{align*}\n{\\overrightarrow{a}}&={\\hat{i}+\\hat{j}+\\hat{k}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"1-0-0-0","type":"align*","ts":1604834425699,"cs":"/CX5aokk6JqGpwIVzKAEwQ==","size":{"width":100,"height":21}}

We know that:-

The magnitude of a vector is

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{a}}&={a_{1}\\hat{i}+a_{2}\\hat{j}+a_{3}\\hat{k}}\t\n\\end{align*}","id":"1-1-2","ts":1604835346314,"cs":"EoJYo1SLtygo7qbQScO+9g==","size":{"width":144,"height":21}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0","type":"align*","code":"\\begin{align*}\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{\\left(a_{1}\\right)^{2}+\\left(a_{2}\\right)^{2}+\\left(a_{3}\\right)^{2}}}}\t\n\\end{align*}","ts":1604834564469,"cs":"AkDh2N/qWyYonRhI4615LA==","size":{"width":202,"height":32}}

{"type":"align*","code":"\\begin{align*}\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{\\left(1\\right)^{2}+\\left(1\\right)^{2}+\\left(1\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{3}}}\t\n\\end{align*}","id":"2-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1604834627521,"cs":"u0hvNhnJ+7eQeaav3vIesQ==","size":{"width":181,"height":64}}

Given vector is:- {"id":"1-0-1-0","type":"align*","code":"\\begin{align*}\n{\\overrightarrow{b}}&={2\\hat{i}-7\\hat{j}-3\\hat{k}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1604834681204,"cs":"4XYXelQmsYMfNZ3OrydqNQ==","size":{"width":124,"height":25}}

{"code":"\\begin{align*}\n{\\left|\\overrightarrow{b}\\right|}&={{\\sqrt[]{\\left(2\\right)^{2}+\\left(-7\\right)^{2}+\\left(-3\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{b}\\right|}&={{\\sqrt[]{4+49+9}}}\\\\\n{\\left|\\overrightarrow{b}\\right|}&={{\\sqrt[]{62}}}\t\n\\end{align*}","id":"2-1-1-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1604834766028,"cs":"SqhymwIJSBXHEfHJwlso+A==","size":{"width":206,"height":116}}

Given vector is:- {"id":"1-0-1-1","type":"align*","code":"\\begin{align*}\n{\\overrightarrow{c}}&={\\frac{1}{{\\sqrt[]{3}}}\\hat{i}+\\frac{1}{{\\sqrt[]{3}}}\\hat{j}-\\frac{1}{{\\sqrt[]{3}}}\\hat{k}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1604834833989,"cs":"QPQZsdfwFL4lNo1R9c3nPg==","size":{"width":185,"height":36}}

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-1-1","code":"\\begin{align*}\n{\\left|\\overrightarrow{c}\\right|}&={{\\sqrt[]{\\left(\\frac{1}{{\\sqrt[]{3}}}\\right)^{2}+\\left(\\frac{1}{{\\sqrt[]{3}}}\\right)^{2}+\\left(\\frac{-1}{{\\sqrt[]{3}}}\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{c}\\right|}&={{\\sqrt[]{\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}}}}\\\\\n{\\left|\\overrightarrow{c}\\right|}&={{\\sqrt[]{1}}=1}\t\n\\end{align*}","ts":1604834903193,"cs":"lLlpTX5vf6IVwvz+5ihuvg==","size":{"width":277,"height":124}}

2. Write two different vectors having same magnitude.

Solun:- Let 

{"code":"\\begin{align*}\n{\\overrightarrow{a}}&={\\hat{i}+\\hat{j}+\\hat{k}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"1-0-0-1","type":"align*","ts":1604834425699,"cs":"jxEef6Ice1A8dszk4MGzxQ==","size":{"width":100,"height":21}}  and {"id":"3-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\overrightarrow{b}}&={\\hat{i}-\\hat{j}-\\hat{k}}\t\n\\end{align*}","ts":1604835084547,"cs":"UP8wSao0F+nIirPMFXMjUQ==","size":{"width":100,"height":25}}

We know that:- 

The magnitude of a vector is

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{a}}&={a_{1}\\hat{i}+a_{2}\\hat{j}+a_{3}\\hat{k}}\t\n\\end{align*}","id":"1-1-2","ts":1604835346314,"cs":"EoJYo1SLtygo7qbQScO+9g==","size":{"width":144,"height":21}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-1-0","type":"align*","code":"\\begin{align*}\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{\\left(a_{1}\\right)^{2}+\\left(a_{2}\\right)^{2}+\\left(a_{3}\\right)^{2}}}}\t\n\\end{align*}","ts":1604834564469,"cs":"+0NKonqxE0Ku2adKcV3uVQ==","size":{"width":202,"height":32}}

{"code":"\\begin{align*}\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{\\left(1\\right)^{2}+\\left(1\\right)^{2}+\\left(1\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{3}}}\t\n\\end{align*}","id":"4-0-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1604835151018,"cs":"JLJxfrR1afRC3/dQIobqxw==","size":{"width":181,"height":64}}and {"id":"4-1-0","type":"align*","code":"\\begin{align*}\n{\\left|\\overrightarrow{b}\\right|}&={{\\sqrt[]{\\left(1\\right)^{2}+\\left(-1\\right)^{2}+\\left(-1\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{b}\\right|}&={{\\sqrt[]{3}}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1604835191731,"cs":"DCuPt/BCci+pilRJQO+mjw==","size":{"width":206,"height":76}}

3. Write two different vectors having same direction.

Solun:- Let 

{"code":"\\begin{align*}\n{\\overrightarrow{a}}&={\\hat{i}+\\hat{j}+\\hat{k}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"1-0-0-1","type":"align*","ts":1604834425699,"cs":"jxEef6Ice1A8dszk4MGzxQ==","size":{"width":100,"height":21}}  and {"code":"\\begin{align*}\n{\\overrightarrow{b}}&={3\\hat{i}+3\\hat{j}+3\\hat{k}}\t\n\\end{align*}","type":"align*","id":"3-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1604835300198,"cs":"qDugg7gYY2/CTvPceQr2Cg==","size":{"width":124,"height":25}}

We know that:- 

The Direction cosine of vector

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{a}}&={a_{1}\\hat{i}+a_{2}\\hat{j}+a_{3}\\hat{k}}\t\n\\end{align*}","id":"1-1-2","ts":1604835346314,"cs":"EoJYo1SLtygo7qbQScO+9g==","size":{"width":144,"height":21}} is

{"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"5-0","code":"\\begin{align*}\n{\\hat{a}}&={\\frac{\\overrightarrow{a}}{\\left|\\overrightarrow{a}\\right|}}\t\n\\end{align*}","ts":1604835507720,"cs":"8l+ucAsEJWG56M4oI8RDIw==","size":{"width":62,"height":56}}

Then 

{"id":"5-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\hat{a}}&={\\frac{\\hat{i}+\\hat{j}+\\hat{k}}{{\\sqrt[]{3}}}}\t\n\\end{align*}","ts":1604835564186,"cs":"ySUKQCHATd8S0b7IiUE9Ew==","size":{"width":98,"height":42}}and {"code":"\\begin{align*}\n{\\hat{b}}&={\\frac{3\\left(\\hat{i}+\\hat{j}+\\hat{k}\\right)}{{\\sqrt[]{27}}}=\\frac{\\hat{i}+\\hat{j}+\\hat{k}}{{\\sqrt[]{3}}}}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"5-1-1","ts":1604835620760,"cs":"kheXumz+EAXmmRO/JBnsqQ==","size":{"width":216,"height":52}}

4. Find the values of x and y so that the vectors {"code":"$2\\hat{i}+3\\hat{j}$","id":"6-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1604835714974,"cs":"kBrpq1O5dtpIDCeMN2avdQ==","size":{"width":50,"height":18}} and {"font":{"family":"Arial","size":10,"color":"#000000"},"code":"$x\\hat{i}+y\\hat{j}$","type":"$","id":"6-1","ts":1604835749003,"cs":"+EGx14MDg5m60mgJbqI/rQ==","size":{"width":52,"height":18}} are equal.

Solun:- We know that for the equal vectors:-

{"id":"7","code":"$\\overrightarrow{a}=\\overrightarrow{b}$","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604835799455,"cs":"3IFNbY8hEV5uqwrdh4Lo5Q==","size":{"width":52,"height":22}}

{"code":"$2\\hat{i}+3\\hat{j}$","id":"6-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1604835714974,"cs":"kBrpq1O5dtpIDCeMN2avdQ==","size":{"width":50,"height":18}} = {"font":{"family":"Arial","size":10,"color":"#000000"},"code":"$x\\hat{i}+y\\hat{j}$","type":"$","id":"6-1","ts":1604835749003,"cs":"+EGx14MDg5m60mgJbqI/rQ==","size":{"width":52,"height":18}}

Compare both sides:

x = 2 and y = 3.

5. Find the scalar and vector components of the vector with the initial point (2, 1) and terminal point (-5, 7).

Solun:- Let A(2, 1) and B(-5, 7) then

{"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\left(x_{2}-x_{1}\\right)\\hat{i}+\\left(y_{2}-y_{1}\\right)\\hat{j}}\\\\\n{\\overrightarrow{\\text{AB}}}&={\\left(-5-2\\right)\\hat{i}+\\left(7-1\\right)\\hat{j}}\\\\\n{\\overrightarrow{\\text{AB}}}&={-7\\hat{i}+6\\hat{j}}\t\n\\end{align*}","id":"8-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604837083168,"cs":"R2f4wOXt85R0W9n1uT7SoA==","size":{"width":201,"height":94}}

Scalar Components are:- -7 and 6

Vector Components are:- {"id":"8-0","aid":null,"type":"gather*","code":"\\begin{gather*}\n{-7\\hat{i}+6\\hat{j}}\t\n\\end{gather*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1605094644928,"cs":"dP3cIkpqomco9otnPbvr7g==","size":{"width":60,"height":16}}

6. Find the sum of vectors {"id":"67","aid":null,"font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\overrightarrow{a}=\\hat{i}-2\\hat{j}+\\hat{k}$","type":"$","ts":1605094759640,"cs":"lJP+2X22fwpmqMTb2omdRw==","size":{"width":104,"height":18}},{"code":"$\\overrightarrow{b}=-2\\hat{i}+4\\hat{j}+5\\hat{k}$","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"69","aid":null,"ts":1605095072589,"cs":"XdB0r83RRhf9H2KWkGE+YQ==","size":{"width":132,"height":22}}, and {"font":{"family":"Arial","size":10,"color":"#000000"},"id":"70","code":"$\\overrightarrow{c}=\\hat{i}-6\\hat{j}-7\\hat{k}$","type":"$","aid":null,"ts":1605094873181,"cs":"K4pAkof8UchBhoZHWbmeZg==","size":{"width":112,"height":18}}.

Solun:- Sum of Vectors = sum of components of vectors

{"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{a}+\\overrightarrow{b}+\\overrightarrow{c}}&={\\left(\\hat{i}-2\\hat{j}+\\hat{k}\\right)+\\left(-2\\hat{i}+4\\hat{j}+5\\hat{k}\\right)+\\left(\\hat{i}-6\\hat{j}-7\\hat{k}\\right)}\\\\\n{\\overrightarrow{a}+\\overrightarrow{b}+\\overrightarrow{c}}&={-4\\hat{j}-\\hat{k}}\t\n\\end{align*}","id":"72","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"ts":1605095201665,"cs":"s7r99ZWyyGTFxyAQsyRppw==","size":{"width":441,"height":60}}

7. Find the unit vector in the direction of the vector {"id":"11","code":"$\\overrightarrow{a}=\\hat{i}+\\hat{j}+2\\hat{k}$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1604839626640,"cs":"HRJ+O/ZRTxUGdg6EA4w6Rg==","size":{"width":108,"height":20}}

Solun:- Given {"id":"11","code":"$\\overrightarrow{a}=\\hat{i}+\\hat{j}+2\\hat{k}$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1604839626640,"cs":"HRJ+O/ZRTxUGdg6EA4w6Rg==","size":{"width":108,"height":20}}

We know that:- 

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"12-0","code":"\\begin{align*}\n{\\hat{a}}&={\\frac{\\overrightarrow{a}}{\\left|\\overrightarrow{a}\\right|}}\\\\\n{\\hat{a}}&={\\frac{\\hat{i}+\\hat{j}+2\\hat{k}}{{\\sqrt[]{\\left(1\\right)^{2}+\\left(1\\right)^{2}+\\left(2\\right)^{2}}}}}\\\\\n{\\hat{a}}&={\\frac{\\hat{i}+\\hat{j}+2\\hat{k}}{{\\sqrt[]{6}}}}\\\\\n{\\hat{a}}&={\\frac{1}{{\\sqrt[]{6}}}\\hat{i}+\\frac{1}{{\\sqrt[]{6}}}\\hat{j}+\\frac{2}{{\\sqrt[]{6}}}\\hat{k}}\t\n\\end{align*}","ts":1604839856296,"cs":"VDpHE8QIPCG+uHyEu10Siw==","size":{"width":176,"height":205}}

8. Find the unit vector in the direction of vector {"type":"$","code":"$\\overrightarrow{\\text{PQ}}$","id":"13","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1604839998035,"cs":"DXIYzkInkGlVAA2d50kCxg==","size":{"width":22,"height":26}}, where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.

Solun:- Given P(1, 2, 3) and Q(4, 5, 6) then

{"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"8-1-1","code":"\\begin{align*}\n{\\overrightarrow{\\text{PQ}}}&={\\left(x_{2}-x_{1}\\right)\\hat{i}+\\left(y_{2}-y_{1}\\right)\\hat{j}+\\left(z_{2}-z_{1}\\right)\\hat{k}}\\\\\n{\\overrightarrow{\\text{PQ}}}&={\\left(4-1\\right)\\hat{i}+\\left(5-2\\right)\\hat{j}+\\left(6-3\\right)\\hat{k}}\\\\\n{\\overrightarrow{\\text{PQ}}}&={3\\hat{i}+3\\hat{j}+3\\hat{k}}\t\n\\end{align*}","ts":1604840246837,"cs":"SprNikggQ8ZyqI92LZVMTw==","size":{"width":288,"height":93}}

We know that:- 

{"type":"align*","id":"12-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\hat{a}}&={\\frac{\\overrightarrow{a}}{\\left|\\overrightarrow{a}\\right|}}\\\\\n{\\hat{\\text{PQ}}}&={\\frac{3\\hat{i}+3\\hat{j}+3\\hat{k}}{{\\sqrt[]{\\left(3\\right)^{2}+\\left(3\\right)^{2}+\\left(3\\right)^{2}}}}}\\\\\n{\\hat{\\text{PQ}}}&={\\frac{3\\hat{i}+3\\hat{j}+3\\hat{k}}{{\\sqrt[]{27}}}}\\\\\n{\\hat{\\text{PQ}}}&={\\frac{3\\left(\\hat{i}+\\hat{j}+\\hat{k}\\right)}{{\\sqrt[]{27}}}}\\\\\n{\\hat{\\text{PQ}}}&={\\frac{\\hat{i}+\\hat{j}+\\hat{k}}{{\\sqrt[]{3}}}}\t\n\\end{align*}","ts":1604840450441,"cs":"9xlYqR7pJH0YRMfT0qV0og==","size":{"width":185,"height":268}}

{"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\hat{\\text{PQ}}}&={\\frac{1}{{\\sqrt[]{3}}}\\hat{i}+\\frac{1}{{\\sqrt[]{3}}}\\hat{j}+\\frac{1}{{\\sqrt[]{3}}}\\hat{k}}\t\n\\end{align*}","id":"14-0","ts":1604840475136,"cs":"YQJcw8JqjXmXNex5cKKkSw==","size":{"width":192,"height":36}}

9. For given vectors, {"code":"\\begin{align*}\n{\\overrightarrow{a}}&={2\\hat{i}-\\hat{j}+2\\hat{k}}\t\n\\end{align*}","type":"align*","id":"15-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1604909242880,"cs":"lLwrdmdJSnTkfloql9X6TQ==","size":{"width":116,"height":21}} and {"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"16","code":"\\begin{align*}\n{\\overrightarrow{b}}&={-\\hat{i}+\\hat{j}-\\hat{k}}\t\n\\end{align*}","ts":1604909322667,"cs":"F0fOIaDAjPvOAkE0hXhb0A==","size":{"width":112,"height":25}}, find the unit vector in the direction of the vector {"code":"$\\overrightarrow{a}+\\overrightarrow{b}$","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"17-0","ts":1604909342205,"cs":"FHm4adzd5wElAte4Lp4+Ug==","size":{"width":50,"height":22}}.

Solun:- Given {"code":"\\begin{align*}\n{\\overrightarrow{a}}&={2\\hat{i}-\\hat{j}+2\\hat{k}}\t\n\\end{align*}","type":"align*","id":"15-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1604909242880,"cs":"Hvg5djrannXxzikR9Pn+tQ==","size":{"width":116,"height":21}} and {"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"16","code":"\\begin{align*}\n{\\overrightarrow{b}}&={-\\hat{i}+\\hat{j}-\\hat{k}}\t\n\\end{align*}","ts":1604909322667,"cs":"F0fOIaDAjPvOAkE0hXhb0A==","size":{"width":112,"height":25}}

{"id":"17-1-0-0-0","code":"\\begin{align*}\n{\\overrightarrow{a}+\\overrightarrow{b}}&={\\hat{i}+\\hat{k}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1604909616026,"cs":"+aH/GkbuhMSkiNV6+tyAGg==","size":{"width":108,"height":25}}

Let {"id":"17-1-0-1","code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\overrightarrow{a}+\\overrightarrow{b}}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1604909921999,"cs":"znvoeyoBaDYy0/RrWGp9XQ==","size":{"width":94,"height":28}}

{"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\hat{i}+\\hat{k}}\t\n\\end{align*}","id":"17-1-0-0-1","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1604909952615,"cs":"BJP65x6uW3shMNKTE0Z/Pw==","size":{"width":80,"height":28}}

We know that:- 

The magnitude of a vector is

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{a}}&={a_{1}\\hat{i}+a_{2}\\hat{j}+a_{3}\\hat{k}}\t\n\\end{align*}","id":"1-1-2","ts":1604835346314,"cs":"EoJYo1SLtygo7qbQScO+9g==","size":{"width":144,"height":21}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0","type":"align*","code":"\\begin{align*}\n{\\left|\\overrightarrow{a}\\right|}&={{\\sqrt[]{\\left(a_{1}\\right)^{2}+\\left(a_{2}\\right)^{2}+\\left(a_{3}\\right)^{2}}}}\t\n\\end{align*}","ts":1604834564469,"cs":"AkDh2N/qWyYonRhI4615LA==","size":{"width":202,"height":32}}

{"id":"17-1-1","code":"\\begin{align*}\n{\\left|\\overrightarrow{\\text{AB}}\\right|}&={{\\sqrt[]{\\left(1\\right)^{2}+\\left(1\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{\\text{AB}}\\right|}&={{\\sqrt[]{2}}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1604909993117,"cs":"HofYeNLIb/EonGpQYldlCg==","size":{"width":144,"height":86}}

We know that:- 

{"code":"\\begin{align*}\n{\\hat{a}}&={\\frac{\\overrightarrow{a}}{\\left|\\overrightarrow{a}\\right|}}\t\n\\end{align*}","id":"12-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1604910007795,"cs":"0kvRg4KkGzLlEaO0oYo1yg==","size":{"width":62,"height":56}}

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\hat{\\text{AB}}}&={\\frac{\\hat{i}+\\hat{k}}{{\\sqrt[]{2}}}}\t\n\\end{align*}","id":"18-0","ts":1604910116163,"cs":"dLjAzw7HBlAlep9VNid+mQ==","size":{"width":86,"height":42}}

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\hat{\\text{AB}}}&={\\frac{1}{{\\sqrt[]{2}}}\\hat{i}+\\frac{1}{{\\sqrt[]{2}}}\\hat{k}}\t\n\\end{align*}","id":"14-1-0","ts":1604910150656,"cs":"9BSOreA03oDJBOOsn8v0zA==","size":{"width":136,"height":36}}

10. Find a vector in the direction of vector {"type":"$","code":"$5\\hat{i}-\\hat{j}+2\\hat{k}$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"19-0","ts":1604914602773,"cs":"IH1UG5FdUxLcKvcmnjUTBg==","size":{"width":78,"height":18}} which has magnitude 8 units.

Solun:- Let given vector is {"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"19-1","code":"\\begin{align*}\n{\\overrightarrow{a}}&={5\\hat{i}-\\hat{j}+2\\hat{k}}\t\n\\end{align*}","ts":1604914818360,"cs":"3s1walupcCzIQErNCVYelw==","size":{"width":116,"height":21}}

We have to find a vector in the direction of {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"20-0","type":"$","ts":1604915030814,"cs":"fJtobm2RBdNRIS669j2IPA==","size":{"width":16,"height":17}} which has magnitude 8 units then

We know that:- 

{"code":"\\begin{align*}\n{\\hat{a}}&={\\frac{\\overrightarrow{a}}{\\left|\\overrightarrow{a}\\right|}}\t\n\\end{align*}","id":"12-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1604910007795,"cs":"0kvRg4KkGzLlEaO0oYo1yg==","size":{"width":62,"height":56}}

{"id":"18-1","code":"\\begin{align*}\n{\\hat{a}}&={\\frac{5\\hat{i}-j+2\\hat{k}}{{\\sqrt[]{\\left(5\\right)^{2}+\\left(1\\right)^{2}+\\left(2\\right)^{2}}}}}\\\\\n{\\hat{a}}&={\\frac{5\\hat{i}-j+2\\hat{k}}{{\\sqrt[]{30}}}}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1604915156236,"cs":"0ExKp/KV+I8eYqSHHMuiiQ==","size":{"width":170,"height":102}}

Let unknown vector is {"type":"align*","id":"21","aid":null,"code":"\\begin{align*}\n{\\overrightarrow{b}}&={\\left|b\\right|\\hat{a}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1604995309171,"cs":"EgtFKP6ldZkaItp541Hmxw==","size":{"width":58,"height":22}} 

{"id":"22","type":"align*","aid":null,"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\overrightarrow{b}}&={8\\left(\\frac{5\\hat{i}-\\hat{j}+2\\hat{k}}{{\\sqrt[]{30}}}\\right)}\t\n\\end{align*}","ts":1604995379262,"cs":"PRw5OUYovoWg+wSrGJ8nIg==","size":{"width":148,"height":45}}

{"code":"\\begin{align*}\n{\\overrightarrow{b}}&={\\frac{40}{{\\sqrt[]{30}}}\\hat{i}-\\frac{8}{{\\sqrt[]{30}}}\\hat{j}+\\frac{16}{{\\sqrt[]{30}}}\\hat{k}}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"id":"23","aid":null,"ts":1605095546080,"cs":"R0JvCQmlxi3qQjnIcAYF7A==","size":{"width":204,"height":36}}

11. Show that the vectors {"code":"$2\\hat{i}-3\\hat{j}+4\\hat{k}$","aid":null,"type":"$","id":"24","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1605002062479,"cs":"4z77YsodftM4IUrXkYEz7A==","size":{"width":84,"height":16}} and {"code":"$-4\\hat{i}+6\\hat{j}-8\\hat{k}$","aid":null,"id":"25-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1605002119065,"cs":"6s/CNnxT2kxGLKZ3FJMNXA==","size":{"width":96,"height":16}} are collinear.

Solun:- We know that:- if two vectors {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"27-0","type":"$","ts":1604915030814,"cs":"WCWljxqJkBqz1Jw73SiJbw==","size":{"width":16,"height":17}} and {"type":"$","code":"$\\overrightarrow{b}$","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"20-1-0-0","ts":1605002577270,"cs":"QLiUUenp5C9JXcAyLU2YeA==","size":{"width":13,"height":20}} are collinear if and only if there exists a non-zero scalar {"aid":null,"id":"26","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\lambda$","ts":1605002452870,"cs":"B1QInrieMWV0b8PPfRxuTw==","size":{"width":8,"height":10}} such that {"id":"28","font":{"color":"#000000","size":10,"family":"Arial"},"aid":null,"code":"$\\overrightarrow{b}=\\lambda \\overrightarrow{a}$","type":"$","ts":1605002618434,"cs":"n4JW/qwFk4QdOwPFsK6M4A==","size":{"width":58,"height":20}}

Let {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"27-1","type":"$","ts":1604915030814,"cs":"LDtCEMnFl+VFxtaplsgcTg==","size":{"width":16,"height":17}} = {"code":"$2\\hat{i}-3\\hat{j}+4\\hat{k}$","aid":null,"type":"$","id":"24","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1605002062479,"cs":"4z77YsodftM4IUrXkYEz7A==","size":{"width":84,"height":16}}  and {"type":"$","code":"$\\overrightarrow{b}$","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"20-1-0-1","ts":1605002577270,"cs":"ow3eXGLaTDAfyNNceS6opQ==","size":{"width":13,"height":20}} = {"code":"$-4\\hat{i}+6\\hat{j}-8\\hat{k}$","aid":null,"id":"25-1","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1605002119065,"cs":"VL33N/uGI6SQzNY4w6Gczw==","size":{"width":96,"height":16}}

{"type":"align*","aid":null,"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\overrightarrow{b}}&={-2\\left(2\\hat{i}-3\\hat{j}+4\\hat{k}\\right)}\\\\\n{\\overrightarrow{b}}&={-2\\overrightarrow{a}}\t\n\\end{align*}","id":"29","ts":1605002837886,"cs":"5hhtrarmO6Vot7UQyJiNDA==","size":{"width":156,"height":56}}

Here {"aid":null,"id":"26","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\lambda$","ts":1605002452870,"cs":"B1QInrieMWV0b8PPfRxuTw==","size":{"width":8,"height":10}} = -2

Hence {"aid":null,"id":"26","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\lambda$","ts":1605002452870,"cs":"B1QInrieMWV0b8PPfRxuTw==","size":{"width":8,"height":10}} is a non-zero scalar then {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"27-2","type":"$","ts":1604915030814,"cs":"Ddn2fPpQa0qrlTlBgN6uzQ==","size":{"width":16,"height":17}} and {"type":"$","code":"$\\overrightarrow{b}$","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"20-1-0-2","ts":1605002577270,"cs":"3J+Zrs7PRhP/znUmU3BIlA==","size":{"width":13,"height":20}} are collinear vectors.

12. Find the direction cosine of the vector {"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\hat{i}+2\\hat{j}+3\\hat{k}$","aid":null,"id":"32-0","ts":1605003377777,"cs":"v8y/HMotKb31vgTFSomYdg==","size":{"width":76,"height":16}}.

Solun:- Let {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"39-0","type":"$","ts":1604915030814,"cs":"nMl8HmpB1e14c+IcXmv7VA==","size":{"width":16,"height":17}} = {"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\hat{i}+2\\hat{j}+3\\hat{k}$","aid":null,"id":"32-1","ts":1605003377777,"cs":"Tyk5ZTZbjf3o7NtApaVqHA==","size":{"width":76,"height":16}}

Direction ratios in {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"40-0","type":"$","ts":1604915030814,"cs":"MkR9QIULbXUdSCYR7PM2wg==","size":{"width":16,"height":17}} is:- a = 1, b = 2, and c = 3

We know that:- direction cosine in the form of direction ratio is:-

{"type":"align*","id":"34-0","code":"\\begin{align*}\n{l}&={\\frac{a}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","aid":null,"font":{"color":"#000000","size":10,"family":"Arial"},"ts":1605003787237,"cs":"tLJaFl7mvm3lRlrIpTq+Jw==","size":{"width":121,"height":32}}, {"code":"\\begin{align*}\n{m}&={\\frac{b}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"35-0","type":"align*","aid":null,"ts":1605003808926,"cs":"lbZMycK7529zm8pAsMQLYg==","size":{"width":132,"height":36}}, and {"id":"36-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{n}&={\\frac{c}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","aid":null,"ts":1605003835284,"cs":"1o3i8WwHjepdLBnkFMo6LA==","size":{"width":126,"height":32}}

{"aid":null,"id":"37-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{l}&={\\frac{1}{{\\sqrt[]{{\\left(1\\right)}^{2}+{\\left(2\\right)}^{2}+{\\left(3\\right)}^{2}}}}}\\\\\n{l}&={\\frac{1}{{\\sqrt[]{14}}}}\t\n\\end{align*}","type":"align*","ts":1605003971671,"cs":"jt5hMIQAs5IfIe2VRbK3sQ==","size":{"width":162,"height":89}}

{"id":"35-1-0","code":"\\begin{align*}\n{m}&={\\frac{2}{{\\sqrt[]{{\\left(1\\right)}^{2}+{\\left(2\\right)}^{2}+{\\left(3\\right)}^{2}}}}}\\\\\n{m}&={\\frac{2}{{\\sqrt[]{14}}}}\t\n\\end{align*}","aid":null,"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1605004033082,"cs":"Nx9AdxH1wSRw4lP3ScdgKQ==","size":{"width":172,"height":89}}

{"aid":null,"id":"36-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{n}&={\\frac{3}{{\\sqrt[]{{\\left(1\\right)}^{2}+{\\left(2\\right)}^{2}+{\\left(3\\right)}^{2}}}}}\\\\\n{n}&={\\frac{3}{{\\sqrt[]{14}}}}\t\n\\end{align*}","type":"align*","ts":1605004062367,"cs":"rk8YF+7HopZ0I+2jI8E8Ow==","size":{"width":168,"height":89}}

13. Find the direction cosine of the vector joining the points A(1, 2, -3) and 

B(-1, -2, 1), directed from A to B.

Solun:- Given A(1, 2, -3) and B(-1, -2, 1)

Then {"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\left(x_{2}-x_{1}\\right)\\hat{i}+\\left(y_{2}-y_{1}\\right)\\hat{j}+\\left(z_{2}-z_{1}\\right)\\hat{k}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","aid":null,"id":"38-0-0-0","ts":1605004505641,"cs":"pYHBvc+qgUW0VWTA12ojCA==","size":{"width":284,"height":24}}

{"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\left(-1-1\\right)\\hat{i}+\\left(-2-2\\right)\\hat{j}+\\left(1+3\\right)\\hat{k}}\\\\\n{\\overrightarrow{\\text{AB}}}&={-2\\hat{i}-4\\hat{j}+4\\hat{k}}\t\n\\end{align*}","id":"38-0-1","font":{"family":"Arial","size":10,"color":"#000000"},"aid":null,"ts":1605004753366,"cs":"ZmYoyU4Tj7CQJgIHxmKIPA==","size":{"width":269,"height":56}}

Direction ratios in {"type":"$","font":{"family":"Arial","color":"#000000","size":10},"aid":null,"id":"41","code":"$\\overrightarrow{\\text{AB}}$","ts":1605004665932,"cs":"OqZgoENgOPY7i42eE1dRig==","size":{"width":20,"height":20}} is:- a = -2, b = -4, and c = 4

We know that:- direction cosine in the form of direction ratio is:-

{"type":"align*","id":"34-0","code":"\\begin{align*}\n{l}&={\\frac{a}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","aid":null,"font":{"color":"#000000","size":10,"family":"Arial"},"ts":1605003787237,"cs":"tLJaFl7mvm3lRlrIpTq+Jw==","size":{"width":121,"height":32}}, {"code":"\\begin{align*}\n{m}&={\\frac{b}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"35-0","type":"align*","aid":null,"ts":1605003808926,"cs":"lbZMycK7529zm8pAsMQLYg==","size":{"width":132,"height":36}}, and {"id":"36-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{n}&={\\frac{c}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","aid":null,"ts":1605003835284,"cs":"1o3i8WwHjepdLBnkFMo6LA==","size":{"width":126,"height":32}}

{"type":"align*","aid":null,"code":"\\begin{align*}\n{l}&={\\frac{-2}{{\\sqrt[]{{\\left(-2\\right)}^{2}+{\\left(-4\\right)}^{2}+{\\left(4\\right)}^{2}}}}}\\\\\n{l}&={\\frac{-2}{6}=\\frac{-1}{3}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"37-1-0","ts":1605004894456,"cs":"wuWr3QGHHbTpLN1d0lIQ1w==","size":{"width":186,"height":85}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"37-1-1-0","type":"align*","code":"\\begin{align*}\n{m}&={\\frac{-4}{{\\sqrt[]{{\\left(-2\\right)}^{2}+{\\left(-4\\right)}^{2}+{\\left(4\\right)}^{2}}}}}\\\\\n{m}&={\\frac{-4}{6}=\\frac{-2}{3}}\t\n\\end{align*}","aid":null,"ts":1605004934819,"cs":"Unax02mCZiDSoUU0CwqOMA==","size":{"width":196,"height":85}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{n}&={\\frac{4}{{\\sqrt[]{{\\left(-2\\right)}^{2}+{\\left(-4\\right)}^{2}+{\\left(4\\right)}^{2}}}}}\\\\\n{n}&={\\frac{4}{6}=\\frac{2}{3}}\t\n\\end{align*}","aid":null,"id":"37-1-1-1","type":"align*","ts":1605005009118,"cs":"jwFfXRcT02gS5KvpeKxwvw==","size":{"width":192,"height":85}}

Direction cosine:- -1/3, -2/3, and 2/3

14. Show that the vector {"id":"42","type":"$","aid":null,"font":{"family":"Arial","color":"#000000","size":10},"code":"$\\hat{i}+\\hat{j}+\\hat{k}$","ts":1605005217423,"cs":"xMqNCsiXX0XY4qn0mfrOcg==","size":{"width":60,"height":16}} is equally inclined to the axes OX, OY, and OZ.

Solun:- Let {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"39-1","type":"$","ts":1604915030814,"cs":"7p6Flxm5oB9k++y//vxQJQ==","size":{"width":16,"height":17}} = {"id":"42","type":"$","aid":null,"font":{"family":"Arial","color":"#000000","size":10},"code":"$\\hat{i}+\\hat{j}+\\hat{k}$","ts":1605005217423,"cs":"xMqNCsiXX0XY4qn0mfrOcg==","size":{"width":60,"height":16}}

Direction ratios in {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"40-1","type":"$","ts":1604915030814,"cs":"usDG2Cx9nKsZnU1PtEQ0xQ==","size":{"width":16,"height":17}} is:- a = 1, b = 1, and c = 1

We know that:- direction cosine in the form of direction ratio is:-

{"type":"align*","id":"34-0","code":"\\begin{align*}\n{l}&={\\frac{a}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","aid":null,"font":{"color":"#000000","size":10,"family":"Arial"},"ts":1605003787237,"cs":"tLJaFl7mvm3lRlrIpTq+Jw==","size":{"width":121,"height":32}}, {"code":"\\begin{align*}\n{m}&={\\frac{b}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"35-0","type":"align*","aid":null,"ts":1605003808926,"cs":"lbZMycK7529zm8pAsMQLYg==","size":{"width":132,"height":36}}, and {"id":"36-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{n}&={\\frac{c}{{\\sqrt[]{a^{2}+b^{2}+c^{2}}}}}\t\n\\end{align*}","aid":null,"ts":1605003835284,"cs":"1o3i8WwHjepdLBnkFMo6LA==","size":{"width":126,"height":32}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"37-0-1-0","aid":null,"type":"align*","code":"\\begin{align*}\n{l}&={\\frac{1}{{\\sqrt[]{{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}}}}}\\\\\n{l}&={\\frac{1}{{\\sqrt[]{3}}}}\t\n\\end{align*}","ts":1605005326788,"cs":"wzStIBF6B5dzAcv4BWgD1w==","size":{"width":162,"height":89}}

{"id":"37-0-1-1","aid":null,"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{m}&={\\frac{1}{{\\sqrt[]{{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}}}}}\\\\\n{m}&={\\frac{1}{{\\sqrt[]{3}}}}\t\n\\end{align*}","ts":1605005353647,"cs":"T7aKTNLlDWOOmGhlhAb/LQ==","size":{"width":172,"height":89}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"aid":null,"code":"\\begin{align*}\n{n}&={\\frac{1}{{\\sqrt[]{{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}}}}}\\\\\n{n}&={\\frac{1}{{\\sqrt[]{3}}}}\t\n\\end{align*}","type":"align*","id":"37-0-1-2","ts":1605005374990,"cs":"rubrl4b7Q032MeJX3WN/mg==","size":{"width":168,"height":89}}

We know that direction cosine is the cosine of direction angles and direction cosines of {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"40-2","type":"$","ts":1604915030814,"cs":"4nGVJ2zSh9ZEwjhSZZMY/A==","size":{"width":16,"height":17}} are equal then {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"40-3","type":"$","ts":1604915030814,"cs":"woJ2T/pMbV4wEfvZUqN6kQ==","size":{"width":16,"height":17}} makes the same angle with all the axes.

Hence given vector is equally inclined to the axes OX, OY, and OZ.

15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are {"font":{"size":10,"color":"#000000","family":"Arial"},"id":"43-0-0","aid":null,"code":"$\\hat{i}+2\\hat{j}-\\hat{k}$","type":"$","ts":1605005930471,"cs":"8LS9JzFeryZDT55sGl/vqQ==","size":{"width":68,"height":20}} and {"id":"43-1","font":{"family":"Arial","color":"#000000","size":10},"code":"$-\\hat{i}+\\hat{j}+\\hat{k}$","aid":null,"type":"$","ts":1605005984050,"cs":"Ipx1LlfaCuAo6SLCatDrvw==","size":{"width":72,"height":16}} respectively, in the ratio 2:1.

(i) internally (ii) externally

Solun:- Let {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"39-2-0","type":"$","ts":1604915030814,"cs":"bV03qDsCtBNnoXWVdH2czQ==","size":{"width":16,"height":17}} = {"font":{"size":10,"color":"#000000","family":"Arial"},"id":"43-0-1","aid":null,"code":"$\\hat{i}+2\\hat{j}-\\hat{k}$","type":"$","ts":1605005930471,"cs":"nieQfP/7LA7AP+JphnW4TA==","size":{"width":68,"height":20}} and {"type":"$","code":"$\\overrightarrow{b}$","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"44","ts":1605002577270,"cs":"zAcA1gZK5EaRDxOKCSabVA==","size":{"width":13,"height":20}} = {"id":"43-1","font":{"family":"Arial","color":"#000000","size":10},"code":"$-\\hat{i}+\\hat{j}+\\hat{k}$","aid":null,"type":"$","ts":1605005984050,"cs":"Ipx1LlfaCuAo6SLCatDrvw==","size":{"width":72,"height":16}}

And point R({"type":"$","id":"45","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\overrightarrow{r}$","aid":null,"ts":1605006387585,"cs":"DqVvHujCiDZrQD5yCQTlKg==","size":{"width":13,"height":16}}) divides line joining by {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"39-2-1","type":"$","ts":1604915030814,"cs":"H4dn3wMQNumccuv7rlKzlQ==","size":{"width":16,"height":17}} and {"type":"$","code":"$\\overrightarrow{b}$","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"20-1-0","ts":1605002577270,"cs":"19RxwjzYeL1ceD2SpAjxSA==","size":{"width":13,"height":20}}

(i) internally in the ration of 2:1

m = 2 and n = 1

{"code":"\\begin{align*}\n{\\overrightarrow{r}}&={\\frac{m\\overrightarrow{b}+n\\overrightarrow{a}}{m+n}}\\\\\n{\\overrightarrow{r}}&={\\frac{2\\left(\\overrightarrow{b}\\right)+1\\left(\\overrightarrow{a}\\right)}{2+1}}\\\\\n{\\overrightarrow{r}}&={\\frac{2\\left(-\\hat{i}+\\hat{j}+\\hat{k}\\right)+\\hat{i}+2\\hat{j}-\\hat{k}}{3}}\\\\\n{\\overrightarrow{r}}&={\\frac{-2\\hat{i}+2\\hat{j}+2\\hat{k}+\\hat{i}+2\\hat{j}-\\hat{k}}{3}}\\\\\n{\\overrightarrow{r}}&={\\frac{-\\hat{i}+4\\hat{j}+\\hat{k}}{3}}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"46-0","ts":1605006824659,"cs":"bm37rzSWvBdoKJI10e60Ew==","size":{"width":229,"height":238}}

(ii) externally in the ration of 2:1

m = 2 and n = 1

{"code":"\\begin{align*}\n{\\overrightarrow{r}}&={\\frac{m\\overrightarrow{b}-n\\overrightarrow{a}}{m-n}}\\\\\n{\\overrightarrow{r}}&={\\frac{2\\left(\\overrightarrow{b}\\right)-1\\left(\\overrightarrow{a}\\right)}{2-1}}\\\\\n{\\overrightarrow{r}}&={\\frac{2\\left(-\\hat{i}+\\hat{j}+\\hat{k}\\right)-\\left(\\hat{i}+2\\hat{j}-\\hat{k}\\right)}{1}}\\\\\n{\\overrightarrow{r}}&={-2\\hat{i}+2\\hat{j}+2\\hat{k}-\\hat{i}-2\\hat{j}+\\hat{k}}\\\\\n{\\overrightarrow{r}}&={-3\\hat{i}+3\\hat{k}}\t\n\\end{align*}","id":"46-1","aid":null,"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1605007009919,"cs":"OG4m7tkfPmVnasH7ZPim+w==","size":{"width":248,"height":204}}

16. Find the position vector of the mid point of the vector joining the points 

P(2, 3, 4) and Q(4, 1, -2).

Solun:- Given points are P(2, 3, 4) and Q(4, 1, -2)

Then

{"code":"\\begin{align*}\n{\\overrightarrow{\\text{OP}}}&={2\\hat{i}+3\\hat{j}+4\\hat{k}}\\\\\n{\\overrightarrow{\\text{OQ}}}&={4\\hat{i}+\\hat{j}-2\\hat{k}}\t\n\\end{align*}","id":"38-0-0-1-1-0","type":"align*","aid":null,"font":{"color":"#000000","family":"Arial","size":10},"ts":1605007505057,"cs":"fNEvrBqnpeOhps7RRqYfUg==","size":{"width":129,"height":56}}

We know that:- 

For midpoint

{"code":"\\begin{align*}\n{\\overrightarrow{r}}&={\\frac{\\overrightarrow{a}+\\overrightarrow{b}}{2}}\t\n\\end{align*}","aid":null,"type":"align*","id":"47","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1605007557355,"cs":"Uurkgdp9fED+Ytopcv3gKA==","size":{"width":90,"height":40}}

{"code":"\\begin{align*}\n{\\overrightarrow{\\text{OR}}}&={\\frac{\\overrightarrow{\\text{OP}}+\\overrightarrow{OQ}}{2}}\\\\\n{\\overrightarrow{\\text{OR}}}&={\\frac{\\left(2\\hat{i}+3\\hat{j}+4\\hat{k}\\right)+\\left(4\\hat{i}+\\hat{j}-2\\hat{k}\\right)}{2}}\\\\\n{\\overrightarrow{\\text{OR}}}&={\\frac{6\\hat{i}+4\\hat{j}+2\\hat{k}}{2}}\\\\\n{\\overrightarrow{\\text{OR}}}&={3\\hat{i}+2\\hat{j}+\\hat{k}}\t\n\\end{align*}","aid":null,"type":"align*","id":"48","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1605007951124,"cs":"xJy7CgGGtr1EEiV/dlOl/w==","size":{"width":268,"height":166}}

17. Show that the points A, B, and C with position vectors, {"aid":null,"code":"\\begin{align*}\n{\\overrightarrow{a}}&={3\\hat{i}-4\\hat{j}-4\\hat{k}}\t\n\\end{align*}","id":"52-0","type":"align*","font":{"color":"#434343","size":10,"family":"Arial"},"ts":1605008448929,"cs":"F1Uzyrx1WUEf4YaCD7mNCQ==","size":{"width":120,"height":18}},{"font":{"size":10,"family":"Arial","color":"#434343"},"type":"align*","id":"50-0","aid":null,"code":"\\begin{align*}\n{\\overrightarrow{b}}&={2\\hat{i}-\\hat{j}+\\hat{k}}\t\n\\end{align*}","ts":1605008417678,"cs":"G95NDWLZRLJf1tcMoIVfhw==","size":{"width":104,"height":22}}, and{"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{c}}&={\\hat{i}-3\\hat{j}-5\\hat{k}}\t\n\\end{align*}","id":"51-0","aid":null,"font":{"size":10,"color":"#434343","family":"Arial"},"ts":1605008332182,"cs":"o+IVx7qBvfvCOs68XPRCeg==","size":{"width":112,"height":18}} respectively form the vertices of a right angled triangle.

Solun:- Given Vectors:-

{"aid":null,"code":"\\begin{align*}\n{\\overrightarrow{a}}&={3\\hat{i}-4\\hat{j}-4\\hat{k}}\t\n\\end{align*}","id":"52-1","type":"align*","font":{"color":"#434343","size":10,"family":"Arial"},"ts":1605008448929,"cs":"6IyzSu7zP9u/jvBzDsVQaA==","size":{"width":120,"height":18}}

{"font":{"size":10,"family":"Arial","color":"#434343"},"type":"align*","id":"50-1","aid":null,"code":"\\begin{align*}\n{\\overrightarrow{b}}&={2\\hat{i}-\\hat{j}+\\hat{k}}\t\n\\end{align*}","ts":1605008417678,"cs":"wYvv7vs+sbNZCP252s6l1A==","size":{"width":104,"height":22}}

{"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{c}}&={\\hat{i}-3\\hat{j}-5\\hat{k}}\t\n\\end{align*}","id":"51-1","aid":null,"font":{"size":10,"color":"#434343","family":"Arial"},"ts":1605008332182,"cs":"gU30TRg3fVjPcg1x36AQFw==","size":{"width":112,"height":18}}

{"aid":null,"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\overrightarrow{b}-\\overrightarrow{a}}\t\n\\end{align*}","type":"align*","id":"53-0","ts":1605082592471,"cs":"4nGIeXCDrJmlfUgMB5XReQ==","size":{"width":92,"height":20}}

{"id":"54-0","type":"align*","aid":null,"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}}&={\\left(2\\hat{i}-\\hat{j}+\\hat{k}\\right)-\\left(3\\hat{i}-4\\hat{j}-4\\hat{k}\\right)}\\\\\n{\\overrightarrow{\\text{AB}}}&={-\\hat{i}+3\\hat{j}+5\\hat{k}}\t\n\\end{align*}","ts":1605082613893,"cs":"1X4fyGGgZKfXhLGPunMUQQ==","size":{"width":252,"height":62}}

{"code":"\\begin{align*}\n{\\left|\\overrightarrow{\\text{AB}}\\right|}&={{\\sqrt[]{\\left(1\\right)^{2}+\\left(3\\right)^{2}+\\left(5\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{\\text{AB}}\\right|}&={{\\sqrt[]{35}}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"57-0","aid":null,"ts":1605084485925,"cs":"yCA+Kd+gwCEHdv6isaQ+Qg==","size":{"width":184,"height":84}}

{"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{\\text{BC}}}&={\\overrightarrow{c}-\\overrightarrow{b}}\t\n\\end{align*}","id":"53-1-0","aid":null,"font":{"family":"Arial","color":"#000000","size":10},"ts":1605082905586,"cs":"lrgzWvOnJO/ugLiQViWTmg==","size":{"width":92,"height":20}}

{"code":"\\begin{align*}\n{\\overrightarrow{\\text{BC}}}&={\\left(\\hat{i}-3\\hat{j}-5\\hat{k}\\right)-\\left(2\\hat{i}-\\hat{j}+\\hat{k}\\right)}\\\\\n{\\overrightarrow{\\text{BC}}}&={-\\hat{i}-2\\hat{j}-6\\hat{k}}\t\n\\end{align*}","type":"align*","id":"54-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"aid":null,"ts":1605083263512,"cs":"R6aWtGeUVM8XsGrkRUvQUw==","size":{"width":244,"height":62}}

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\left|\\overrightarrow{\\text{BC}}\\right|}&={{\\sqrt[]{\\left(1\\right)^{2}+\\left(2\\right)^{2}+\\left(6\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{\\text{BC}}\\right|}&={{\\sqrt[]{41}}}\t\n\\end{align*}","aid":null,"id":"57-1-0","ts":1605084998747,"cs":"rX8h9PtTKc+2TosHcbnTIQ==","size":{"width":184,"height":84}}

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"aid":null,"code":"\\begin{align*}\n{\\overrightarrow{\\text{AC}}}&={\\overrightarrow{c}-\\overrightarrow{a}}\t\n\\end{align*}","id":"53-1-1","ts":1605083373162,"cs":"cwAbQJhzk9BeiFoz6M633A==","size":{"width":93,"height":20}}

{"aid":null,"code":"\\begin{align*}\n{\\overrightarrow{\\text{AC}}}&={\\left(\\hat{i}-3\\hat{j}-5\\hat{k}\\right)-\\left(3\\hat{i}-4\\hat{j}-4\\hat{k}\\right)}\\\\\n{\\overrightarrow{\\text{AC}}}&={-2\\hat{i}+\\hat{j}-\\hat{k}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"54-1-1","ts":1605083570812,"cs":"I9hHg2fs0lm44AcB3r7JNw==","size":{"width":260,"height":62}}

{"type":"align*","id":"57-1-1","aid":null,"code":"\\begin{align*}\n{\\left|\\overrightarrow{\\text{AC}}\\right|}&={{\\sqrt[]{\\left(-2\\right)^{2}+\\left(1\\right)^{2}+\\left(-1\\right)^{2}}}}\\\\\n{\\left|\\overrightarrow{\\text{AC}}\\right|}&={{\\sqrt[]{6}}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1605085368596,"cs":"NmXlm6dnXNAJksrz6aPiHw==","size":{"width":208,"height":84}}

{"id":"58","font":{"color":"#000000","family":"Arial","size":10},"aid":null,"code":"\\begin{align*}\n{\\left|\\overrightarrow{\\text{BC}}\\right|^{2}}&={\\left|\\overrightarrow{\\text{AB}}\\right|^{2}+\\left|\\overrightarrow{\\text{AC}}\\right|^{2}}\t\n\\end{align*}","type":"align*","ts":1605085505724,"cs":"1YGKhlYczOqHNt4i89L8+Q==","size":{"width":153,"height":42}}

41 = 35 + 6

41 = 41

So, {"id":"58","font":{"color":"#000000","family":"Arial","size":10},"aid":null,"code":"\\begin{align*}\n{\\left|\\overrightarrow{\\text{BC}}\\right|^{2}}&={\\left|\\overrightarrow{\\text{AB}}\\right|^{2}+\\left|\\overrightarrow{\\text{AC}}\\right|^{2}}\t\n\\end{align*}","type":"align*","ts":1605085505724,"cs":"1YGKhlYczOqHNt4i89L8+Q==","size":{"width":153,"height":42}}

Hence ABC is right angle triangle.

18. In triangle ABC, which of the following is not true:

{"id":"59-0","aid":null,"code":"\\begin{align*}\n{\\left(\\text{A}\\right)\\,\\overrightarrow{\\text{AB}}+\\overrightarrow{\\text{BC}}+\\overrightarrow{\\text{CA}}}&={\\overrightarrow{0}}\\\\\n{\\left(\\text{B}\\right)\\,\\overrightarrow{\\text{AB}}+\\overrightarrow{\\text{BC}}-\\overrightarrow{\\text{AC}}}&={\\overrightarrow{0}}\\\\\n{\\left(\\text{C}\\right)\\,\\overrightarrow{\\text{AB}}+\\overrightarrow{\\text{BC}}-\\overrightarrow{\\text{CA}}}&={\\overrightarrow{0}}\\\\\n{\\left(\\text{D}\\right)\\,\\overrightarrow{\\text{AB}}-\\overrightarrow{\\text{C}\\text{B}}+\\overrightarrow{\\text{CA}}}&={\\overrightarrow{0}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1605086342231,"cs":"CtVWhvqm5OIswOsXrVBswg==","size":{"width":166,"height":122}}

Solun:- According to triangle law of addition:-

{"aid":null,"id":"60-0","type":"align*","code":"\\begin{align*}\n{\\overrightarrow{\\text{AC}}}&={\\overrightarrow{\\text{AB}}+\\overrightarrow{\\text{BC}}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1605088300300,"cs":"82bjx6IWO6mIEYLcYgsxWA==","size":{"width":106,"height":21}}

{"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}+\\overrightarrow{\\text{BC}}-\\overrightarrow{\\text{AC}}}&={0}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"aid":null,"id":"60-1-0","type":"align*","ts":1605088466867,"cs":"hImcBoCqZZ05BaTRjPSnAQ==","size":{"width":133,"height":21}}

So option B is correct.

We know that:- {"code":"\\begin{align*}\n{\\overrightarrow{\\text{AC}}}&={-\\overrightarrow{\\text{C}\\text{A}}}\t\n\\end{align*}","type":"align*","aid":null,"id":"61-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1605088546397,"cs":"PALr3fnmPgre1uv8meqUqw==","size":{"width":78,"height":20}}

{"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}+\\overrightarrow{\\text{BC}}+\\overrightarrow{\\text{C}\\text{A}}}&={0}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"60-1-1-0","aid":null,"ts":1605088588945,"cs":"m67SgbJpkRVyoW8FSF7RoA==","size":{"width":133,"height":21}}

So option A is correct.

Also:- {"type":"align*","code":"\\begin{align*}\n{\\overrightarrow{\\text{BC}}}&={-\\overrightarrow{\\text{C}\\text{B}}}\t\n\\end{align*}","aid":null,"font":{"family":"Arial","size":10,"color":null},"id":"62","ts":1605088750736,"cs":"OoeHHwkDWSWd0mx4bzgkow==","size":{"width":76,"height":20}}

{"aid":null,"code":"\\begin{align*}\n{\\overrightarrow{\\text{AB}}-\\overrightarrow{\\text{CB}}+\\overrightarrow{\\text{C}\\text{A}}}&={0}\t\n\\end{align*}","id":"63","type":"align*","font":{"family":"Arial","size":10,"color":null},"ts":1605088776996,"cs":"EO2O6cpvSjWdN682F3RqfA==","size":{"width":133,"height":21}}

So option D is also correct.

Hence option C is not true.

The correct answer is C.

19. If {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"64","type":"$","ts":1604915030814,"cs":"ePJHD0FvqiLSBDkGP+zGhQ==","size":{"width":16,"height":17}} and {"id":"65","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\overrightarrow{b}$","aid":null,"ts":1605089080042,"cs":"6vfM1jJGymGjXwQlkhmiqQ==","size":{"width":13,"height":20}} are two collinear vectors, then which of the following are incorrect:

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\left(\\text{A}\\right)\\,\\overrightarrow{b}}&={\\lambda\\overrightarrow{a}}\\\\\n{\\left(\\text{B}\\right)\\,\\overrightarrow{a}}&={\\pm\\overrightarrow{b}}\t\n\\end{align*}","id":"59-1-0","type":"align*","aid":null,"ts":1605089027014,"cs":"x7C2xlqCxzZCClFaP7rNmg==","size":{"width":88,"height":52}}

(C) the respective components of {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"64","type":"$","ts":1604915030814,"cs":"ePJHD0FvqiLSBDkGP+zGhQ==","size":{"width":16,"height":17}} and {"id":"65","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\overrightarrow{b}$","aid":null,"ts":1605089080042,"cs":"6vfM1jJGymGjXwQlkhmiqQ==","size":{"width":13,"height":20}} are proportional

(D) both the vectors {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"64","type":"$","ts":1604915030814,"cs":"ePJHD0FvqiLSBDkGP+zGhQ==","size":{"width":16,"height":17}} and {"id":"65","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\overrightarrow{b}$","aid":null,"ts":1605089080042,"cs":"6vfM1jJGymGjXwQlkhmiqQ==","size":{"width":13,"height":20}} have same direction, but different magnitude

Solun:- According to the definition of collinear vector:- Parallel vectors are said to be collinear vectors irrespective of their magnitude.

Such that {"font":{"family":"Arial","color":"#000000","size":10},"aid":null,"id":"59-1-1","type":"align*","code":"\\begin{align*}\n{\\overrightarrow{b}}&={\\lambda\\overrightarrow{a}}\t\n\\end{align*}","ts":1605089333028,"cs":"c2zFAybbwHTmlG/4EsIUlw==","size":{"width":58,"height":20}}, {"id":"66","code":"$\\lambda$","aid":null,"type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1605089366683,"cs":"0aaU2oJnmFWWZ+6SJqWDRA==","size":{"width":8,"height":10}} is non-zero scalar.

So, option A is true.

{"id":"59-1-0","code":"\\begin{align*}\n{\\overrightarrow{a}}&={\\pm\\overrightarrow{b}}\t\n\\end{align*}","aid":null,"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1605089428529,"cs":"xLcT08FjfnpPdgyHyLO0pw==","size":{"width":61,"height":20}}Here {"id":"66","code":"$\\lambda$","aid":null,"type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1605089366683,"cs":"0aaU2oJnmFWWZ+6SJqWDRA==","size":{"width":8,"height":10}} is -1 or 1 that is a non-zero scalar.

So, option B is correct.

And option C is also correct because {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"64","type":"$","ts":1604915030814,"cs":"ePJHD0FvqiLSBDkGP+zGhQ==","size":{"width":16,"height":17}} and {"id":"65","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\overrightarrow{b}$","aid":null,"ts":1605089080042,"cs":"6vfM1jJGymGjXwQlkhmiqQ==","size":{"width":13,"height":20}} are proportional so components of {"code":"$\\overrightarrow{a}$","font":{"color":"#000000","family":"Arial","size":10},"id":"64","type":"$","ts":1604915030814,"cs":"ePJHD0FvqiLSBDkGP+zGhQ==","size":{"width":16,"height":17}} and {"id":"65","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\overrightarrow{b}$","aid":null,"ts":1605089080042,"cs":"6vfM1jJGymGjXwQlkhmiqQ==","size":{"width":13,"height":20}} are also proportional.

Option D is incorrect because vectors having the opposite direction are also collinear.

The correct answer is D.


See Also:-

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