Vector Algebra Test

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Vector Algebra Test - 53

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Question 1
1 / -0

$$ABC$$ is a triangle, the point $$P$$ is on side $$BC$$ such that $$3\vec {BP}=2\vec {PC}$$, the point $$Q$$ is on the line $$\vec {CA}$$ such that $$4\vec {CQ}=\vec {QA}$$. If $$R$$ is the common point of $$\vec {AP}$$ & $$\vec {BQ}$$, then the ratio in which the line joining $$CR$$ divides $$\vec {AB}$$ is

A
$$2:5$$

B
$$3:8$$

C
$$4:1$$

D
$$6:1$$

Solution

We have,
$$\begin{matrix} \frac { { BP } }{ { PC } } =\frac { 2 }{ 3 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, PV\, \, of\, P=\frac { { 2\overrightarrow { c } +3\overrightarrow { b } } }{ 5 } \\ \frac { { QA } }{ { CQ } } =4\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, PV\, \, of\, Q=\frac { { 4\overrightarrow { c } } }{ 5 } \\ \end{matrix}$$
Let R divides $$BQ$$ in $$\lambda :1$$ and AP in $$\mu :1$$
$$\begin{matrix} \frac { { \lambda \frac { { 4\overrightarrow { c } } }{ 5 } +\overrightarrow { b } } }{ { \lambda +1 } } =\frac { { \mu \left( { \frac { { 2\overrightarrow { c } } }{ 5 } +\frac { 3 }{ 5 } \overrightarrow { b } } \right) } }{ { \mu +1 } } \\ \frac { { 4\lambda } }{ { 5\left( { \lambda H } \right) } } \overrightarrow { c } +\frac { 1 }{ { \lambda H } } \overrightarrow { b } =\frac { { 2\mu } }{ { 5\left( { \mu H } \right) } } \overrightarrow { c } +\frac { { 3\mu } }{ { 5\left( { \mu H } \right) } } \overrightarrow { b } \\ \frac { { 4\lambda } }{ { 5\left( { \lambda H } \right) } } =\frac { { 2\mu } }{ { 5\left( { \mu H } \right) } } \\ \frac { \lambda }{ { \lambda H } } =\frac { { 3\mu } }{ { 5\left( { \mu H } \right) } } \\ \frac { { 4\lambda } }{ 5 } \frac { { 3\mu } }{ { 5\left( { \mu H } \right) } } =\frac { { 2\mu } }{ { 5\left( { \mu H } \right) } } \\ \frac { { 4\lambda } }{ 5 } \times 3=2 \\ 2\lambda =\frac { 5 }{ 3 } \\ \lambda =\frac { 5 }{ 6 } \\ PV\, \, of\, \, R=\frac { { \frac { 4 }{ 5 } \times \frac { 5 }{ 6 } \overrightarrow { c } +\overrightarrow { b } } }{ { \frac { { 11 } }{ 6 } } } =\frac { { \frac { 2 }{ 3 } \overrightarrow { c } +\overrightarrow { b } } }{ { \frac { { 11 } }{ 6 } } } \\ PV\, \, \, ofR=\frac { { 12 } }{ { 33 } } \overrightarrow { c } +\frac { 6 }{ { 11 } } \overrightarrow { b } \\ \end{matrix}$$
Let E divides BA in $$r:1$$ and CR in $$\beta :1$$
$$\begin{matrix} \frac { { \overrightarrow { b } } }{ { rH } } =\frac { { \beta \overrightarrow { c } -\frac { { 12 } }{ { 33 } } \overrightarrow { c } -\frac { 6 }{ { 11 } } \overrightarrow { b } } }{ { \beta -1 } } \\ \frac { { \overrightarrow { b } } }{ { rH } } =\frac { { \left( { \beta \overrightarrow { c } -\frac { 4 }{ { 33 } } } \right) \overrightarrow { c } -\frac { 6 }{ { 11 } } \overrightarrow { b } } }{ { \beta -1 } } \\ \beta =\frac { 4 }{ { 11 } } \\ \frac { 1 }{ { rH } } =\frac { { -6 } }{ { 11 } } \times \frac { 1 }{ { \frac { 4 }{ { 11 } } -1 } } \\ =\frac { { -6 } }{ { 11 } } \times \frac { { 11 } }{ { -7 } } \\ \frac { 1 }{ { rH } } =\frac { 6 }{ 7 } \\ rH=\frac { 7 }{ 6 } \\ e=\frac { 1 }{ 6 } \\ \frac { { BE } }{ { AE } } =\frac { 1 }{ 6 } \\ \frac { { AE } }{ { BE } } =\frac { 6 }{ 1 } \\ \end{matrix}$$
Then,
Option $$D$$ is correct answer.
Question 2
1 / -0

If $$\hat {i},\hat {j},\hat {k}$$ are positive vectors of $$A,B,C$$ and $$\vec {AB}=\vec {CX}$$, then positive vector of $$X$$ is

A
$$-\hat {i}+\hat {j}+\hat {K}$$

B
$$\hat {i}-\hat {j}+\hat {K}$$

C
$$\hat {i}+\hat {j}-\hat {K}$$

D
$$\hat {i}+\hat {j}+\hat {K}$$

Solution
Question 3
1 / -0

Four persons $$P,\ Q,\ R$$ and $$S$$ are initially at the four corners of a square side $$d$$. Each person now moves with a constant speed $$v$$ in such a way that $$P$$ always moves directly towards $$Q,\ Q$$ towards $$R,\ R$$ towards $$S,$$ and $$S$$ towards $$P$$. The four persons will meet after time

A
$$\dfrac {d}{2v}$$

B
$$\dfrac {d}{v}$$

C
$$\dfrac {3d}{2v}$$

D
$$They\ will\ never\ meet$$

Solution

Here, velocity components will be $$v\cos 45=\dfrac{v}{\sqrt{2}}$$

And, displacement will be $$\dfrac{d}{\sqrt{2}}$$

So time taken will be

$$ t=\dfrac{d}{v} $$

$$ =\dfrac{\dfrac{d}{\sqrt{2}}}{\dfrac{v}{\sqrt{2}}}=\dfrac{d}{v} $$
Question 4
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If $$\bar a + \bar b$$ is perpendicular to $$\bar b$$ and $$\bar a + 2\bar b$$ is perpendicular to $$\bar a$$ then.

A
$$\left| {\bar a} \right| = \left| {\bar b} \right|$$

B
$$\left| {\bar a} \right| = \sqrt 2 \left| {\bar b} \right|$$

C
$$\left| {\bar b} \right| = \sqrt 2 \left| {\bar a} \right|$$

D
$$\left| {\bar a} \right| = \left| {\bar b} \right|\sqrt 3 $$

Solution

Since $$\vec a+\vec b$$ is $$\bot $$ to $$\vec b$$
So,
$$\vec b\cdot (\vec a+\vec b)=0$$

$$\vec b.\vec a+\vec b.\vec b=0$$

$$\vec a.\vec b+|\vec b|^2=0$$

$$\vec a.\vec b=-|\vec b|^2$$ ---- $$(1)$$

Also $$\vec a+2\vec b$$ is perpendicular to $$\vec a$$

So,

$$\vec a.(\vec a+2\vec b)=0$$

$$\vec a.\vec a+2\vec a.\vec b=0$$

$$|\vec a|^2+2\vec a.\vec b=0$$

$$|\vec a|^2+2 \times -|\vec b|^2=0$$ from $$(1)$$

$$|\vec a|=\sqrt 2|\vec b|$$
Question 5
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Directions For Questions

In a parallelogram OABC, vectors $$\vec{a}, \vec{b}, \vec{c}$$ are respectively the position vectors of vertices A, B, C with reference to O as origin. A point E is taken on the side BC which divides it in the ratio of $$2:1$$ internally. Also, the line segment AE intersect the line bisecting the angle O internally in point P. If CP, when extended meets AB in point F. Then?

...view full instructions

The position vector of point P, is?

A
$$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

B
$$\dfrac{|\vec{a}||\vec{c}|}{3|\vec{c}+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

C
$$\dfrac{2|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\bar{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

D
$$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}-\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

Solution

According to question:

$$\begin{array}{l} B=\overrightarrow { a } +\overrightarrow { c } \\ \overrightarrow { E } =\dfrac { { 2\overrightarrow { c } +\overrightarrow { a } +\overrightarrow { c } } }{ 3 } =\dfrac { { 3\overrightarrow { c } +\overrightarrow { a } } }{ 3 } \\ then, \\ Position\, vector\, is: \\ \, \, \, \, \, \overrightarrow { OP } =\lambda \, (\overrightarrow { a } +\overrightarrow { c } )-----(i) \\ \, \, \, \, Now, \\ \, \, \, \, \, \, \, positon\, \, vector\, of\overrightarrow { \, p } : \\ \, \, \, \, \, \overrightarrow { p } =\overrightarrow { a } +\mu \left( { \dfrac { { 3\overrightarrow { c } +\overrightarrow { a } } }{ 3 } -\overrightarrow { a } } \right) \\ \, \, \, \overrightarrow { p } =\overrightarrow { a } +\mu \left( { \dfrac { { 3\overrightarrow { c } -2\overrightarrow { a } } }{ 3 } } \right) -----(ii) \\ Now,\, equating\, \, eqn.\, \, (i)\, \, \& \, (ii) \\ \lambda \, \left( { \dfrac { { \overrightarrow { a } } }{ { \left| { \overline { a } } \right| } } +\dfrac { { \overrightarrow { c } } }{ { \left| { \overline { c } } \right| } } } \right) =\overrightarrow { a } +\dfrac { \mu }{ 3 } \left( { 3\overrightarrow { c } -2\overrightarrow { a } } \right) \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =1-\dfrac { { 2\mu } }{ 3 } \, \, \, \, \, \, \, (cofficient\, compare) \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =\dfrac { { 3\mu } }{ 3 } ,\, \, \, \, \, \, \, \, \, \mu =\dfrac { \lambda }{ { \left| { \overline { a } } \right| } } \\ And, \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =1-\dfrac { 2 }{ 3 } .\, \dfrac { \lambda }{ { \left| { \overline { c } } \right| } } \\ \lambda \left( { \dfrac { 1 }{ { \left| { \overline { a } } \right| } } +\dfrac { 2 }{ 3 } .\, \dfrac { 1 }{ { \left| { \overline { c } } \right| } } } \right) =1 \\ \lambda =\dfrac { { 3\left| { \overline { a } } \right| \, \left| { \overline { c } } \right| } }{ { 3\left| { \overline { c } } \right| +2\left| { \overline { a } } \right| } } \\ \, positon\, \, vector\, of\overrightarrow { \, p } : \\ \overrightarrow { p } =\dfrac { { 3\left| { \overline { a } } \right| \, \left| { \overline { c } } \right| } }{ { 3\left| { \overline { c } } \right| +2\left| { \overline { a } } \right| } } \, \, \, \left( { \dfrac { { \overrightarrow { a } } }{ { \left| { \overline { a } } \right| } } +\dfrac { { \overrightarrow { c } } }{ { \left| { \overline { c } } \right| } } } \right) \\ so\, ,\, the\, correct\, \, option\, is\, A. \end{array}$$
Question 6
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If $$\overline { a } =\cfrac { 1 }{ \sqrt { 10 } } \left( 3\overline { i } +\overline { k } \right) ;\overline { a } =\cfrac { 1 }{ 7 } \left( 2\overline { i } +3\overline { j } -6\overline { k } \right) $$ then the value of
$$(2\overline { a } -\overline { b } ).[(\overline { a } \times \overline { b } )\times (\overline { a } +2\overline { b } )]$$

A
$$5$$

B
$$3$$

C
$$-5$$

D
$$-3$$

Solution
Question 7
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If $$\vec a,\vec b,\vec c$$ non-zero vectors such that $$\vec a$$ is perpendicular to $$\vec b$$ and $$\vec c$$ and non-zero vector coplanar with $$\vec a + \vec b$$ and $$2\vec b - \vec c$$ and $$\vec d.\vec a = 1$$ , then the minimum value of $$\left| {\vec d} \right|$$

A
$$\frac{2}{{\sqrt {13} }}$$

B
$$\frac{1}{{\sqrt {13} }}$$

C
$$\frac{3}{{\sqrt {13} }}$$

D
$$\frac{4}{{\sqrt {13} }}$$
Question 8
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Let $$\vec{a},\vec{b},\vec{c}$$ be three non-zero vectors such that $$\vec{a}+\vec{b}+\vec{c}=0$$ and $$\lambda \vec{b}\times \vec{a}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=0$$, then $$\lambda$$ is

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$-2$$

Solution
Question 9
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If $$a,b,c \in N$$, the number of points having position vectors $$a\hat i + b\hat j + c\hat k$$ such that $$6 \le a + b + c \le 10$$ is

A
110

B
116

C
120

D
127

Solution

Given:$$6\le\,a+b+c\le \,10$$

$$\because\,a,b,c,\in N$$

$$\therefore\,a+b+c=6\,or\,7\,or\,8\,or\,9\,or\,10$$
and $$a,b,c\ge 1$$

$$\therefore\,x+y+x=3\,or\,4\,or\,5\,or\,6\,or\,7$$ where $$x,y,z\ge 1$$

Let $$x=a-1,\,y=b-1,\,$$ and $$z=c-1$$

$$\therefore\,$$Required number of points$$=^{3-1+3}C_{3}+^{3-1+4}C_{4}+^{3-1+5}C_{5}+^{3-1+6}C_{6}+^{3-1+7}C_{7}$$

$$=^{5}C_{3}+^{6}C_{4}+^{7}C_{5}+^{8}C_{6}+^{9}C_{7}$$

$$=\dfrac{5!}{3!2!}+\dfrac{6!}{2!4!}+\dfrac{7!}{5!2!}+\dfrac{8!}{6!2!}+\dfrac{9!}{7!2!}$$

$$=\dfrac{20}{2}+\dfrac{30}{2}+\dfrac{42}{2}+\dfrac{56}{2}+\dfrac{72}{2}$$

$$=10+15+21+28+36=110$$
Question 10
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If $$\vec{a}$$ be any vector, then the value of $$|\vec{a} \times \hat{ i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2$$

A
$$|\vec{a}|^2$$

B
$$0$$

C
$$2|\vec{a}|^2$$

D
$$3|\vec{a}|^2$$

Solution

Let $$\bar{a}=x\hat{i}+y\hat{j}+z\hat{k}$$
$$(\bar{a}\times \hat{i})^{2}+(\bar{a}\times \hat{j})+|\bar{a}\times \hat{k}|^{2}$$
$$\Rightarrow |-y\hat{k}+z\hat{i}|^{2}+|x\hat{k}-z\hat{i}|^{2}+|-x\hat{j}+y\hat{i}|^{2}$$
$$\Rightarrow y^{2}+z^{2}+x^{2}+z^{2}+x^{2}+y^{2}$$
$$\Rightarrow 2(x^{2}+y^{2}+z^{2})$$
$$\Rightarrow 2|\bar{a}|^{2}$$
$$\therefore (\bar{a}\times \hat{i})^{2}+|\bar{a}\times \hat{j}|^{2}+|\bar{a}\times \hat{k}|^{2}=2|\bar{a}|^{2}$$

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Vector Algebra Test - 53

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Vector Algebra Test - 53

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SHARING IS CARING

Question 1

$$2:5$$

$$3:8$$

$$4:1$$

$$6:1$$

Solution

Question 2

$$-\hat {i}+\hat {j}+\hat {K}$$

$$\hat {i}-\hat {j}+\hat {K}$$

$$\hat {i}+\hat {j}-\hat {K}$$

$$\hat {i}+\hat {j}+\hat {K}$$

Solution

Question 3

$$\dfrac {d}{2v}$$

$$\dfrac {d}{v}$$

$$\dfrac {3d}{2v}$$

$$They\ will\ never\ meet$$

Solution

Question 4

$$\left| {\bar a} \right| = \left| {\bar b} \right|$$

$$\left| {\bar a} \right| = \sqrt 2 \left| {\bar b} \right|$$

$$\left| {\bar b} \right| = \sqrt 2 \left| {\bar a} \right|$$

$$\left| {\bar a} \right| = \left| {\bar b} \right|\sqrt 3 $$

Solution

Question 5

$$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

$$\dfrac{|\vec{a}||\vec{c}|}{3|\vec{c}+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

$$\dfrac{2|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\bar{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

$$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}-\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

Solution

Question 6

$$5$$

$$3$$

$$-5$$

$$-3$$

Solution

Question 7

$$\frac{2}{{\sqrt {13} }}$$

$$\frac{1}{{\sqrt {13} }}$$

$$\frac{3}{{\sqrt {13} }}$$

$$\frac{4}{{\sqrt {13} }}$$

Question 8

$$1$$

$$2$$

$$-1$$

$$-2$$

Solution

Question 9

110

116

120

127

Solution

Question 10

$$|\vec{a}|^2$$

$$0$$

$$2|\vec{a}|^2$$

$$3|\vec{a}|^2$$

Solution

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