Vector Algebra Test

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Vector Algebra Test - 36

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Question 1
1 / -0

If $$ \displaystyle \bar{a}\times \bar{b}=\bar{b}\times \bar{c}=\bar{c}\times \bar{a} $$ then $$ \displaystyle \bar{a}+\bar{b}+\bar{c}=? $$

A
$$abc$$

B
$$-1$$

C
$$0$$

D
$$2$$

Solution

Let us consider $$\overline { a } +\overline { b } +\overline { c } =k$$
Multiply with $$\overline { b } $$ on both sides , we get $$\overline { a }\times \overline { b } +\overline { b } \times\overline { b } +\overline { c }\times \overline { b } =k\times\overline { b } $$
$$\Rightarrow k \times \overline { b } =\overline { b } \times \overline { c }+0+\overline { c }\times \overline { b } =0 $$
Similarly we get $$k \times \overline { a } =0$$ and $$k \times \overline { c } =0$$
therefore $$k=0$$ for equations to satisfy
therefore correct option is $$C$$
Question 2
1 / -0

The vector $$\left ( \bar{a }\times\bar{b } \right )\times \left ( \bar{c }\times\bar{b } \right )$$ is

A
At right angle to $$\bar{b}$$

B
Is parallel to $$\bar{c}$$

C
Is parallel to $$\bar{b}$$

D
None of these

Solution

$$(a\times b)$$ will be a vector perpendicular to $$\vec b$$.
$$(c\times b)$$ will be a vector perpendicular to $$\vec b$$.
Hence
$$(a\times b)\times (c\times b)$$ will be a vector perpendicular to both $$(a\times b)$$ and $$(c\times b)$$.
Thus it will be a vector parallel to $$\vec b$$.
Question 3
1 / -0

In a trapezium the vector $$\overline{BC} = \alpha \overline{AD}$$. We will then find that $$\bar{p}= \overline{AC}+\overline{BD}$$ is collinear with $$\overline{AD}$$. if $$\bar{p}= \mu \overline{AD}$$ then

A
$$\mu =\alpha +2$$

B
$$\mu +\alpha = 2$$

C
$$\alpha = \mu +1$$

D
$$\mu = \alpha+1$$

Solution

Let $$A$$ be the origin.
So,
$$\alpha \vec{d} = \vec{c} - \vec{b}$$
$$p = \vec{c} + \vec{d} - \vec{b} = \mu{\vec{d}}$$

Combining two equations, we get $$(1 + \alpha)\vec{d} = \mu \vec{d}$$
$$\Rightarrow \mu = \alpha + 1$$
Question 4
1 / -0

If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point outside $$AB$$, then

A
$$ \displaystyle \overline{PA}+\overline{PB}+\overline{PC}=0 $$

B
$$ \displaystyle \overline{PA}+\overline{PB}+2\overline{PC}=\bar{0} $$

C
$$ \displaystyle \overline{PA}+\overline{PB}=\overline{PC} $$

D
$$ \displaystyle \overline{PA}+\overline{PB}=2\overline{PC} $$

Solution

Let $$\vec{p}, \vec{a}, \vec{b}, \vec{c}$$ be the position vectors of points $$P, A, B, C$$ respectively.
$$L.H.S = \overline{PA} + \overline{PB} = \vec{p} - \vec{a} + \vec{p} - \vec{b} = 2\left(\vec{p} - \frac{\vec{a} + \vec{b}}{2}\right) = 2\vec{PC}$$
Question 5
1 / -0

Given $$A=ai+bj+ck, \ B=di+3j+4k$$ and $$C=3i+j-2k$$. If the vectors $$A,B$$ and $$C$$ form a triangle such that $$A=B+C$$ and $$area(\Delta ABC)=5\sqrt6$$, then

A
$$ a=-8, \ b=-4, \ c=2, \ d=-11 $$

B
$$a=-8, \ b=4, \ c=-2, \ d=-11$$

C
$$a=-8,\ b=4, \ c=2, \ d=-11$$

D
None of the above

Solution

Here $$A,B,C$$ are the vectors which represent the sides of the triangle $$ABC$$ where
$$A=ai+bj+ck\\ B=di+3j+4k\\ C=3i+j-2k$$
Given that $$A=B+C$$
$$\displaystyle \therefore ai+bj+ck=\left( d+3 \right) i+4j+2k\\ \Rightarrow a=d+3,b=4,c=2\\ B\times C=\begin{vmatrix} i & j & k \\ d & 3 & 4 \\ 3 & 1 & -2 \end{vmatrix}=-10i+\left( 2d+12 \right) j+\left( d-9 \right) k$$
$$\therefore$$ Area of the $$\displaystyle \triangle ABC=\frac { 1 }{ 2 } \left| B\times C \right| =\frac { 1 }{ 2 } \sqrt { \left[ 100+{ \left( 2d+12 \right) }^{ 2 }+{ \left( d-9 \right) }^{ 2 } \right] } =5\sqrt { 6 } $$
$$\Rightarrow \sqrt { 5{ d }^{ 2 }+30d+325 } =10\sqrt { 6 } \Rightarrow 5{ d }^{ 2 }+30d+325=600\\ \Rightarrow 5{ d }^{ 2 }+30d-275=0\Rightarrow { d }^{ 2 }+6d-55=0\\ \Rightarrow \left( d+11 \right) \left( d-5 \right) =0\Rightarrow d=5,-11$$
When $$d=5,a=8,b=4,c=2$$ and when $$d=-11,a=-8,b=4,c=2$$
Question 6
1 / -0

A unit vector perpendicular to each of the vectors $$\displaystyle 2\hat{i}+4\hat{j}-\hat{k} $$ and $$\displaystyle \hat{i}-2\hat{j}+3\hat{k} $$ forming a right handed system is

A
$$\displaystyle 7\hat{i}-10\hat{j}+8\hat{k}$$

B
$$\displaystyle \frac{10\hat{i}-7\hat{j}-8\hat{k}}{\sqrt{213}}$$

C
$$\displaystyle -7\hat{i}+10\hat{j}+8\hat{k}$$

D
$$\displaystyle \frac{-10\hat{i}-7\hat{j}-8\hat{k}}{\sqrt{213}}$$

Solution

Unit vector perpendicular to the given 2 vectors is $$(2\hat{i} + 4\hat{j} - \hat{k}) \times (\hat{i} - 2\hat{j} + 3\hat{k})$$
$$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -1 \\ 1 & -2 & 3 \end{vmatrix}$$
$$= \hat{i}(12 - 2) - \hat{j}(6 + 1) + \hat{k}(- 4 - 4)$$
$$= 10 \hat{i} - 7\hat{j} - 8\hat{k}$$
The unit vector will be obtained by dividing the obtained vector with its modulus.
So, we get the answer as $$\dfrac{10 \hat{i} - 7\hat{j} - 8\hat{k}}{\sqrt{100 + 49 + 64}} = \dfrac{10 \hat{i} - 7\hat{j} - 8\hat{k}}{\sqrt{213}}$$
Question 7
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If $$ M $$ and $$ N $$ are the midpoints of the diagonals $$AC$$ and $$BD$$, respectively, of a quadrilateral $$ABCD$$, then $$\overrightarrow {AB} + \overrightarrow {AD} + \overrightarrow {CB} + \overrightarrow {CD} $$ is equal to

A
$$4\overrightarrow {NM}$$

B
$$4\overrightarrow {MN}$$

C
$$2\overrightarrow {MN}$$

D
None of these

Solution
Question 8
1 / -0

Two identical particles are located at $$\overrightarrow{x}$$ and $$ \overrightarrow{y}$$ with reference to the origin of three dimensional co-ordinate system. The position vector of centre of mass of the system is given by

A
$$\overrightarrow { x } - \overrightarrow { y }$$

B
$$\displaystyle \frac {\overrightarrow { x } + \overrightarrow { y }} {{2}}$$

C
$$(\overrightarrow { x } - \overrightarrow { y })$$

D
$$\displaystyle \frac{\overrightarrow { x } - \overrightarrow { y }} {{2}}$$

Solution
Question 9
1 / -0

If $$b \neq 0,$$ then every vector $$a$$ can be written in a unique manner as the sum of a vector $$a_{||}$$ parallel to b and a vector $$a_{\perp}$$ perpendicular to $$b$$. If $$a$$ is parallel to $$ b,$$ then $$a_{||} = a $$ and $$ a_{\perp } =0$$. If $$a$$ is perpendicular to $$b$$, then $$a_{||} = 0$$ and $$a_{\perp} = a$$. The vector $$a_{||}$$ is called the projection of $$a$$ on $$b$$ and is denoted by $$proj_{b}a$$. Since $$proj_{b} a$$ is parallel to $$b$$, it is a scalar multiple of the unit vector in the direction of $$b$$, i.e., $$proj _{b} a =\lambda u_{b}$$
The scalar $$\lambda $$ is called the component of $$a$$ in the direction of $$b$$ and is denoted by $$comp _{b} a $$. In fact, $$proj _{b} a = (a \cdot u_{b})u_{b}$$ and $$comp _{b} a = a \cdot u_{b}$$

If $$a = -2i + j + k $$ and $$ b =4i -3j + k $$, then $$proj_{b} a$$ is equal to

A
$$ 4i -3j + 2k$$

B
$$\displaystyle -\frac{ 5}{13} (4i- 3j + k)$$

C
$$\displaystyle \frac{ 5}{13} (4i -3j+ k)$$

D
$$\displaystyle -\frac{ 4}{11} (2i -j + 2k)$$

Solution

$$\displaystyle \left| b \right| =\sqrt { 26 } ,{ u }_{ b }=\frac { b }{ \left| b \right| } =\frac { 1 }{ \sqrt { 26 } } \left( 4i-3j+k \right) $$
$$\displaystyle { comp }_{ b }a=a.{ u }_{ b }=\frac { -10}{ \sqrt{26} } $$
$$\displaystyle { Proj }_{ b }a=\left( { comp }_{ b }a \right) { u }_{ b }=-\frac { 5 }{ 13 } \left( 4i-3j+k \right) $$
Question 10
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If the points $$P, \ Q, \ R, \ S$$ have position vectors $$\overrightarrow p, \ \overrightarrow q, \ \overrightarrow r, \ \overrightarrow s$$ such that $$\overrightarrow p - \overrightarrow q = 2(\overrightarrow s - \overrightarrow r)$$, then find out the correct one:

A
$$PQ$$ and $$RS$$ bisect each other

B
$$PQ$$ and $$PR$$ bisect each other

C
$$PQ$$ and $$RS$$ trisect each other

D
$$PQ$$ and $$PR$$ trisect each other

Solution

We have, $$p-q=2(s-r)$$
$$\displaystyle \Rightarrow p+2r=q+2s\Rightarrow \frac { p+2r }{ 1+2 } =\frac { q+2s }{ 1+2 } $$
$$\therefore$$ Point dividing $$PR$$ in the ratio $$2:1$$ is same as the point dividing $$QS$$ in the ratio $$2:1$$
Hence $$QS$$ and $$PR$$ trisect each other.

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Vector Algebra Test - 36

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Vector Algebra Test - 36

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SHARING IS CARING

Question 1

$$abc$$

$$-1$$

$$0$$

$$2$$

Solution

Question 2

At right angle to $$\bar{b}$$

Is parallel to $$\bar{c}$$

Is parallel to $$\bar{b}$$

None of these

Solution

Question 3

$$\mu =\alpha +2$$

$$\mu +\alpha = 2$$

$$\alpha = \mu +1$$

$$\mu = \alpha+1$$

Solution

Question 4

$$ \displaystyle \overline{PA}+\overline{PB}+\overline{PC}=0 $$

$$ \displaystyle \overline{PA}+\overline{PB}+2\overline{PC}=\bar{0} $$

$$ \displaystyle \overline{PA}+\overline{PB}=\overline{PC} $$

$$ \displaystyle \overline{PA}+\overline{PB}=2\overline{PC} $$

Solution

Question 5

$$ a=-8, \ b=-4, \ c=2, \ d=-11 $$

$$a=-8, \ b=4, \ c=-2, \ d=-11$$

$$a=-8,\ b=4, \ c=2, \ d=-11$$

None of the above

Solution

Question 6

$$\displaystyle 7\hat{i}-10\hat{j}+8\hat{k}$$

$$\displaystyle \frac{10\hat{i}-7\hat{j}-8\hat{k}}{\sqrt{213}}$$

$$\displaystyle -7\hat{i}+10\hat{j}+8\hat{k}$$

$$\displaystyle \frac{-10\hat{i}-7\hat{j}-8\hat{k}}{\sqrt{213}}$$

Solution

Question 7

$$4\overrightarrow {NM}$$

$$4\overrightarrow {MN}$$

$$2\overrightarrow {MN}$$

None of these

Solution

Question 8

$$\overrightarrow { x } - \overrightarrow { y }$$

$$\displaystyle \frac {\overrightarrow { x } + \overrightarrow { y }} {{2}}$$

$$(\overrightarrow { x } - \overrightarrow { y })$$

$$\displaystyle \frac{\overrightarrow { x } - \overrightarrow { y }} {{2}}$$

Solution

Question 9

$$ 4i -3j + 2k$$

$$\displaystyle -\frac{ 5}{13} (4i- 3j + k)$$

$$\displaystyle \frac{ 5}{13} (4i -3j+ k)$$

$$\displaystyle -\frac{ 4}{11} (2i -j + 2k)$$

Solution

Question 10

$$PQ$$ and $$RS$$ bisect each other

$$PQ$$ and $$PR$$ bisect each other

$$PQ$$ and $$RS$$ trisect each other

$$PQ$$ and $$PR$$ trisect each other

Solution

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