Vector Algebra Test

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Vector Algebra Test - 21

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Question 1
1 / -0

The vector
$$\vec a + \vec b,\vec a - k\vec b$$ where $$k$$ scalar are collinear, for

A
$$k=0$$

B
$$k=-1$$

C
$$k=1$$

D
$$k=2$$

Solution

Say $$\vec{A}=\vec{a}+\vec{b}, \vec{B}=\vec{a}-k\vec{b}$$

If two vectors are collinear then
$$\vec{B}=\lambda\vec{A}$$ ......(1)

Now,
$$\vec{B}=1(\vec{a}-k\vec{b})$$
$$\lambda=1$$

From equation (1), we get
$$\vec{B}=1\times \vec{A}$$
$$\vec{a}+\vec{b}=\vec{a}-k\vec{b}$$
$$k=-1$$

Hence for $$k=-1$$, $$\vec{A}$$ and $$\vec{B}$$ are collinear.
Question 2
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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?

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The position vector of point F, is?

A
$$\dfrac{{3\left| {\overline a } \right|\,\left| {\overline c } \right|}}{{3\left| {\overline c } \right| + 2\left| {\overline a } \right|}}\,\,\,\left( {\dfrac{{\overrightarrow a }}{{\left| {\overline a } \right|}} + \dfrac{{\overrightarrow c }}{{\left| {\overline c } \right|}}} \right)$$

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

C
$$\vec{a}+\dfrac{2|\vec{a}|}{|\vec{c}|}\vec{c}$$

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

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 3
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If $$ | \overline{a} | = 1 , | \overline{b} | = 2, | \overline{a} - \overline{b} |^2 + | \overline{a} + 2 \overline{b} |^2 = 20, $$ then $$ ( \overline{a} , \overline{b} ) = $$

A
$$ \dfrac {\pi}{3} $$

B
$$ \dfrac {\pi}{4} $$

C
$$ \dfrac {\pi}{6} $$

D
$$ \dfrac {2 \pi}{3} $$

Solution

Given
$$|\bar{a}|=1 , |\bar{b}|=2, |\bar{a}-\bar{b}|^{2}+|\bar{a}+2\bar{b}|^{2}=20$$

$$|\bar{a}|^{2}+|\bar{b}|^{2}-2|\bar{a}||\bar{b}| \cos (\bar{a},\bar{b})+|\bar{a}|^{2}+4|\bar{b}|^{2}+4|\bar{a}||\bar{b}| \cos (\bar{a},\bar{b})=20$$

$$\Rightarrow 2|\bar{a}|^{2}+5|\bar{b}|^{2}+2|\bar{a}||\bar{b}| \cos (\bar{a},\bar{b})=20$$

$$\Rightarrow 2(1)^{2}+5(2)^{2}+2\times 1\times 2\times \cos (\bar{a},\bar{b})=20$$

$$\Rightarrow 22+4 \cos (\bar{a},\bar{b})=20$$

$$\cos (\bar{a},\bar{b})=\dfrac{-1}{2}$$
$$\Rightarrow \pi -(\bar{a},\bar{b})=\dfrac{\pi }{3}$$
$$\Rightarrow (\bar{a},\bar{b})=(\pi -\dfrac{\pi }{3})=\dfrac{2\pi }{3}$$
Question 4
1 / -0

If $$\bar{a}=2\bar{i}+\bar{j}+\bar{k}, \bar{b}=\bar{i}+5\bar{j}, \bar{c}=4\bar{i}+4\bar{j}-2\bar{k}$$ then the length of the projection of $$(3\bar{a}-2\bar{b})$$ in the direction of $$\bar{c}$$.

A
$$3$$

B
$$-3$$

C
$$33$$

D
$$-33$$

Solution

$$3\vec{a} - 2\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k} - 2\hat{i} - 10\hat{j}$$
$$\implies 4\hat{i} - 7\hat{j} +3\hat{k}$$
Scalar projection = $$\cfrac{(3\vec{a} -2\vec{b}).\vec{c}}{|\vec{c}|}$$
$$\implies \cfrac{16-28-6}{\sqrt{16+16+4}}$$
$$\implies \cfrac{18}{\sqrt{36}}$$
$$\implies 3$$
Question 5
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In a triangle ABC, if $$ 2\vec { AC } =3\vec { CB }$$, then $$2\vec { OA } +3\vec { OB }$$ equals ?

A
$$5\vec { OC }$$

B
$$-\vec { OC }$$

C
$$\vec { OC }$$

D
None of these

Solution

$$\begin{array}{l} Given\, in\, \, \, \Delta ABC, \\ \, \, \, \, \, \, \, \, \, \, \, \, 2\, \overrightarrow { AB } =3\overrightarrow { CB, } \\ \, \, \, \, \, \, \, \, \, \, \, \, 2\overrightarrow { OA } =3\overrightarrow { OB } , \\ We\, know\, that, \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \overrightarrow { AC } =\overrightarrow { OC } \, -\overrightarrow { OA } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \overrightarrow { CB } =\overrightarrow { OB } -\overrightarrow { OC } \\ then, \\ \, \, \, 2(\overrightarrow { OC } -\overrightarrow { OA } )=3(\overrightarrow { OB } -\overrightarrow { OC } ) \\ 2\, \overrightarrow { OC } +2\overrightarrow { \, OA } =3\overrightarrow { OB } -3\overrightarrow { OC } \\ 2\, \, \overrightarrow { OC } -3\, \overrightarrow { \, OC } =3\, \overrightarrow { \, OB } +2\overrightarrow { OA } \\ 5\, \overrightarrow { OC } =\, \, 2\overrightarrow { OA } \, +3\, \overrightarrow { \, OB } \end{array}$$
$$\,So,\,\,\,\,2\overrightarrow {OA} \, + 3\,\overrightarrow {\,OB} = 5\,\overrightarrow {OC} $$
Question 6
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If $$|\overrightarrow{a}| = 5, |\overrightarrow{a} - \overrightarrow{b}|=8$$ and $$|\overrightarrow{a} + \overrightarrow{b}| = 10$$, then $$|\overrightarrow{b}|$$ is equal to:

A
$$1$$

B
$$\sqrt{57}$$

C
$$3$$

D
$$57$$

Solution

Given,
$$\left| {\overline a } \right| = 5\,$$
$$\,\left| {\overline a - \overline b } \right| = 8\, \, equation - - - (1$$
$$\,\left| {\overline a + \overline b } \right| = 10\,\, equation - - - (2$$

Squaring and adding both equation (1) & (2)

$$\underline { \begin{array}{l} { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }-2\left| a \right| \left| b \right| \cos \theta =64 \\ { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }+2\left| a \right| \left| b \right| \cos \theta =100 \end{array} } $$
$$\begin{array}{l} 2\left( { { { \left| a \right| }^{ 2 } }+{ { \left| b \right| }^{ 2 } } } \right) =164 \\ { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }=82 \\ we\, have, \\ \left| a \right| =5 \\ then, \\ { \left( 5 \right) ^{ 2 } }+{ \left| b \right| ^{ 2 } }=82 \\ { \left| b \right| ^{ 2 } }=82-25 \\ { \left| b \right| ^{ 2 } }=57 \\ \left| b \right| =\sqrt { 57 } \end{array}$$

So the correct Ans is option B.
Question 7
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If the vector $$OP$$ in $$XY$$ plane whose magnitude is $$\sqrt3$$ makes an angle $$60^o$$ with $$Y-$$ axis, the length of the component of the vector in direction of $$X-$$ axis is :

A
$$1$$

B
$$\sqrt3$$

C
$$\dfrac {1}{2}$$

D
$$\dfrac {3}{2}$$

Solution

Vector $$V=x \uparrow + y \uparrow$$
$$\tan 30^0=\dfrac{y}{x} \Rightarrow \dfrac{1}{\sqrt{3}}\dfrac{y}{x} \Rightarrow \boxed{x=\sqrt{3}y}$$
given $$|v|=\sqrt{3}$$
$$\Rightarrow \sqrt{x^2+y^2}=\sqrt{3}$$
$$\Rightarrow \sqrt{3y^2+y^2}=\sqrt{3}$$ (using (1)
$$\Rightarrow 4y^2=3$$
$$\Rightarrow \boxed{y=\dfrac{\sqrt{3}}{2}}$$
$$x=\sqrt{3}y=\sqrt{3}\times \dfrac{\sqrt{3}}{2}$$
$$\boxed{x=\dfrac{3}{2}}$$
$$\therefore$$ component of vector in $$x-axis$$ direction is $$\dfrac{3}{2}$$.
Question 8
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Let $$\overrightarrow{a}=\hat{i}-\hat{k}, \overrightarrow{b}=x\hat{i}+\hat{j}+\left(1-x\right)\hat{k},\overrightarrow{c}=y\hat{i}+x\hat{j}+\left(1+x-y\right)\hat{k}$$. then $$\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]$$ depends on

A
only $$x$$

B
only $$y$$

C
both $$x$$ and $$y$$

D
neither $$x$$ nor $$y$$

Solution

$$\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]=\left[\begin{matrix} 1 & 0 & -1 \\ x &1 &1-x \\ y & x & 1+x-y \end{matrix}\right]$$
$${c}_{1}+{c}_{2}\rightarrow{c}_{3}$$
$$=\left[\begin{matrix} 1 & 0 & -1+1 \\ x & 1 & 1-x+x +1\\y &x &1+x-y+x+y \end{matrix}\right]$$
$$=\left[\begin{matrix} 1 & 0 & 0 \\ x & 1 & 2 \\y &x &1+2x \end{matrix}\right]$$
$$=1\left(1+2x-2x\right)-0+0=1$$
Question 9
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For three vectors $$\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$$ which of the following expressions is not equal to any of remaining is

A
$$\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right)$$

B
$$\left(\overrightarrow{v}\times \overrightarrow{w}\right).\overrightarrow{u} $$

C
$$\overrightarrow{v}.\left(\overrightarrow{u}\times \overrightarrow{w}\right)$$

D
$$\left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$$

Solution

In a scalar triple product dot and cross can be interchanged.
$$\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right) =\left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$$
the vectors can be changed cyclically.
$$\therefore \left(\overrightarrow{v}\times \overrightarrow{w}\right).\overrightarrow{u}= \left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$$
$$\Rightarrow \overrightarrow{v}.\left(\overrightarrow{u}\times \overrightarrow{w}\right) =-\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right)$$
Question 10
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If $$\bar{a}$$ and $$\bar{b}$$ are non-collinear unit vectors and $$\left|\bar{a}+\bar{b}\right|=\sqrt{3}$$ then $$\left(2\bar{a}+5\bar{b}\right)\cdot \left(3\bar{a}-\bar{b}\right)=?$$

A
$$\dfrac{15}{4}$$

B
$$\dfrac{15}{2}$$

C
$$15$$

D
$$16$$

Solution

$$\left| \bar { a } +\bar { b } \right| =\sqrt { { \left| \bar { a } \right| }^{ 2 }+{ \left| \bar { b } \right| }^{ 2 }+2\bar { a } .\bar { b } } =\sqrt { 3 } $$
Since $$\left| \bar { a } \right| =\left| \bar { b } \right| =1$$
$$\bar { a } .\bar { b } =\cfrac { 1 }{ 2 } $$
So, $$(2\bar { a } +5\bar { b } ).(3\bar { a } -\bar { b } )=6{ \left| \bar { a } \right| }^{ 2 }-2\bar { a } .\bar { b } +15\bar { b } .\bar { a } -5{ \left| \bar { b } \right| }^{ 2 }$$
as $$\bar { a } .\bar { b } =\bar { b } .\bar { a } $$, we can say that
$$(2\bar { a } +5\bar { b } ).(3\bar { a } -\bar { b } )=6+13(\cfrac { 1 }{ 2 } )-5=\cfrac { 15 }{ 2 } $$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 21

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Vector Algebra Test - 21

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SHARING IS CARING

Question 1

$$k=0$$

$$k=-1$$

$$k=1$$

$$k=2$$

Solution

Question 2

$$\dfrac{{3\left| {\overline a } \right|\,\left| {\overline c } \right|}}{{3\left| {\overline c } \right| + 2\left| {\overline a } \right|}}\,\,\,\left( {\dfrac{{\overrightarrow a }}{{\left| {\overline a } \right|}} + \dfrac{{\overrightarrow c }}{{\left| {\overline c } \right|}}} \right)$$

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

$$\vec{a}+\dfrac{2|\vec{a}|}{|\vec{c}|}\vec{c}$$

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

Solution

Question 3

$$ \dfrac {\pi}{3} $$

$$ \dfrac {\pi}{4} $$

$$ \dfrac {\pi}{6} $$

$$ \dfrac {2 \pi}{3} $$

Solution

Question 4

$$3$$

$$-3$$

$$33$$

$$-33$$

Solution

Question 5

$$5\vec { OC }$$

$$-\vec { OC }$$

$$\vec { OC }$$

None of these

Solution

Question 6

$$1$$

$$\sqrt{57}$$

$$3$$

$$57$$

Solution

Question 7

$$1$$

$$\sqrt3$$

$$\dfrac {1}{2}$$

$$\dfrac {3}{2}$$

Solution

Question 8

only $$x$$

only $$y$$

both $$x$$ and $$y$$

neither $$x$$ nor $$y$$

Solution

Question 9

$$\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right)$$

$$\left(\overrightarrow{v}\times \overrightarrow{w}\right).\overrightarrow{u} $$

$$\overrightarrow{v}.\left(\overrightarrow{u}\times \overrightarrow{w}\right)$$

$$\left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$$

Solution

Question 10

$$\dfrac{15}{4}$$

$$\dfrac{15}{2}$$

$$15$$

$$16$$

Solution

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