Vector Algebra Test

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Vector Algebra Test - 66

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Question 1
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The position vector of the point which divides the join of points $$2 \vec a -3\vec b$$ and $$\vec a+\vec b$$ in the ratio $$3:1$$ is

A
$$\dfrac{3\vec a-2\vec b}{2}$$

B
$$\dfrac{7\vec a-8\vec b}{4}$$

C
$$\dfrac{3\vec a}{4}$$

D
$$\dfrac{5\vec a}{4}$$

Solution

Let the position vector of the point $$R$$ divides the join of points $$2\vec a-3\vec b$$ and $$\vec a+\vec b$$.
$$\therefore $$ Position vector $$R=\dfrac{3( \vec a+\vec b)+1( 2\vec a-3\vec b)}{3+1}$$
Since, the position vector of a point $$R$$ divides the line segment joining the points $$P$$ and $$Q$$, whose position vectors are $$\vec p$$ and $$\vec q$$ in the ration $$m:n$$ internally, is given by $$\dfrac{m\vec q +n\vec p}{m+n}$$
$$\therefore R=\dfrac{5\vec a}{4}$$
Question 2
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The value of $$\hat { i }. (\hat { j } \times \hat { k }) + \hat { j }. (\hat {
i } \times \hat { k })+\hat { k }. (\hat { i } \times \hat { j })$$ is

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$3$$

Solution
Question 3
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The value of $$ \hat {i} .( \hat {j} \times \hat {k}) + \hat {j} . ( \hat {i} \times \hat {k}) + \hat {k} .( \hat {i} \times \hat {j}) $$

A
$$ 0 $$

B
$$ -1 $$

C
$$ 1 $$

D
$$ 3 $$

Solution

$$ \overrightarrow {i} . ( \overrightarrow { j} \times \overrightarrow {k} ) + \overrightarrow {j} .( \overrightarrow {i} \times \overrightarrow {k}) + \overrightarrow {k} ( \overrightarrow {i} \times \overrightarrow {j}) $$
$$ \overrightarrow {i} . \overrightarrow {i'} + \overrightarrow {j} .( - \overrightarrow {j}) + \overrightarrow {k} . \overrightarrow {k} = | i|^2 - | j|^2 + |k|^2 $$
$$ = 1 - 1 +1 = +1 $$
hence answer is ( C).
Question 4
1 / -0

If $$ \overrightarrow {a} $$ is non zero vector of magnitude 'a ' and $$ \lambda $$ a nonzero scalar then $$ \lambda \overrightarrow {a} $$ is unit vector

A
$$ \lambda =1 $$

B
$$ \lambda = -1 $$

C
$$ a = | \lambda| $$

D
$$ a = 1 / | \lambda | $$

Solution

$$|\lambda\bar{a}| \space is \space a \space unit \space vector$$
$$So \space its \space magnitude \space is \space 1$$
$$ |\lambda \bar{a}|=|\lambda||\bar{a}|=|\lambda|a=1 $$
$$ \Rightarrow a=\frac{1}{|\lambda|} $$
$$Hence\space answer\space is\space (D)$$
Question 5
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Three vectors of magnitudes $$a,\ 2a,3a$$ meeting a point and three directions are along the diagonals of three adjacent faces of a cube. The magnitude of their resultant is

A
$$3a$$

B
$$5a$$

C
$$2a$$

D
$$4a$$

Solution

Solution:
Let the vectors of magnitudes $$a,2a,3a$$ act along $$OP,OQ,OR$$ respectively.Then vectors are $$OP,OQ,OR$$ are
$$a\left(\cfrac{\vec i+\vec j}{\sqrt2}\right),$$$$2a\left(\cfrac{\vec j+\vec k}{\sqrt2}\right),$$$$3a\left(\cfrac{\vec k+\vec i}{\sqrt2}\right)$$ respectively.
Their resultant say $$R$$ is given by
$$\vec R =$$$$a\left(\cfrac{\vec i+\vec j}{\sqrt2}\right)+$$$$2a\left(\cfrac{\vec j+\vec k}{\sqrt2}\right)+$$$$3a\left(\cfrac{\vec k+\vec i}{\sqrt2}\right)$$
$$=\cfrac{a}{\sqrt2}(4\vec i+3\vec j+5\vec k)$$
$$\therefore |\vec R|=\sqrt{\cfrac{a^2}2(16+9+25)}=5a$$
Hence, B is the correct answer.
Question 6
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If $$\vec {x}$$ is a vector whose initial point divides the line joining $$5\hat{i}$$, and $$ 5\hat{j}$$ in the ratio $$\lambda :1$$ and the terminal point is the origin. Also given $$\left | \vec {x} \right |\leq \sqrt{37}$$, then $$\lambda $$ belongs to

A
$$\left [ -\dfrac{1}{6} ,\dfrac{1}{6}\right ]$$

B
$$(-\infty ,-6)\cup \left ( -\dfrac{1}{6} ,\infty \right )$$

C
$$(-\infty ,-8)$$

D
$$(1,\infty )$$

Solution

From the condition given for $$\vec x$$ we have

$$\displaystyle \frac{|\lambda(5j)+5i|}{\lambda+1}\leq\sqrt{37}$$

$$25(\lambda^{2}+1)\leq 37(\lambda+1)^{2}$$

$$\Rightarrow(6\lambda+1)(\lambda+6)\geq 0$$

$$\displaystyle \lambda\in(-\infty, -6)\mathrm\cup (-\frac{1}{6}, \infty)$$
Question 7
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A scooterist follows a track on a ground that turns to his left by an angle 60$$^{0}$$ after every 400 m. Starting from the given point displacement of the scooterist at the third turn and eighth turn are :

A
$$800\mathrm{ m}; 0\mathrm{ m}$$

B
$$800\mathrm{m},\ 800\sqrt{3}\mathrm{m}$$

C
$$800\mathrm{m};400\sqrt{3}\mathrm{m}$$

D
$$800; 800\sqrt{3}\mathrm{m}$$

Solution

Let $$\bar { { x }_{ 1 } } ,\bar { { x }_{ 2 } } ,\bar { { x }_{ 3 } } ,.....\bar { { x }_{ 8 } } $$ be the positions of scooterist after $${ 1 }^{ st },{ 2 }^{ nd },{ 3 }^{ rd }......,{ 8 }^{ th }$$ turn respectively.
Thus its motion can be visualized as hexagon inscribed in circle of radius $$400m$$.
Let $$\bar { { x }_{ 0 } } $$ be its initial position.
$$\therefore $$ After $${ 3 }^{ rd }$$ turn
$${ x }_{ 3 } -{ x }_{ 0 }=800m$$
After $${ 8 }^{ th }$$ turn
$${ x }_{ 0 } - { x }_{ 2 } = 800\cos30^{0} = 400\sqrt 3m$$
Question 8
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$$\mathrm{l}\mathrm{n}$$ a triangle O$$\mathrm{A}\mathrm{B},\ \mathrm{E}$$ is the mid-point of $$\mathrm{O}\mathrm{B}$$ and $$\mathrm{D}$$ is a point on $$\mathrm{A}\mathrm{B}$$ such that $$\mathrm{A}\mathrm{D}$$: $$\mathrm{D}\mathrm{B}=2: 1$$. lf $$\mathrm{O}\mathrm{D}$$ and $$\mathrm{A}\mathrm{E}$$ interesect at $$\mathrm{P}$$, then the ratio $$\displaystyle\frac{OP}{PD}$$ is

A
$$1:2$$

B
$$3:2$$

C
$$8:3$$

D
$$4:3$$

Solution

$$\Delta OAB$$ let $$O=(0,0),A=(x_1,y_1),B=(x_2,y_2)$$
Then, $$E=(\dfrac{x_2}{2},\dfrac{y_2}{2})$$,$$D=(\dfrac{2x_2+x_1}{3},\dfrac{2y_2+y_1}{3})$$
Equation of OD is: $$y-0=\dfrac{2y_2+y_1}{2x_2+x_1}(x-0)$$
$$y=\dfrac{2y_2+y_1}{2x_2+x_1}x$$........(1)
Equation of AE is: $$(y-y_1)=\dfrac{y_2-2y_1}{x_2-2x_1}(x-x_1)$$
From eq(1)
$$(\dfrac{2y_2+y_1}{2x_2+x_1})(x)-y_1=\dfrac{y_2-2y_1}{x_2-2x_1}(x-x_1)$$
so $$ x=\dfrac{(2x_2+x_1)}{5} =\dfrac{3(\dfrac{2x_2+x_1}{3})+2(0)}{(3+2)}$$
so ratio $$OP:PD=3:2$$
Question 9
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In a quadrilateral $$PQRS,\ \vec{PQ}=\vec{a}, \vec{QR}=\vec{b}, \vec{SP}=\vec{a} - \vec{b}.\ M$$ is the mid-point of $$QR$$ and $$X$$ is a point on $$SM$$ such that $$\vec{SX}=\dfrac{4}{5}\vec{SM}$$, then $$\vec{PX}$$ is

A
$$\dfrac{1}{5}\vec{PR}$$

B
$$\dfrac{3}{5}\vec{PR}$$

C
$$\dfrac{2}{5}\vec{PR}$$

D
None of these

Solution

Choose $$P$$ as the origin of reference.
$$\vec{PR}=\vec{a}+\vec{b}$$,
$$\vec{PS}=-(\vec{a}-\vec{b})=\vec{b}-\vec{a}$$

$$\vec{PM}=\vec{a}+\dfrac{\vec{b}}{2}$$
Given :
$$\vec{SX}=\dfrac{4}{5}\vec{SM}$$
$$\vec{PX}-(\vec{b}-\vec{a})=\dfrac{4}{5}\left \{ \vec{a} +\dfrac{\vec{b}}{2}-(\vec{b}-\vec{a})\right \}$$
$$\Rightarrow \vec{PX}=\dfrac{3}{5}(\vec{a}+\vec{b})$$
$$=\dfrac{3}{5}\vec{PR}$$

Hence, option $$B.$$
Question 10
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The position vectors of $$A$$ and $$B$$ are $$2\hat{i}+2\hat{j}+\hat{k}$$ and $$2\hat{i}+4\hat{j}+4\hat{k}.$$ The length of the internal bisector of $$\angle BOA$$ of the triangle $$AOB$$ is

A
$$\displaystyle \sqrt { \dfrac { 136 }{ 9 } } $$

B
$$\displaystyle \sqrt { \dfrac { 139 }{ 9 } } $$

C
$$\displaystyle \dfrac { 20 }{ 3 } $$

D
$$\displaystyle \sqrt { \dfrac { 217 }{ 9 } } $$

Solution

Here $$OA=3,OB=6$$
$$\therefore$$ Internal bisector $$OD$$ divides $$AB$$ in the ratio $$1:2$$
$$\Rightarrow$$ Position vector of $$D$$ is $$\displaystyle \left( 2,\frac { 8 }{ 3 } ,2 \right) $$

$$\displaystyle \therefore \left| OD \right| =\sqrt { 4+\dfrac { 64 }{ 9 } +4 } =\sqrt { \dfrac { 136 }{ 9 } } $$