Vector Algebra Test

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Vector Algebra Test - 59

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Question 1
1 / -0

If $$M$$ and $$N$$ are the mid-points of the diagonals $$AC$$ and $$BD$$ respectively of a quadrilateral $$ABCD$$, then the value of $$\overline { AB } +\overline { AD } +\overline { CB } +\overline { CD } $$

A
$$2\overline { MN } $$

B
$$2\overline { NM } $$

C
$$4\overline { NM } $$

D
$$4\overline { MN } $$

Solution

$$\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}$$
$$=\vec{b}-\vec{a}+\vec{d}-\vec{a}+\vec{b}+\vec{c}+\vec{d}-\vec{c}$$
$$=2\vec{b}+2\vec{d}-2\vec{a}-2\vec{c}$$
$$=2(\vec{b}+\vec{d}-\vec{a}-\vec{c})$$
$$=4\left(\dfrac{\vec{b}+\vec{d}}{2}-\dfrac{\vec{a}+\vec{c}}{2}\right)$$
$$=4(\vec{N}-\vec{M})$$
$$=4\vec{MN}$$.
Question 2
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If $$\overline { a } $$ and $$\overline { b } $$ include an angle of $${120}^{o}$$ and their magnitudes are $$2$$ and $$\sqrt{3}$$ then $$\overline { a } .\overline { b } $$ is

A
$$3$$

B
$$-\sqrt{3}$$

C
$$\sqrt{3}$$

D
$$-3$$

Solution

Given, $$|\vec{a}| = 2, \, |\vec{b}| = \sqrt{3}$$
$$\theta = 120^o$$
$$\vec{a} . \vec{b} = |\vec{a}| |\vec{b}| cos \theta$$
$$= 2 \times \sqrt{3} \times cos \, 120^o$$
$$= 2 \sqrt{3} \times \left(\dfrac{-1}{2} \right)$$
$$\therefore \vec{a} . \vec{b} = - \sqrt{3}$$
Question 3
1 / -0

The perimeter of the triangle whose vertices are the points $$2\bar{i}-\bar{j}+\bar{k}, \bar{i}-3\bar{j}-5\bar{K},3\bar{i}-4\bar{j}-4\bar{j}-4\bar{k}$$ is

A
$$\sqrt{3}+\sqrt{35}+\sqrt{7}$$

B
$$\sqrt{6}+\sqrt{35}+\sqrt{7}$$

C
$$\sqrt{6}+\sqrt{30}+\sqrt{41}$$

D
$$\sqrt{6}+\sqrt{35}+\sqrt{41}$$

Solution

Given position vector

$$A=2i-j+k$$

$$B=i-3j+k$$

$$C=3i-4j-4k$$

Perimeter$$=|\overline{AB}|+|\overline{BC}|+|\overline{CA}|$$

$$|\overline{AB}|=\sqrt{(2-1)^2+(-1+3)^2+(1+5)^2}=\sqrt{1+4+36}=\sqrt{41}$$

$$|\overline{BC}|=\sqrt{(3-1)^2+(-4+3)^2+(-4+5)^2}=\sqrt{4+1+1}=\sqrt{6}$$

$$|\overline{CA}|=\sqrt{(3-2)^2+(-4+1)^2+(-4-1)^2}=\sqrt{1+9+25}=\sqrt{35}$$

$$\therefore$$ Perimeter$$=\sqrt{6}+\sqrt{35}+\sqrt{41}$$.
Question 4
1 / -0

If a,b,c are all non-zero, then the number of values of $$\lambda $$ such that $$a(-4\bar{i}+5\bar{j})+b(3\bar{i}-3\bar{j}+\bar{k})+c(\bar{i}+\bar{j}+3\bar{k})=\lambda (a\bar{i}+b\bar{j}+c\bar{k})$$ is :

A
3

B
2

C
1

D
0

Solution
Question 5
1 / -0

If $$S$$ is the circumcentre, $$O$$ is the orthocentre of $$\triangle{ABC}$$, then $$\overline { SA } +\overline { SB } +\overline { SB } $$ equals

A
$$\overline { SO } $$

B
$$2\overline { SO } $$

C
$$\overline { OS } $$

D
$$2\overline { OS } $$

Solution

Given, 'S' is the circumcentre
'O' is the orthocentre
$$\vec{SA}=\vec{a}$$
$$\vec{SB}=\vec{b}$$
$$\vec{SC}=\vec{c}$$
$$\vec{SO}=\vec{s}$$
$$\vec{SP}||\vec{AO}$$
$$\vec{SP}=\dfrac{\vec{SB}+\vec{SC}}{2}=\dfrac{\vec{b}+\vec{c}}{2}$$
$$\vec{SP}=\lambda \vec{AO}$$
$$\Rightarrow \dfrac{\vec{b}+\vec{c}}{2}=\lambda (\vec{s}-\vec{a})$$
$$\Rightarrow \vec{s}=\dfrac{\vec{b}+\vec{c}}{2\lambda}+\vec{a}$$ …….$$(1)$$
$$\vec{SQ}=\mu \vec{BO}$$
$$\dfrac{\vec{a}+\vec{c}}{2}=\mu (\vec{s}-\vec{b})$$
$$\Rightarrow \vec{s}=\dfrac{\vec{a}+\vec{c}}{2\mu}+\vec{b}$$ ……….$$(2)$$
$$\vec{a}+\dfrac{\vec{b}+\vec{c}}{2\lambda}=\vec{b}+\dfrac{\vec{a}+\vec{c}}{2\mu}$$
$$\Rightarrow \vec{a}+\dfrac{\vec{b^2}}{2\lambda}+\dfrac{\vec{c}}{2\lambda}=\vec{b}+\dfrac{\vec{a}}{2\mu}+\dfrac{\vec{c}}{2\mu}$$
$$\therefore 1=\dfrac{1}{2\mu}$$
$$\Rightarrow \mu =\dfrac{1}{2}$$
$$\therefore \vec{s}=\vec{a}+\vec{c}+\vec{b}$$
$$\therefore \vec{SO}=\vec{SA}+\vec{SB}+\vec{SC}$$.
Question 6
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If $$\overline { a } =\left( 2\overline { i } -10\overline { j } +6\overline { k } \right) ;\overline { b } =\left( 5\overline { i } -3\overline { j } +\overline { k } \right) $$. The ratio of projection of $$\overline { a } $$ on $$\overline { b } $$ to projection of $$\overline { b } $$ on $$\overline { a } $$ is

A
$$2:1$$

B
$$1:2$$

C
$$2:3$$

D
$$3:2$$

Solution

$$\cfrac { Projection\quad of\quad \vec { a } \quad on\quad \vec { b } \quad }{ Projection\quad of\quad \vec { b } \quad on\quad \vec { a } } $$
$$=\cfrac { \cfrac { \vec { a } .\vec { b } }{ \left| \vec { b } \right| } }{ \cfrac { \vec { b } .\vec { a } }{ \left| \vec { a } \right| } } =\cfrac { \left| \vec { a } \right| }{ \left| \vec { b } \right| } =\cfrac { \sqrt { 4+100+76 } }{ \sqrt { 25+9+1 } } =\cfrac { 2 }{ 1 } $$
Question 7
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Let $$\bar{a},\bar{b}$$ be two noncollinear vectors. If $$A=(x+4y)\bar{a}+(2x+y+1)\bar{b},$$
$$B=(y-2x+2)\bar{a}+(2x-3y-1)\bar{b} \quad and \quad 3A=2B$$ then (x,y) =

A
(1,2)

B
(1,-2)

C
(2,-1)

D
(-2,-1)

Solution

$$3A=(3x+12y)\bar{a}+(6x+3y+3)\bar{b}$$
$$2B=(2y-4x+4)\bar{a}+(4x-6y-2)\bar{b}$$
$$3A=2B$$
$$\Rightarrow 3x+12y=2y-4x+4$$ & $$6x+3y+3=4x-6y-2$$
$$7x+10y=4$$ & $$2x+9y=-5$$
Solving $$7x+10y=4$$ & $$2x+9y=-5$$
We get $$x=2$$, $$y=-1$$
$$\therefore (x, y)=(2, -1)$$.
Question 8
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If $$\overline { a } $$ is a vector of magnitude $$\sqrt{3}$$ and $$\overline { b } $$ is unit vector making an angle $$\tan ^{ -1 }{ \left( 1/\sqrt { 2 } \right) } $$ with $$\overline { a } $$ then projection of $$\overline { a } $$ on $$\overline { b } $$ is

A
$$\cfrac{\sqrt{3}}{2}$$

B
$$\sqrt{2}$$

C
$$\sqrt{3}$$

D
$$\sqrt{6}$$

Solution

$$\left| \vec { a } \right| =\sqrt 3$$, $$\left| \vec { b } \right| =1$$
$$\tan \theta =\cfrac{1}{\sqrt 2}$$
$$\vec { a } .\vec { b } =\left| \vec { a } \right| \left| \vec { b } \right| \cos \theta$$
$$=\sqrt 3 \times 1\times \cfrac{\sqrt 2}{\sqrt 3}$$
$$=\sqrt 2$$
Projection of $$\vec { a } $$ on $$\vec { b } $$ $$=\cfrac { \vec { a } .\vec { b } }{ \left| \vec { b } \right| } $$
$$=\cfrac{\sqrt 2}{1}=\sqrt 2$$
Question 9
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Given $$\vec{\alpha} = 3\hat{i} + \hat{j} + 2\hat{k}\ ,\ \vec{\beta} = \hat{i} - 2\hat{j} - 4\hat{k}$$ are the position vectors of the points $$A$$ and $$B$$. Then the distance of the point $$-\hat{i} + \hat{j} + \hat{k}$$ from the passing through $$B$$ and perpendicular to $$AB$$ is

A
$$-7$$

B
$$10$$

C
$$15$$

D
$$20$$

Solution

We have,

$$ \overrightarrow{\alpha }=3\widehat{i}+\widehat{j}+2\widehat{k} $$

$$ \overrightarrow{\beta }=\widehat{i}-2\widehat{j}-4\widehat{k} $$

So,
$$ \overrightarrow{AB}=\overrightarrow{\beta }-\overrightarrow{\alpha } $$

$$ \overrightarrow{AB}=\left( \widehat{i}-2\widehat{j}-4\widehat{k} \right)-\left( 3\widehat{i}+\widehat{j}+2\widehat{k} \right) $$

$$ \overrightarrow{AB}=\widehat{i}-2\widehat{j}-4\widehat{k}-3\widehat{i}-\widehat{j}-2\widehat{k} $$

$$ \overrightarrow{AB}=-2\widehat{i}-3\widehat{j}-6\widehat{k} $$

Now, equation of plane passing through the point and perpendiular vector is

So,

$$ \left( -2\widehat{i}-3\widehat{j}-6\widehat{k} \right).\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)=\left( -2\widehat{i}-3\widehat{j}-6\widehat{k} \right).\left( -\widehat{i}+\widehat{j}+\widehat{k} \right) $$

$$ \Rightarrow 2x+3y+6z=2-3-6 $$

$$ \Rightarrow 2x+3y+6z=-7 $$
Hence, this is the answer.
Question 10
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A (1,-1,-1) , B (2,1,-2) and C (-5,2,-6) are the position vectors of the vertices of triangle ABC The length of the bisector of its internal angle at A is:

A
$$\dfrac{\sqrt{10}}{4}$$

B
$$\dfrac{3\sqrt{10}}{4}$$

C
$$\sqrt{10}$$

D
none

Solution

$$\dfrac {AB}{AC}=\dfrac {BE}{EC}$$
$$\dfrac {\sqrt 6}{3\sqrt 6}=\dfrac {BE}{EC}=\dfrac {1}{3}$$
So point $$E=\left (\dfrac {1}{4}+\dfrac {5}{4},-3\right)$$
So $$AE=\dfrac {3\sqrt {10}}{4}$$