Vector Algebra Test

Result

Vector Algebra Test - 62

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Question 1
1 / -0

A unit vector $$d$$ is equally inclined at an angle $$\alpha$$ with the vectors $$a=\cos \theta. i+ \sin \theta. j , b=-\sin \theta.i+\cos =\theta. j$$ and $$c=k$$. Then $$\alpha$$ is equal to

A
$$\cos^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$$

B
$$\cos^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$$

C
$$\cos^{-1} \dfrac{1}{3}$$

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

Solution

$$\begin{array}{l} a=\cos \theta \, \, i+\sin \theta \, \, j \\ b=-\sin \theta \, \, i+\cos \theta \, \, j \\ c=\hat { k } \\ a.b=\left( { \cos \theta \, \, i+\sin \theta \, \, j } \right) \left( { -\sin \theta \, \, i+\cos \theta \, \, j } \right) \\ \, \, \, \, \, \, \, \, \, =-\cos \theta .\sin \theta +\sin \theta .\cos \theta \\ a.b=\left| a \right| \left| b \right| \, \, \cos \alpha \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ \begin{array}{l} 0=1\times 1\, \cos \alpha \\ 0=\cos \alpha \\ \alpha =\frac { \pi }{ 2 } \end{array} \right] \\ 0=\sqrt { \cos^2 \theta +\sin^2 \theta } .\sqrt { \sin^2 \theta +\cos^2 \theta } \, \, \cos \alpha \end{array}$$
Question 2
1 / -0

In the vectors $$\bar { AB } =3\hat { i } +4\hat { k } $$ and $$\bar { AC } =5\hat { i } -2\hat { j } +4\hat { k } $$ are the series of a triangle ABC, then the length of the median through A is

A
$$\sqrt { 18 } $$

B
$$\sqrt { 72 } $$

C
$$\sqrt { 33 } $$

D
$$\sqrt { 45 } $$

Solution

Take $$A$$ as the origin $$(0,0)$$
So vector $$AB = 3i + 4k$$
$$AC= 5i -2j+ 4k$$ will become position vector
So cordinate of $$B$$ is $$(3,0,4)$$ and $$C$$ is $$(5, -2, 4)$$
The midpoint of $$BC$$ can be easily found by midpoint formula of two points as $$D(4,-1,4)$$
So length of median $$A$$ is
$$AD = \sqrt{(4-0)^2 + (-1-0)^2 + (4-0)^2}$$
$$= \sqrt{33}$$
Question 3
1 / -0

The foot of the perpendicular drawn from a point with position vector $$\hat { i } +4\hat { k } $$ on the line joining the points $$\hat { j } +3\hat { k } $$, $$2\hat { i } -3\hat { j } +\hat { k } $$ is

A
$$4\hat { i } +5\hat { j } +5\hat { k } $$

B
$$\frac { 1 }{ 3 } (\hat { i }+\hat { j } +\hat { 8k } )$$

C
$$4\hat { i } +4\hat { j } -5\hat { k } $$

D
$$4\hat { i } -5\hat { j } +5\hat { k } $$

Solution

$$\begin{array}{l} \frac { x }{ 2 } =\frac { { y-1 } }{ { -4 } } =\frac { { z-3 } }{ { -2 } } \\ \frac { x }{ 1 } =\frac { { y-1 } }{ { -2 } } =\frac { { z-3 } }{ { -1 } } \\ Direction\, ratio\, of\, \overrightarrow { AB } :\, r-1,\, -2r+1,\, -r+1 \\ \left( { r-1 } \right) -2\left( { -2r+1 } \right) -\left( { -r-1 } \right) =0 \\ r=\frac { 1 }{ 3 } \\ foot\, of\, perpendicular\left( { \frac { 1 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 8 }{ 3 } } \right) \\ Now, \\ \frac { 1 }{ 3 } \left( { \hat { i } +\hat { j } +8\hat { k } } \right) \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$$
Question 4
1 / -0

The distance of the point $$ \text{P} $$ with position vector $$3\hat{i}+6 \hat{j}+8\hat{k} $$ from $$ y $$ - axis

A
$$\sqrt{62}$$

B
$$10$$

C
$$3\sqrt{5}$$

D
$$\sqrt{73}$$

Solution
Question 5
1 / -0

If $$\vec { a } , \vec { b } , \vec { c }$$ are unit vectors such that $$\vec { a } + \vec { b } + \vec { c } = 0 ,$$ the value of $$\vec { a } \cdot \vec { b } + \vec { b } \cdot \vec { c } + \vec { c } \cdot \vec { a }$$ is

A
$$1$$

B
$$3$$

C
$$\dfrac{-3}{2}$$

D
None of these

Solution
Question 6
1 / -0

If $$\left( \bar { a } -\bar { b } \right) =\bar { \left( a \right) } =\bar { \left( b \right) } $$ where $$\bar { a } $$ and $$\bar { b } $$ are non zero vectors then the angle between $$\bar { a } -\bar { b } $$

A
$${ 120 }^{ 0 }$$

B
$${ 45 }^{ 0 }$$

C
$${ 60 }^{ 0 }$$

D
$${ 90 }^{ 0 }$$

Solution
Question 7
1 / -0

Line passing through $$ (3,4,5) $$ and $$ (4,5,6) $$ has direction ratios $$ \ldots $$

A
$$ 1,1,1 $$

B
$$ \sqrt{3}, \sqrt{3}, \sqrt{3} $$

C
$$ \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} $$

D
$$ 7,9,11 $$

Solution

Given points $$(3,4,5)$$ and $$(4,5,6)$$
The drs are given as $$(4-3,5-4,6-5)=(1,1,1)$$
Question 8
1 / -0

If radius vector of a point varies with time $$t$$ as $$\vec { r } =\vec { b } t\left( 1-\alpha t \right) $$ where $$\vec { b } $$ is a constant vector and a is a positive constant, then its

A
Velocity is constant

B
Acceleration is constant

C
Magnitude of acceleration is constant but direction of acceleration is changing

D
Magnitude of velocity is changing but its direction is unchanged

Solution
Question 9
1 / -0

let $$\bar { a } ,\bar { b } ,\bar { c } $$ are three mutually perpendicular unit vectors and a unit vector $$ \bar { r } $$ satisfying the equation $$\left( \bar { b } -\bar { c } \right) \times \left( \bar { r } \times \bar { a } \right) +\left( \bar { c } -\bar { a } \right) \times \left( \bar { r } \times \bar { b } \right) +\left( \bar { a } -\bar { b } \right) \times \left( \bar { r } \times \bar { c } \right) =0$$ then $$\bar { r } $$ is __________________.

A
$$\dfrac { 1 }{ \sqrt { 3 } } \left( \bar { a } +\bar { b } +\bar { c } \right) $$

B
$$\dfrac { 1 }{ \sqrt { 14 } } \left( 2\bar { a } +3\bar { b } +\bar { c } \right) $$

C
$$-\dfrac { 1 }{ \sqrt { 14 } } \left( 2\bar { a } +3\bar { b } +\bar { c } \right) $$

D
$$-\dfrac { 1 }{ \sqrt { 3 } } \left( \bar { a } +\bar { b } +\bar { c } \right) $$
Question 10
1 / -0

Let $$\overline { a } =4\hat { i } +3\hat { j } -\hat { k } $$ and$$\overline { b } =2\hat { i } -6\hat { j } -3\hat { k } .$$ Then a unit vector $$\bot $$ to both $$\overline { a }$$ and $$\overline { b } $$is.

A
$$\dfrac { 1 }{ 7 } \left( -3\hat { i } -2\hat { j } +3\hat { k } \right) $$

B
$$\dfrac { 1 }{ 7 } \left( 3\hat { i } +2\hat { j } -6\hat { k } \right) $$

C
$$\dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right) $$

D
$$\dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right) $$

Solution