MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 10

Let $$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}, b=\hat{i}+2\hat{j}-2\hat{k}$$. Then a unit vector perpendicular to both $$\vec{a}-\vec{b}$$ and $$\vec{a}+\vec{b}$$ is :

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

If $$\vec a,\vec b, \vec c$$ are three coplanar vectors, then $$v\left[ 2\vec { a } +3\vec { b } ,2\vec { b- } 5\vec { c } ,2\vec { c } +3\vec { a } \right]$$ is

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$$0$$
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$$1$$
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$$-\sqrt 3$$
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$$\sqrt 3$$

If $$\vec { x }$$ and $$\vec { y } $$ be unit vectors and $$\left| \vec { z } \right| =\frac { 2 }{ \sqrt { 7 } } $$ such that $$\vec { z } +\vec { z } \times \vec { x } =\vec { y } ,$$ then the angle $$\theta $$ between $$\vec { x }$$ and $$\vec { z } $$ can be

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$${ 30 }^{ \circ }$$
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$${ 60 }^{ \circ }$$
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$${ 90 }^{ \circ }$$
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None of these

The position vector of point C with respect to B is $$\bar { i } +\bar { j } $$ . and that of B with respect to A is $$\bar { i } +\bar { j } $$.The position vector of C with respect to A is ____________.

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$$2\bar { i } $$
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$$-2\bar { i } $$
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$$2\bar { j } $$
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$$-2\bar { j } $$

If the vectors $$3 \overline { p } + \overline { q } : 5 \overline { p } - 3 \overline { q }$$ and $$2 \overline { p } + \overline { q } ; 4 \overline { p } - 2 \overline { q }$$ are pairs of mutually perpendicular vectors then $$\sin ( \overline { p } \overline { q } )$$ is:

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$$\frac { \sqrt { 55 } } { 4 }$$
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$$\frac { \sqrt { 55 } } { 8 }$$
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$$\frac { 3 } { 16 }$$
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$$\frac {{ \sqrt 247 }} { 16 }$$

Let $$\hat {a}$$ and $$\hat {b}$$ two unit vector such that $${ \left( \hat { a } .\hat { b } \right) }^{ 2 }-\left| \hat { a } \times \hat { b } \right| $$ is maximum then $$\left| \hat { a } .\hat { b } \right|$$ is equal to

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$$1$$
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$$\dfrac{1}{3}$$
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$$0$$
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$$-\dfrac{1}{3}$$

The cartesian equation of the plane perpendicular to vector $$3\bar {i}-2\bar {j}-2\bar {k}$$ and passing through the point $$2\bar {i}+3\bar {j}-\bar {k}$$ is

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$$3x+2y+2z=2$$
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$$3x-2y+2z=2$$
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$$3x+2y-2z=2$$
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$$3x-2y-2z=2$$

The position vectors of two vertices and the centroid of a triangle are $$\overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } ,\overset { \rightarrow }{ 2i } -\overset { \rightarrow }{ j } +\overset { \rightarrow }{ k } $$ and $$\overset { \rightarrow }{ k } $$ respectively. The position vector of the third vertex of the triangle is :

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$$\overset { \rightarrow }{ -3i } +\overset { \rightarrow }{ 2k } $$
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$$\overset { \rightarrow }{ 3i } -\overset { \rightarrow }{ 2k } $$
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$$\overset { \rightarrow }{ i } +\frac { 2 }{ 3 } \overset { \rightarrow }{ k } $$
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none of these

Unit vector perpendicular to the plane of the triangle $$ABC$$ with position vectors of the vertices $$A , B , C ,$$ is $$($$ where $$\Delta$$ is the area of the triangle $$A B C$$ ) .

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$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { \Delta }$$
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$$\dfrac { ( \vec { a } \times \vec { b } + \overline { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 2 \Delta }$$
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$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 3\Delta }$$
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$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 4\Delta }$$

$$\bar { a } ,\bar { b } $$ and $$\bar { c } $$ are unit vector such that $$\bar { a } +\bar { b } -\bar { c } =0$$. then the angle between $$\bar { a } $$ and $$\bar { b } $$ is :-

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$$\dfrac { \pi }{ 6 } $$
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$$\dfrac { \pi }{ 3 } $$
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$$\dfrac { \pi }{ 2 } $$
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$$\dfrac { 2\pi }{ 3 }$$

Let $$a=(1,-2,3)$$ and $$b=(2,7,4)$$ then

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$$a.b=0$$
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$$a.b=-9$$
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$$a.b=4$$
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$$a.b=-4$$

If $$u,\ v,\ w$$ are non-coplanar vector and $$p,\ q$$ are real numbers, then the equality $$[3u\ pv\ pw]-[pv\ w\ qw]-[2w\ qv\ qu]=0$$ holds for

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Exactly two values of $$(p,\ q)$$
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More than but not all values of $$(p,\ q)$$
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All values of $$(p,\ q)$$
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Exactly one values of $$(p,\ q)$$

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

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$$\sqrt { 18 } $$
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$$\sqrt { 72 } $$
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$$\sqrt { 33 } $$
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$$\sqrt { 45 } $$

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

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$$\sqrt { 18 } $$
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$$\sqrt { 72 } $$
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$$\sqrt { 33 } $$
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$$\sqrt { 45 } $$

Let $$\overrightarrow { a } ,\overrightarrow { b }$$ and $$\overrightarrow { c } $$ be three non-zero vectors such that no two of them are collinear and $$(\overrightarrow { a } \times \overrightarrow { b } )\times \overrightarrow { c } =\frac { 1 }{ 3 } \left| \overrightarrow { b } \right| \left| \overrightarrow { c } \right| \overrightarrow { a } $$. If $$\theta$$ is the angle between vectors $$\overrightarrow { b }$$ and $$\overrightarrow { c }$$, then a value of $$\sin { \theta } $$ is

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$$\frac 23$$
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$$\frac { -2\sqrt { 3 } }{ 3 } $$
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$$\frac { 2\sqrt { 2 } }{ 3 } $$
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$$\frac { -\sqrt { 2} }{ 3 } $$

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

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$$\cos^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$$
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$$\cos^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$$
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$$\cos^{-1} \dfrac{1}{3}$$
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$$\dfrac{\pi}{2}$$

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

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$$4\hat { i } +5\hat { j } +5\hat { k } $$
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$$\frac { 1 }{ 3 } (\hat { i }+\hat { j } +\hat { 8k } )$$
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$$4\hat { i } +4\hat { j } -5\hat { k } $$
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$$4\hat { i } -5\hat { j } +5\hat { k } $$

Let $$ABCD$$ is a triangular pyramid with base vectors $$\vec {AB}= 2\bar {i}+3\bar {j}-\bar {k}$$ and $$\vec {AC}=\bar {i}-2\bar {k}$$, If volume of the triangular pyramid is $$\sqrt{150}$$ unit then its height is

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$$10$$
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$$20$$
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$$18$$
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$$23$$

If $$\left | \vec{c} \right |$$ = 60 and $$\vec{c}$$ ($$\hat{i}$$ + 2$$\hat{j}$$ + 5$$\hat{k}$$) = 0 , then a value of $$\vec{c}$$.(7$$\hat{i}$$ + 2$$\hat{i}$$ + 3$$\hat{k}$$) is :

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$$4\sqrt{2}$$
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12
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24
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$$12\sqrt{2}$$

For any vectors $$\vec{a}$$, the value of $$(\vec{a}\times \hat{i})^2+(\vec{a}\times \hat{j})^2+(a\times \hat{k})^2$$ is equal to?

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$$3\vec{a^2}$$
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$$\vec{a^2}$$
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$$2\vec{a^2}$$
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None of these

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

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$$\sqrt{62}$$
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$$10$$
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$$3\sqrt{5}$$
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$$\sqrt{73}$$

The adjacent sides of a parallelogram are $$ \vec{A}=2 \hat{i}-3 \hat{j}+\hat{k} $$ and $$ \vec{B}=-2 \hat{i}+4 \hat{j}-\hat{k} $$ What is the area of the parallelogram?

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$$4$$ units
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$$7$$ units
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$$ \sqrt{5} $$ units
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$$ \sqrt{8} $$ units

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

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$$1$$
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$$3$$
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$$\dfrac{-3}{2}$$
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None of these

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 $$\bar { a } -\bar { b } $$

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$${ 120 }^{ 0 }$$
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$${ 45 }^{ 0 }$$
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$${ 60 }^{ 0 }$$
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$${ 90 }^{ 0 }$$

If the vectors $$2\hat i + 3 \hat j , 5 \hat i + 6 \hat j , 8 \hat i +$$ $$\lambda$$j have their initial point at $$(1 , 1)$$$ then the value of $$\lambda$$ so that the vectors terminated on one line is

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5
0%
9
0%
4
0%
0

If $$\bar { OP } =2\hat { i } +3\hat { j } -\hat { k } $$ and $$\bar { OQ } =3\hat { i } -4\hat { j } -2\hat { k } $$ then the modulus $$\bar { PQ } $$ is

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$$\sqrt { 13 } $$
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$$\sqrt { 51 } $$
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$$\sqrt { 39 } $$
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$$\sqrt { 67 } $$

From the figure the correct relation is :

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$$\bar { A } +\bar { B } +\bar { E } =\bar { 0 } $$
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$$\bar { C } +\bar { D } =\bar { -A } $$
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$$\bar { B } +\bar { E } +\bar { C } =\bar { -D } $$
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All of these

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

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$$ 1,1,1 $$
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$$ \sqrt{3}, \sqrt{3}, \sqrt{3} $$
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$$ \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} $$
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$$ 7,9,11 $$

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 __________________.

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

$$ If \bar { A } = 2 \hat {i} + \hat {j} + \hat {k}\ and\ \bar { B } = \hat {i} + \hat {j} + \hat {k} $$ two vectors,then the unit vector is

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Perpendicular to $$ \bar { A }\ is\ \dfrac { -\hat { j } +\hat { k } }{ \sqrt { 2 } } $$
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Parallel to $$ \bar { A }\ is\ \dfrac { 2 +\hat { j } +\hat { k } }{ \sqrt { 6 } } $$
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Parallel to $$ \bar { B }\ is\ \dfrac { -\hat { j } +\hat { k } }{ \sqrt { 2 } } $$
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Parallel to $$ \dfrac { \hat { i } +\hat { j } +\hat { k } }{ 3 } $$

If $$\vec { x } $$ is a vector in the direction of $$(2,-2,1)$$ of magnitude $$6$$ and $$\vec { y } $$ is a vector in the direction of $$(1,1,-1)$$ of magnitude $$\sqrt{3}$$, then $$\left| \vec { x } +2\vec { y } \right| =...$$

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$$40$$
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$$\sqrt { 35 } $$
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$$\sqrt { 17 } $$
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$$2\sqrt { 10 } $$

The position vector of a point P is $$\overrightarrow r=x \overrightarrow i + y \overrightarrow j+x \overrightarrow k$$, Where $$x,y,z,\epsilon N$$ and $$\overrightarrow a= \overrightarrow i+ \overrightarrow j+\overrightarrow k$$. If $$\overrightarrow r. \overrightarrow a=10$$, then the number of possible positions of P is ___________.

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$$30$$
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$$72$$
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$$66$$
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$$36$$

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.

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$$\dfrac { 1 }{ 7 } \left( -3\hat { i } -2\hat { j } +3\hat { k } \right) $$
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$$\dfrac { 1 }{ 7 } \left( 3\hat { i } +2\hat { j } -6\hat { k } \right) $$
0%
$$\dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right) $$
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$$\dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right) $$

If $$\vec{a}=\hat{i}+2\hat{j}+2\hat{k}$$ and $$\vec{b}=2\hat{i}+\hat{j}+2\hat{k}$$. Find the projection vector of $$\vec{b}$$ on $$\vec{a}$$.

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$$\displaystyle\frac{8}{9}\left(\hat{i}+2\hat{j}+2\hat{k}\right)$$
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$$\displaystyle\frac{8}{9}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$$
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$$\displaystyle\frac{9}{8}\left(\hat{i}+2\hat{j}+2\hat{k}\right)$$
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$$\displaystyle\frac{9}{8}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$$

Unit vector perpendicular to vector $$A=-3\hat { i } -2\hat { j } -3\hat { k } $$ and $$B=2\hat { { i } } +4\hat { { j } } +6\hat { { k } } $$ both is

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$$\dfrac { 3\hat { { j } } -2\hat { { k } } }{ \sqrt { 13 } } $$
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$$\dfrac { 3\hat { { k } } -2\hat { { j } } }{ \sqrt { 13 } } $$
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$$\dfrac { - \hat { { j } } +2\hat { { k } } }{ \sqrt { 13 } } $$
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$$\dfrac { \hat { { i } } + 3\hat { { j } } -\hat { { k } } }{ \sqrt { 13 } } $$

If the position vector $$\vec{a}$$ of point $$(12, n) $$ is such that $$\left | \vec{a} \right | = 13$$, then find the value (s) of $$n$$.

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$$\pm 6$$
0%
$$\pm 4$$
0%
$$\pm 5$$
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$$\pm 7$$

Express $$ \vec{AB}$$ in terms of unit vectors $$ \hat{i} $$ and $$\hat{j}$$, when the points are:
A(4,-1), B(1,3)
Find $$ \left | \vec{AB} \right |$$ in each case.

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$$ \vec{AB} = -3\hat{i}-4\hat{j}, \left | \vec{AB} \right | = 5 $$
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$$ \vec{AB} = +3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5 $$
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$$ \vec{AB} = -3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5 $$
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none of these

What is the scalar projection of
$$\vec{a}=\hat{i}+2\hat{j}+\hat{k}$$ on $$\vec{b}=4\hat{i}+4\hat{j}+7\hat{k}
$$ ?

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$$\dfrac{\sqrt{6}}{9}$$
0%
$$\dfrac{19}{9}$$
0%
$$\dfrac{9}{19}$$
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$$\dfrac{\sqrt{6}}{19}$$

If $$a, b, c$$ are vectors such that $$a+b+c = 0$$ and $$|a| = 7, |b| = 5, |c| = 3$$, then the angle between $$c$$ and $$b$$ is

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$$\dfrac{\pi}{3}$$
0%
$$\dfrac{\pi}{6}$$
0%
$$\dfrac{\pi}{4}$$
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$$\pi$$

If a unit vector $$ \vec{a} $$ makes an angle $$ \dfrac{\pi }{3} $$ with $$ \hat{i},\dfrac{\pi }{4} $$ with $$ \hat{j} $$ and an accute angle $$ \theta $$ with $$ \hat{k}, $$ then find $$ \theta $$ and hence, the components of $$ \vec{a} $$.

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$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$
0%
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{-1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$
0%
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$
0%
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}-\dfrac{1}{2}\hat{k} $$

The adjacent sides of a parallelogram are represented by the vectors $$ \vec{a} = \hat{i}+\hat{j}+\hat{k} $$ and $$ \vec{b} = 2\hat{i}+\hat{j}+2\hat{k}.$$ Find unit vectors parallel to the diagonals of the parallelogram.

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$$ \dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}+\hat{k}) $$
0%
$$ \dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k}) $$
0%
$$ \dfrac{1}{\sqrt{22}}(3\hat{i}+2\hat{j}+3\hat{k}),\dfrac{1}{\sqrt{2}}(\hat{i}+\hat{k}) $$
0%
$$ \dfrac{1}{\sqrt{2}}(+\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k}) $$

The unit vector normal to the plane containing $$\vec{a}=(\hat{i}-\hat{j}-\hat{k})$$ and $$\vec{b}=(\hat{i}+\hat{j}+\hat{k})$$ is?

0%
$$(\hat{j}-\hat{k})$$
0%
$$(-\hat{j}+\hat{k})$$
0%
$$\dfrac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$$
0%
$$\dfrac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$$

Let $$\vec{a}=2\hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-\hat{k}$$ and $$\vec{c}=\hat{i}+\hat{j}-2\hat{k}$$ be three vectors. A vector in the plane of $$\vec{b}$$ and $$\vec{c}$$ whose projection on $$\vec{a}$$ is of magnitude $$\sqrt{(2/3)}$$ is

0%
$$2\hat{i}+3\hat{j}-3\hat{k}$$
0%
$$2\hat{i}+3\hat{j}+3\hat{k}$$
0%
$$-2\hat{i}-\hat{j}+5\hat{k}$$
0%
$$2\hat{i}+\hat{j}+5\hat{k}$$

If $$\bar{a}$$ and $$\bar{b} = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$$ are collinear and $$\bar{a} . \bar{b} = 27$$, then $$\bar{a}$$ is equal to

0%
$$3 (\hat{i} + \hat{j} + \hat{k})$$
0%
$$\hat{i} + 2\hat{j} + 2 \hat{k}$$
0%
$$2 \hat{i} + 2\hat{j} + 2 \hat{k}$$
0%
$$\hat{i} + 3\hat{j} + 3 \hat{k}$$
0%
$$\hat{i} - 3 \hat{j} + 2 \hat{k}$$

Let $$O$$ be the circumcentre, $$G$$ be the centroid and $$O$$ be the orthocentre of a $$\triangle ABC$$. Three vectors are taken through $$O$$ and are represented by $$\vec{a}=\vec{OA}, \vec{b}=\vec{OB}$$ and $$\vec{c}=\vec{OC}$$ then $$\vec{a}+\vec{b}+\vec{c}$$ is

0%
$$\vec{OG}$$
0%
$$2\vec{OG}$$
0%
$$\vec{OO}$$
0%
None of them

If $$(\vec a\times \vec b)^2 +(\vec a. \vec b)^2 =144$$ and $$|\vec a|=4$$, then $$|\vec b|=$$

0%
$$16$$
0%
$$8$$
0%
$$3$$
0%
$$12$$

A parallelogram is constructed on the vectors
$$\vec{a}=3\vec{\alpha}-\vec{\beta}, \vec{b}=\vec{\alpha}+3\vec{\beta}$$ if $$|\vec{\alpha}|=|\vec{\beta}|=2$$ and angle between $$\vec{\alpha}$$ and $$\vec{\beta}$$ is $$\pi/3$$ then the length of a diagonal of the parallelogram is

0%
$$4\sqrt{5}$$
0%
$$4\sqrt{3}$$
0%
$$4\sqrt{7}$$
0%
$$None\ of\ these$$

$$A, B, C$$ and $$D$$ have position vectors $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$ respectively, such that $$\vec{a} - \vec{b} = 2 (\vec{d} - \vec{c})$$. Then

0%
$$AB$$ and $$CD$$ bisect each other
0%
$$BD$$ and $$AC$$ bisect each other
0%
$$AB$$ and $$CD$$ trisect each other
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$$BD$$ and $$AC$$ trisect each other

$$p\hat{i}+3\hat{j}+4\hat{k}$$ and $$\sqrt{q}\hat{i}+4\hat{k}$$ are two vectors, where $$p,q>0$$ are two scalars, then the length of the vectors is equal to

0%
All value of $$(p,q)$$
0%
Only finite number of values of $$(p,q)$$
0%
Infinite number of values of $$(p,q)$$
0%
No value fo $$(p,q)$$

$$(\vec r. \hat i)(\vec r \times \hat i)+ (\vec r. \hat j)(\vec r \times \hat j) +(\vec r. \hat k)(\vec r \times \hat k)$$ is equal to

0%
$$3\ \vec r$$
0%
$$\vec r$$
0%
$$\vec 0$$
0%
$$None\ of\ these$$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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