Vector Algebra Test

Result

Vector Algebra Test - 61

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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

A
$${ 30 }^{ \circ }$$

B
$${ 60 }^{ \circ }$$

C
$${ 90 }^{ \circ }$$

D
None of these

Solution

We have,
$$ \vec { z } +\vec { z } \times \vec { x } =\vec { y } \\ \left| { \vec { z } +\vec { z } \times \vec { x } } \right| =\left| { \vec { y } } \right| =1 \\ { \left| { \vec { z } +\vec { z } \times \vec { x } } \right| ^{ 2 } }=1 \\ \left( { \vec { z } +\vec { z } \times \vec { x } } \right) .\left( { \vec { z } +\vec { z } \times \vec { x } } \right) =1 \\ \Rightarrow { \left| { \vec { z } } \right| ^{ 2 } }+\left| { \vec { z } \times \vec { x } } \right| =1$$
$$ \\ \dfrac { 4 }{ 7 } +\dfrac { 4 }{ 7 } { \sin ^{ 2 } }\theta =1 \\ { \sin ^{ 2 } }\theta =\dfrac { 3 }{ 4 } \\ \sin \theta =\dfrac { { \sqrt { 3 } } }{ 2 } \\ \therefore \theta ={ 60^{ o } }$$

$$ \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer.$$
Question 2
1 / -0

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

A
$$a.b=0$$

B
$$a.b=-9$$

C
$$a.b=4$$

D
$$a.b=-4$$

Solution

Given, $$a=(1,-2,3)$$ and $$b=(2,7,4)$$.
Now,
$$a.b$$
$$=(1,-2,3).(2,7,4)$$
$$=2-14+12$$
$$=0$$.
Question 3
1 / -0

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

A
$$1$$

B
$$\dfrac{1}{3}$$

C
$$0$$

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

Solution

We have,
$$\begin{array}{l} { \left( { \vec { a } .\vec { b } } \right) ^{ 2 } }-\left| { \vec { a } \times \vec { b } } \right| \\ ={ \cos ^{ 2 } }\theta -\sin \theta \\ =1-{ \sin ^{ 2 } }\theta -\sin \theta \\ =1-\left[ { { { \sin }^{ 2 } }\theta +\sin \theta +\dfrac { 1 }{ 4 } -\dfrac { 1 }{ 4 } } \right] \\ =\dfrac { 5 }{ 4 } -{ \left( { \sin \theta +\dfrac { 1 }{ 2 } } \right) ^{ 2 } } \\ Max\, value\, =\dfrac { 5 }{ 4 } -\dfrac { 1 }{ 4 } \, \, at\, \, \sin \theta =0 \\ =1 \\ \left| { \vec { a } .\vec { b } } \right| =\left| { \cos \theta } \right| =1 \\ Hence,\, the\, option\, A\, is\, correct\, answer. \end{array}$$
Question 4
1 / -0

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

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

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

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

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

Solution

$$\begin{array}{l} \left( { \overrightarrow { r } -\overrightarrow { a } } \right) .\overrightarrow { N } =0 \\ \Rightarrow \overrightarrow { a } =2\hat { i } +3\hat { j } -\hat { k } \\ \Rightarrow \overrightarrow { N } =3\hat { i } -2\hat { j } -2\hat { k } \\ \Rightarrow \left( { \overrightarrow { r } -2\hat { i } -3\hat { j } +\hat { k } } \right) .\left( { 3\hat { i } -2\hat { j } -2\hat { k } } \right) =0 \\ \Rightarrow \overrightarrow { r } =x\hat { i } +y\hat { j } +\hat { k } \\ \Rightarrow \left\{ { \left( { x-2 } \right) i+\left( { y-3 } \right) \hat { j } +\left( { z+1 } \right) \hat { k } } \right\} .\left( { 3i-2\hat { j } -2\hat { k } } \right) =0 \\ \Rightarrow 3\left( { x-2 } \right) -2\left( { y-3 } \right) -2\left( { z+1 } \right) =0 \\ \Rightarrow 3x-6-2y+6-2z-2=0 \\ 3x-2y-2z=2\, \, \, \, Ans. \end{array}$$
Question 5
1 / -0

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 :

A
$$\overset { \rightarrow }{ -3i } +\overset { \rightarrow }{ 2k } $$

B
$$\overset { \rightarrow }{ 3i } -\overset { \rightarrow }{ 2k } $$

C
$$\overset { \rightarrow }{ i } +\frac { 2 }{ 3 } \overset { \rightarrow }{ k } $$

D
none of these

Solution

Let the third vertex be $$C=x\widehat{i}+y\widehat{j}+z\widehat{k}$$
and $$A=(1,1,0), B=(2,-1,1), G=(0,0,1)$$
Now,
$$(0,0,1)=(\dfrac{1+2+x}{3}, \dfrac{1-1+y}{3}, \dfrac{0+1+z}{3})$$

Equating this, we get,
$$x=-3,y=0,z=2$$
Hence, $$C=-3\widehat{i}+2\widehat{k}$$
Question 6
1 / -0

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

A
Exactly two values of $$(p,\ q)$$

B
More than but not all values of $$(p,\ q)$$

C
All values of $$(p,\ q)$$

D
Exactly one values of $$(p,\ q)$$

Solution

$$\begin{array}{l} Since, \\ \left[ { 3u\, \, pv\, \, pw } \right] -\left[ { pv\, \, wq\, \, qv } \right] =0 \\ \therefore 3{ p^{ 2 } }\left[ { u.\left( { v\times w } \right) } \right] -pq\, \left[ { v.\left( { w\times u } \right) } \right] \\ -2{ q^{ 2 } }\left[ { w.\left( { v\times u } \right) } \right] =0 \\ \Rightarrow \left( { 3{ p^{ 2 } }-pq+2{ q^{ 2 } } } \right) \left[ { u.\left( { v\times w } \right) } \right] =0 \\ But\, \, \left[ { u\, \, \, v\, \, \, w } \right] \ne 0 \\ \Rightarrow 3{ p^{ 2 } }-pq+2{ q^{ 2 } }=0 \\ \therefore p=q=0 \end{array}$$
Question 7
1 / -0

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 $$A B C$$ ) .

A
$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { \Delta }$$

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

C
$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 3\Delta }$$

D
$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 4\Delta }$$

Solution

$$\begin{array}{l} Vector\, perpendicular\, to\, the\, plane\, of\, \Delta ABC \\ is\, \, \overrightarrow { AB } \times \, \overrightarrow { AC } \\ =\left( { \, \overrightarrow { b } -\, \overrightarrow { a } } \right) \times \left( { \, \overrightarrow { c } -\, \overrightarrow { a } } \right) \\ =\, \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } \\ Area\, of\, triangle\, =\frac { 1 }{ 2 } \left| { \, \overrightarrow { AB } \times \, \overrightarrow { BC } } \right| \\ 2\Delta =\left| { \, \overrightarrow { AB } \times \, \overrightarrow { AC } } \right| \\ Unit\, vector=\frac { { \left( { \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } } \right) } }{ { \left| { \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } } \right| } } \\ =\frac { { \left( { \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } } \right) } }{ { 2\Delta } } \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$$
Question 8
1 / -0

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

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

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

C
$$\dfrac { \pi }{ 2 } $$

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

Solution

$$ \bar{a} +\bar{b}-\bar{c} = 0$$
$$ \bar{a}+\bar{b} = \bar{c}$$
$$ |\bar{a}+\bar{b}| = |\bar{c}| = 1$$
$$ a^{2} +b^{2}+2ab cos\theta =1$$
$$ 1+1+2cos\theta = 1$$
$$ 2cos\theta = -1$$
$$ cos\theta = -1/2$$
So $$ \theta = 2\pi /3$$
Question 9
1 / -0

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

A
$$10$$

B
$$20$$

C
$$18$$

D
$$23$$
Question 10
1 / -0

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:

A
$$\sqrt { 18 } $$

B
$$\sqrt { 72 } $$

C
$$\sqrt { 33 } $$

D
$$\sqrt { 45 } $$

Solution

We have,

$$ AB=3\overrightarrow{i}+4\overrightarrow{k} $$

$$ AC=5\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k} $$

So, the coordinate of

$$ B\,is\,\left( 3,0,4 \right) $$

$$ C\,is\,\left( 5,-2,4 \right) $$

The mid point of $$BC$$ can be easily found by midpoint formula of two points as

$$\begin{align}

$$ D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( \dfrac{3+5}{2},\dfrac{0-2}{2},\dfrac{4+4}{2} \right) $$

$$ D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 5,-2,4 \right) $$

So, the length of median through A is

$$ AD=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( -1-0 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}} $$

$$ AD=\sqrt{33} $$
Hence, this is the answer

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 61

Result

Vector Algebra Test - 61

Score

-

Rank

-

SHARING IS CARING

Question 1

$${ 30 }^{ \circ }$$

$${ 60 }^{ \circ }$$

$${ 90 }^{ \circ }$$

None of these

Solution

Question 2

$$a.b=0$$

$$a.b=-9$$

$$a.b=4$$

$$a.b=-4$$

Solution

Question 3

$$1$$

$$\dfrac{1}{3}$$

$$0$$

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

Solution

Question 4

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

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

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

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

Solution

Question 5

$$\overset { \rightarrow }{ -3i } +\overset { \rightarrow }{ 2k } $$

$$\overset { \rightarrow }{ 3i } -\overset { \rightarrow }{ 2k } $$

$$\overset { \rightarrow }{ i } +\frac { 2 }{ 3 } \overset { \rightarrow }{ k } $$

none of these

Solution

Question 6

Exactly two values of $$(p,\ q)$$

More than but not all values of $$(p,\ q)$$

All values of $$(p,\ q)$$

Exactly one values of $$(p,\ q)$$

Solution

Question 7

$$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { \Delta }$$

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

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

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$10$$

$$20$$

$$18$$

$$23$$

Question 10

$$\sqrt { 18 } $$

$$\sqrt { 72 } $$

$$\sqrt { 33 } $$

$$\sqrt { 45 } $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.