Vector Algebra Test

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Vector Algebra Test - 20

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Question 1
1 / -0

If $$\bar{a}$$ is unit vector, then $$|\bar{a}\times \hat{i}|^2+|\bar{a}\times \hat{j}|^2+|\bar{a}\times \hat{k}|^2=$$ _____________.

A
$$2$$

B
$$1$$

C
$$0$$

D
$$3$$

Solution

Let $$x, y$$ and $$z$$ be the direction angles of the vector $$\overrightarrow{a}$$
Then,
$$|\overrightarrow{a} \times \widehat{i}| =|\overrightarrow{a}||\widehat{i}|sin(x)=sinx$$
Similarly,
$$|\overrightarrow{a} \times \widehat{j}| =siny$$
$$|\overrightarrow{a} \times \widehat{k}| =sinz$$

Therefore, $$|\overrightarrow{a} \times \widehat{i}|^2+|\overrightarrow{a} \times \widehat{j}|^2+|\overrightarrow{a} \times \widehat{k}|^2=sin^2x+sin^2y+sin^2z$$
$$=(1-cos^2x)+(1-cos^2y)+(1-cos^2z)$$
$$=3-(cos^2x+cos^2y+cos^2z)$$
$$=3-1$$
$$=2$$

Hence, the answer is option (A).
Question 2
1 / -0

If $$\vec { a } $$ and $$\vec { b } $$ are non-zero non-collinear vectors, then $$\left[ \vec { a } \quad \vec { b } \quad \hat { i } \right] \hat { i } +\left[ \vec { a } \quad \vec { b } \quad \hat { j } \right] \hat { j } +\left[ \vec { a } \quad \vec { b } \quad \hat { k } \right] \hat { k } $$ is equal to

A
$$\vec { a } +\vec { b } $$

B
$$\vec { a } \times \vec { b } $$

C
$$\vec { a } -\vec { b } $$

D
$$\vec { b } \times \vec { a } $$

Solution

Let $$ \vec{a}= a_1\hat{i} + a_2 \hat{j}+ a_3 \hat{k}$$
$$ \vec{b} = b_{1} \hat{i}+ b_2 \hat{j} + b_3 \hat{k}$$
$$ [\vec{a}\,\vec{b} \,\vec{c}]. \hat{i}+ [\vec{a}\, \vec{b}] \hat{j} + [\vec{a}\,\vec{b}\, \hat{k}] \hat{k}$$
$$ = \left ( \left ( \vec{a} \times \vec{b} \right ) \hat{i}\right )\hat{i}+ \left ( \left ( \vec{a} \times \vec{b} \right )\hat{j} \right ) \hat{j}+ \left ( \left ( \vec{a}\times \vec{b} \right ).\hat{k} \right )\hat{k}$$
$$ = \vec{a} \times \vec{b}$$
Question 3
1 / -0

The vector $$z = 3 - 4i$$ is turned anticlockwise through an angle of $$180^{\circ}$$ and stretched $$\dfrac{5}{2}$$ times. The complex number corresponding to the newly obtained vector is ....

A
$$-\dfrac{15}{2}-10i$$

B
$$-\dfrac{15}{2}+10i$$

C
$$\dfrac{15}{2}+10i$$

D
$$\dfrac{15}{2}-10i$$

Solution

$$z=3-4i$$ lie in fourth quadrant in complex plane, after turned anticlockwise through $$180^0$$ this will lie in II quadrant, therefore, the number will be $$-3+4i$$ now after stretching it $$2.5$$ times i.e., multiplying by $$2.5$$, the required complex number will be
$$\Rightarrow 2.5\times (-3+4i)$$
$$\Rightarrow -\dfrac{15}{2}+10i$$

Hence, this is the answer.
Question 4
1 / -0

If A and B are the points $$(2,1,-2),(3,-4,5)$$, then the angle that $$OA$$ makes with $$OB$$ is:

A
$${\cos ^{ - 1}}\left( {\dfrac{{4\sqrt 2 }}{{15}}} \right)$$

B
$${\cos ^{ - 1}}\left( {\dfrac{4}{{15\sqrt 2 }}} \right)$$

C
$${\cos ^{ - 1}}\left( {\dfrac{{18\sqrt 2 }}{{15}}} \right)$$

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

Solution

$$0\left( {0,0,0} \right)$$

$$OA = - 2\widehat i - \widehat j + 2\widehat k$$, $$\left( {OA} \right) = \sqrt {4 + 1 + 4} = 3$$

$$OB = - 3\widehat i + 4\widehat j - 5\widehat k$$, $$\left( {OB} \right) = \sqrt { 9 + 16 + 25} = \sqrt 50 $$

Now $$\cos \theta = \left| {{{\mathop {OA}\limits^ \to .\mathop {OB}\limits^ \to } \over {\left( {\mathop {OA}\limits^ \to } \right).\left( {\mathop {OB}\limits^ \to } \right)}}} \right|$$

$$ = \left| {{{6 - 4 - 10} \over {3 \times \sqrt {50} }}} \right|$$

$$ = {8 \over {3 \times 5\sqrt 2 }}$$

$$\cos \theta = {8 \over {15\sqrt 2 }}$$

$$\cos \theta = {{8 \times \sqrt 2 } \over {15 \times 2}}$$

$$\cos \theta = {{4\sqrt 2 } \over {15}}$$

$$\theta = {\cos ^{ - 1}}\left( {\dfrac{{4\sqrt 2 }}{{15}}} \right)$$
Question 5
1 / -0

If $$\vec {a}.\vec {b}.\vec {c}$$ represents the vectors $$\vec {BC}.\vec {CA}.\vec {AB}$$ respectively, then which one is correct

A
$$\vec {a} \times \vec {b}= \vec {b} \times \vec {c}$$

B
$$\vec {b} \times \vec {c}=\vec {c} \times \vec {a}$$

C
$$\vec {c} \times \vec {a}=\vec {a} \times \vec {b}$$

D
$$\vec {a} \times \vec {b}=\vec {b} \times \vec {c}=\vec {c} \times \vec {a}$$

Solution

Given that $$\vec{a}, \vec{b}, \vec{c}$$ represents $$\vec{BC}, \vec{CA}, \vec{AB}$$ respectively.
$$\therefore$$ ABC is a triangle and $$\vec{a}, \vec{b}, \vec{c}$$ are the sides of $$\triangle{ABC}$$, respectively.
$$\therefore \; ar{\left( \triangle{ABC} \right)} = \cfrac{1}{2} \times \vec{a} \times \vec{b} = \cfrac{1}{2} \times \vec{b} \times \vec{c} = \cfrac{1}{2} \vec{c} \times \vec{a}$$
Hence, $$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$$.
Question 6
1 / -0

If $$\left| {\widehat a - \widehat b} \right| = \sqrt 3 $$ , then $$\left| {\widehat a + \widehat b} \right|$$ may be:-

A
$$1$$

B
$${{\sqrt 3 } \over 2}$$

C
Either (1) and (2)

D
None of these

Solution

$$\left| \overset { \wedge }{ a } -\overset { \wedge }{ b } \right| =\sqrt { 3 } $$
$$ Squaring\quad both\quad sides$$
$$\implies\quad { \left| \overset { \wedge }{ a } \right| }^{ 2 }+{ \left| \overset { \wedge }{ b } \right| }^{ 2 }-2(\overset { \wedge }{ a } .\overset { \wedge }{ b } )={ (\sqrt { 3 } ) }^{ 2 }$$
$$ \because \overset { \wedge }{ a } \& \overset { \wedge }{ b } \quad represent\quad unit\quad vectors.$$
$$\implies\quad 1+1-2(\overset { \wedge }{ a } .\overset { \wedge }{ b } )=3$$
$$\implies\quad -2\overset { \wedge }{ a } .\overset { \wedge }{ b } =1$$
$$\implies\quad 2\overset { \wedge }{ a } .\overset { \wedge }{ b } =-1\quad -(i)$$
$$ \left| \overset { \wedge }{ a } +\overset { \wedge }{ b } \right| =\sqrt { { \left| \overset { \wedge }{ a } \right| }^{ 2 }+{ \left| \overset { \wedge }{ b } \right| }^{ 2 }+2\overset { \wedge }{ a } .\overset { \wedge }{ b } } $$
$$ =\sqrt { 1+1-1 } $$
$$ =\sqrt { 2-1 } =1$$
Question 7
1 / -0

A line passes through the points whose position vectors $$ \hat { i } +\hat { j } -2\hat { k }$$ and $$\hat { i } -3\hat { j } +\hat { k }$$. Then the position vector of a point on it at a unit distance from the first point is

A
$$\dfrac { 1 }{ 5 } \left( 5\hat { i } +\hat { j } -7\hat { k } \right)$$

B
$$\dfrac { 1 }{ 5 } \left( 5\hat { i } +9\hat { j } -13\hat { k } \right)$$

C
$$\left( \hat { i } -4\hat { j } +3\hat { k } \right)$$

D
$$\left( \hat { i } +4\hat { j } +3\hat { k } \right)$$

Solution

Let points of the position vector $$\hat i + \hat j - 2\hat k$$ and $$\hat i - 3\hat j + \hat k$$ are $$A(1,1-2)$$ and $$(1,-3,1)$$
$$AB=\sqrt{(1-1)^2+(1+3)^2+(-2-1)^2}=5$$
Required points which divide $$AB$$ in ratio $$1:4$$
The required position vector of the point
$$=\dfrac{1 \times (\vec i -3 \vec j + \vec k)+4(\vec i+\vec j-2 \vec k)}{1+4}=\dfrac{1}{5}(5 \vec i + \vec j -7 \vec k)$$
Question 8
1 / -0

If $$\vec{a}+\vec{b}\perp \vec{a}$$ and $$|\vec{b}|=\sqrt{2}|\vec{a}|$$, then?

A
$$(2\vec{a}+\vec{b})||\vec{b}$$

B
$$(2\vec{a}+\vec{b})\perp \vec{b}$$

C
$$(2\vec{a}-\vec{b})\perp \vec{b}$$

D
$$(2\vec{a}+\vec{b})\perp\vec{a}$$

Solution

Consider the problem ,

$$\vec a + \vec b \bot \vec a$$
therefore,
$$(\vec a+\vec b) \cdot \vec a=0$$

$$\vec a\cdot \vec a+\vec b \cdot \vec a=0$$

$$|\vec a|^2+\vec a \cdot \vec b=0$$

$$\vec a \cdot \vec b=-|\vec a|^2$$ --- $$(1)$$

$$(2\vec a+\vec b).\vec b=2\vec a \cdot \vec b+\vec b\cdot \vec b$$

$$=2\vec a\cdot \vec b+|\vec b|^2$$

$$=2 \times -|\vec a|^2+(\sqrt 2 |\vec a|)^2$$

$$=-2|\vec a|^2+2|\vec a|^2$$

$$=0$$

Therefore,
$$(2\vec a + \vec b) \bot \vec b$$

Hence option $$B$$ is the correct answer.
Question 9
1 / -0

Let $$a=\hat{i}+2\hat j+3\hat k$$ and $$b=3\hat i+\hat j$$. Find the unit vector in the direction of the $$a+b$$.

A
$$\dfrac{4\hat i+3\hat j+3\hat k}{\sqrt{34}}$$

B
$$\dfrac{4\hat i+3\hat j+3\hat k}{\sqrt{44}}$$

C
$$\dfrac{5\hat i+3\hat j+3\hat k}{\sqrt{34}}$$

D
None of these

Solution

Given, $$a=\hat{i}+2\hat j+3\hat k$$ and $$b=3\hat i+\hat j$$

$$a + b = \left( {\hat i + 2\hat j + 3\hat k} \right) + \left( {3\hat i + \hat j} \right)$$
$$ = 4\hat i + 3\hat j + 3\hat k$$

$$\left| {a + b} \right| = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 3 \right)}^2}} $$
$$ = \sqrt {16 + 9 + 9} $$
$$ = \sqrt {34} $$

So, unit vector in the direction of $$a+b $$
$$ = \dfrac{{a + b}}{{\left| {a + b} \right|}}$$

$$ = \dfrac{{4\hat i + 3\hat j + 3\hat k}}{{\sqrt {34} }}$$
Question 10
1 / -0

The set of values of $$c$$ for which the angle between the vectors $$cx\hat{i}-6\hat{j}+3\hat{k}$$ and $$x\hat{i}-2\hat{j}+2cx\hat{k}$$ is acute for every $$x\in R$$ is

A
$$(0,4/3)$$

B
$$[0,2/3]$$

C
$$(11/9,4/3)$$

D
$$[0,1/3)$$

Solution

$$cx^{2}+12+6cx>0\implies c(3c-4)<0\implies c\in(0,\dfrac{4}{3})$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 20

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Vector Algebra Test - 20

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SHARING IS CARING

Question 1

$$2$$

$$1$$

$$0$$

$$3$$

Solution

Question 2

$$\vec { a } +\vec { b } $$

$$\vec { a } \times \vec { b } $$

$$\vec { a } -\vec { b } $$

$$\vec { b } \times \vec { a } $$

Solution

Question 3

$$-\dfrac{15}{2}-10i$$

$$-\dfrac{15}{2}+10i$$

$$\dfrac{15}{2}+10i$$

$$\dfrac{15}{2}-10i$$

Solution

Question 4

$${\cos ^{ - 1}}\left( {\dfrac{{4\sqrt 2 }}{{15}}} \right)$$

$${\cos ^{ - 1}}\left( {\dfrac{4}{{15\sqrt 2 }}} \right)$$

$${\cos ^{ - 1}}\left( {\dfrac{{18\sqrt 2 }}{{15}}} \right)$$

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

Solution

Question 5

$$\vec {a} \times \vec {b}= \vec {b} \times \vec {c}$$

$$\vec {b} \times \vec {c}=\vec {c} \times \vec {a}$$

$$\vec {c} \times \vec {a}=\vec {a} \times \vec {b}$$

$$\vec {a} \times \vec {b}=\vec {b} \times \vec {c}=\vec {c} \times \vec {a}$$

Solution

Question 6

$$1$$

$${{\sqrt 3 } \over 2}$$

Either (1) and (2)

None of these

Solution

Question 7

$$\dfrac { 1 }{ 5 } \left( 5\hat { i } +\hat { j } -7\hat { k } \right)$$

$$\dfrac { 1 }{ 5 } \left( 5\hat { i } +9\hat { j } -13\hat { k } \right)$$

$$\left( \hat { i } -4\hat { j } +3\hat { k } \right)$$

$$\left( \hat { i } +4\hat { j } +3\hat { k } \right)$$

Solution

Question 8

$$(2\vec{a}+\vec{b})||\vec{b}$$

$$(2\vec{a}+\vec{b})\perp \vec{b}$$

$$(2\vec{a}-\vec{b})\perp \vec{b}$$

$$(2\vec{a}+\vec{b})\perp\vec{a}$$

Solution

Question 9

$$\dfrac{4\hat i+3\hat j+3\hat k}{\sqrt{34}}$$

$$\dfrac{4\hat i+3\hat j+3\hat k}{\sqrt{44}}$$

$$\dfrac{5\hat i+3\hat j+3\hat k}{\sqrt{34}}$$

None of these

Solution

Question 10

$$(0,4/3)$$

$$[0,2/3]$$

$$(11/9,4/3)$$

$$[0,1/3)$$

Solution

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