Vector Algebra Test

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Vector Algebra Test - 44

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Question 1
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If $$\vec A=\vec B + \vec C$$ and the magnitude of $$A,B$$ and $$C$$ are $$5,4$$ and $$3 $$ minutes respectively,then the angle between $$A$$ and $$C$$ is:

A
$$cos^{-1} (4/5)$$

B
$$cos^{-1} (3/5)$$

C
$$tan^{-1} (3/4)$$

D
$$sin^{-1} (3/5)$$

Solution

$$\bar A= \bar B+ \bar C$$ forms a right angle triangle. Where $$A= 5$$, $$B=4$$, $$C=3$$
From the figure, Let Angle between $$\bar A$$ and $$\bar C$$ is $$\theta$$.
$$cos(\theta)=\frac{3}{5}$$ $$\Rightarrow \theta = cos^{-1}(\frac{3}{5})$$
Question 2
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The resultant of two forces acting at an angle of $$150^\circ$$ is $$10$$ N and its perpendicular to one of the forces.The two other force is:

A
$$\dfrac{20}{\sqrt 3}\ N$$

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

C
$$20\ N$$

D
$$\dfrac{20}{ 3}\ N$$

Solution

Let forces are $$ A$$ and $$B$$.
$$R$$ is the resultant force of $$A$$ and $$B$$. As shown in figure, $$ R$$ is perpendicular to the $$B$$.
From the figure, $$Acos(60^o)=R=10$$ $$\Rightarrow A=20 N$$,
$$\Rightarrow $$ $$Asin(60^o)=B$$ $$\Rightarrow B=10\sqrt{3} N$$,
$$\therefore$$ Forces are $$20 N$$ and $$10\sqrt{3}$$.
Question 3
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Three forces of magnitude $$6\ N$$,$$6\ N$$ and $$\sqrt {72}\ N$$ act at a corner of a cube along three sides as shown in the figures.The resultant of these forces is:

A
$$12\ N$$ along OB

B
$$18\ N$$ along OA

C
$$18\ N$$ along OC

D
$$12\ N$$ along OE

Solution

Given : $$\vec{F_1} = \sqrt{72} N$$ along OG, $$\vec{F_2} = 6 N$$ along OA and $$\vec{F_3} = 6 N$$ along OB
Since the three force are mutually perpendicular to each other, so magnitude resultant force is given by
$$|F_R| = \sqrt{F_1^2+F_2^2 + F_3^2}$$
$$\therefore$$ $$|F_R| = \sqrt{(\sqrt{72})^2+6^2 + 6^2} = 12 N$$ along OE
Question 4
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If $$\vec { a }$$ and $$\vec { b }$$ are two unit vector and $$\theta$$ is the angle between them, then $$\left( \vec { a } +\vec { b } \right)$$ is a unit vector if $$\theta =$$

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

B
$$\dfrac { \pi }{ 4 } $$

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

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

Solution

Given $$|\vec a+\vec b|=1$$
$$\Rightarrow |\vec a|^2+2\vec a\cdot \vec b+|\vec b|^2=1$$, square both sides
$$\Rightarrow 1+2|\vec a||\vec {b}|\cos \theta+1=1$$
$$\Rightarrow \cos\theta=-\dfrac{1}{2}$$
$$\therefore \theta=\dfrac{2\pi}{3}$$
Question 5
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The projection of $$\vec {i} - \vec {j}$$ on $$z$$-axis is

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

The projection of $$\vec{a}=i-j$$ on $$\vec{b}=k$$ is given by
$$a_{1}=\dfrac{\vec{a}.\vec{b}}{|\vec{b}|}$$
$$=\dfrac{(i-j).(k)}{1}$$
$$=0$$
Question 6
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In the given polygon the $$\vec {AD} $$ is equal to :

A
$$\vec P +\vec Q + \vec S$$

B
$$\vec {AC} +\vec {CD}$$

C
Both A and B

D
$$\vec P +\vec Q + \vec S+\vec R$$

Solution

Polygon law of vector addition states that if a number of vectors can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.
Using Polygon law of vector addition,

$$\bar {AD}= \bar P +\bar Q+ \bar S$$,
$$\bar {AD}= \bar {AC} + \bar {CD}$$
Question 7
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If $$\overrightarrow { a } $$ is vector of magnitude $$x$$ , $$m$$ is non-zero scalar and $$m\overrightarrow { a } $$ is a unit vector then x in terms of m is:

A
$$m=x$$

B
$$x=\left| m \right| $$

C
$$x=\cfrac { 1 }{ \left| m \right| } $$

D
$$x=m/2$$

Solution

Given $$|m\vec{a}|=1$$
$$\Rightarrow |m||\vec{a}|=1$$
$$\Rightarrow |m|x=1$$
$$\Rightarrow x=\dfrac{1}{|m|}$$

Remember: modulus can never be negative
Question 8
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Which of the following diagram is correct?

A
Only figure (i) and (ii)

B
Only figure (i) ,(ii) and (iii)

C
Only figure (iv)

D
All the given diagrams are correct

Solution

In the given figures we have to see the direction of vectors.
In first figure, figure is closed with any one adjacent vectors in same direction. Hence all figures are correct having same rule.
Hence option $$D$$ is correct.
Question 9
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If $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$ are the position vectors of the points B and C respectively, then the position vector of the point D such that $$\overrightarrow{BD}=4\overrightarrow{BC}$$ is

A
$$4(\overrightarrow{c}-\overrightarrow{b})$$

B
$$-4(\overrightarrow{c}-\overrightarrow{b})$$

C
$$4\overrightarrow{c}-3\overrightarrow{b}$$

D
$$4\overrightarrow{c}+3\overrightarrow{b}$$

Solution

Solution:
Given that: $$\overrightarrow{BD}=4\overrightarrow{BC}$$
It means $$D$$ divides the join of $$BC$$ externally in the ratio $$4:3.$$
$$\therefore$$ Position vector of $$D=\cfrac{4\overrightarrow c-3\overrightarrow b}{4-3}=4\overrightarrow c-3\overrightarrow b$$
Hence, C is the correct option.
Question 10
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Consider the vectors $$\bar{a}=\hat{i}-2\hat{j}+\hat{k}$$ and $$ \bar{b}=4\hat{i}-4\hat{j}+7\hat{k}$$
Find the scalar projection of $$\bar{a}$$ on $$\bar{b}.$$

A
1

B
$$\dfrac{19}{9}$$

C
$$\dfrac{17}{9}$$

D
$$\dfrac{23}{9}$$

Solution

$$\bar{a}=\hat{i}-2\hat{j}+\hat{k}$$ and $$ \bar{b}=4\hat{i}-4\hat{j}+7\hat{k}$$

Scalar projection of $$ a$$ on $$b$$
$$ P=\cfrac { \vec { a } .\vec { b } }{ \left| \vec { b } \right| } \\ P=\cfrac { (\hat { i } -2\hat { j } +\hat { k } ).(4\hat { i } -4\hat { j } +7\hat { k } ) }{ \sqrt { { 4 }^{ 2 }+{ (-4) }^{ 2 }+{ 7 }^{ 2 } } } \\ \hat { i } .\hat { i } =\hat { j } .\hat { j } =\hat { k } .\hat { k } =1\\ P=\cfrac { 4+8+7 }{ 9 } \\ P=\cfrac { 19 }{ 9 } $$