Vector Algebra Test

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Vector Algebra Test - 15

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Question 1
1 / -0

Position vectors of mid point of the vector joining the points $$P(2, 3, 4)$$ and $$Q (4, 1, -2)$$ is

A
$$2\widehat{i}+\widehat{j}+\widehat{k}$$

B
$$3\widehat{i} + 2\widehat{j} + \widehat{k}$$

C
$$-3\widehat{i} - \widehat{j} + \widehat{k}$$

D
None

Solution

Let $$\overrightarrow{OP} = 2 \widehat{i}+3\widehat{j}+4\widehat{k}$$ and $$\overrightarrow{OQ} = 4 \widehat{i}+\widehat{j}-2\widehat{k}$$
Let mid-point of $$PQ$$ be $$R$$
$$\therefore \overrightarrow {OR} = \displaystyle \frac{\overrightarrow {OP} + \overrightarrow {OQ}}{2}$$
$$\overrightarrow{OR} = \displaystyle \frac{6 \widehat{i} + 4\widehat{j}+2\widehat{k}}{2}$$
Hence, $$\overrightarrow{OR} = 3\widehat{i} + 2 \widehat{j} + \widehat{k}$$
Question 2
1 / -0

The projection $$\displaystyle 2\hat{i}+3\hat{j}-2\hat{k}$$ on the vector $$\displaystyle \hat{i}+2\hat{j}-3\hat{k}$$ is

A
$$\displaystyle \frac{1}{\sqrt{14}}$$

B
$$\displaystyle \sqrt{14}$$

C
$$\displaystyle \frac{2}{\sqrt{14}}$$

D
none of these

Solution

Projection of $$\bar{a}$$ on $$\bar{b}$$ is a cos$${\theta}$$ where $$\theta$$ is the angle between the two.
Let $$\bar{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$$ and $$\bar{b} = \hat{i} + 2\hat{j} - 3\hat{k}$$
$$cos{\theta} = \dfrac{2 + 6 + 6}{\sqrt{17} \times \sqrt{14}}$$
Thus projection = $$(2 + 6 + 6)/\sqrt{14} = \sqrt{14}$$
Question 3
1 / -0

$$\vec{A} \ and \ \vec{B}$$ are two vectors, find the angle between them, if
$$\left | \vec{A}\times \vec{B} \right |=\sqrt{3}(\vec{A.}\vec{B})$$ the value of is :-

A
$$90^{0}$$

B
$$60^{0}$$

C
$$45^{0}$$

D
$$30^{0}$$

Solution

$$\because \left | \vec{A\times }\vec{B} \right |=\sqrt{3}(\vec{A.}\vec{B})$$

$$|A||B| \sin\theta= \sqrt{3}|A||B|\cos\theta$$

$$\Rightarrow \tan\theta =\sqrt{3}\Rightarrow\theta =60^{0}$$
Question 4
1 / -0

Let $$\vec{AB}= 3\widehat {i} + \widehat{j}- \widehat{k}$$ and $$\vec{AC}=\widehat{i} -\widehat{j} +3\widehat{k}$$ and a point P on the line segment BC is equidistant from $$AB$$ and $$AC$$, then $$\vec{AP}$$ is

A
$$2\widehat{i}- \widehat{k}$$

B
$$\widehat{i}- 2\widehat{k}$$

C
$$2\widehat{i} + \widehat{k}$$

D
None of these

Solution

Given : $$\vec{AB}=3\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{AC}=\hat{i}-\hat{j}+3\hat{k}$$
$$P$$ is a point on $$BC$$ such that it is equidistant from $$AB$$ and $$AC$$
$$\implies P$$ is a midpoint of $$BC$$
$$\implies \vec{AP} = \cfrac{\vec{AB}+\vec{AC}}{2}$$
$$=\dfrac{4\hat{i}+2\hat{k}}{2}$$
$$ = 2\hat{i}+\hat{k}$$
Hence, option C is correct.
Question 5
1 / -0

The vector sum of (N) coplanar forces, each of magnitude F, when each force is making an angle of
$$\frac{2\pi }{N}$$ with that preceding one, is :

A
F

B
$$\frac{NF}{2}$$

C
NF

D
Zero

Solution

Given that:
Total forces$$ = N$$
An angle between forces$$=\dfrac{2\pi}{N}$$

Therefore,
Total angle between $$N$$ co-planer force $$=\dfrac{2\pi}{N}\times N = 2\pi$$

Thus, the forces represent a polygon of $$N$$ sides.

As, by the polygon method of vector addition, the vectors can be added by the placing the vectors in order as the sides of the polygon and the resultant of these vectors will join them to the tail of the first vector to head of the last vector. This will close the polygon and will give the magnitude and the direction of the resultant vector.

As the $$N$$ number forces mentioned above, represents a closed polygon, the resultant will be zero.
Question 6
1 / -0

Four point charges $${q}_A=2\mu C$$, $${q}_B=5\mu C$$, $${q}_C=2\mu C$$, $${q}_D=5\mu C$$ are located at the four corners of a square ABCD of side 10 cm. The force on a charge of $$1\mu C$$ placed at the centre of the square is

A
Zero

B
Towards AB

C
Towards BC

D
Towards AD

Solution

$$q_B$$ will attract q along OB while $$q_D$$ towards OD having equal magnitude both will cancel each other's effect.
Similarly $$q_A$$ and $$q_C$$ will cancel each other's effect.
Thus net force can q will be zero
Question 7
1 / -0

If $$\vec{a}=3\hat{i}-2\hat{j}+\hat{k},\vec{b}=2\hat{i}-4\hat{j}-3\hat{k}$$, $$\vec{c}=-1\hat{i}+2\hat{j}+2\hat{k}$$ then $$\vec{a}+\vec{b}+\vec{c}=$$

A
$$3\hat{i}-4\hat{j}$$

B
$$3\hat{i}+4\hat{j}$$

C
$$4\hat{i}-4\hat{j}$$

D
$$4\hat{i}+4\hat{j}$$

Solution

$$\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$$
$$\vec{b}=2\hat{i}-4\hat{j}-3\hat{k}$$
$$\vec{c}=-1\hat{i}+2\hat{j}+2\hat{k}$$

$$\vec{a}+\vec{b}+\vec{c} = (3+2-1)\hat{i} + (-2-4+2)\hat{j} + (1-3+2)\hat{k} = 4\hat{i} - 4\hat{j}$$
Question 8
1 / -0

Namita walks 14 metres towards west, then turns to her right and walks 14 metres and then turns to her left and walks 10 metres. Again turning to her left she walks 14 metres. What is the shortest distance (in metres) between her starting point and the present position ?

A
10

B
24

C
28

D
38

Solution

so, shortest distance $$=24$$
Hence,
option $$B$$ is correct answer.
Question 9
1 / -0

The projection of $$\displaystyle a=3i-j+5k$$ on $$\displaystyle b=2i+3j+k$$ is

A
$$\displaystyle 8/\sqrt{\left ( 35 \right )}$$

B
$$\displaystyle 8/\sqrt{\left ( 39 \right )}$$

C
$$\displaystyle 8/\sqrt{\left ( 14 \right )}$$

D
$$\displaystyle \sqrt{\left ( 14 \right )}$$

Solution

Projection of $$a$$ on $$b$$ is $$=\cfrac{a\cdot b}{|b|} =\cfrac{3.2-1.3+5.1}{\sqrt{4+9+1}}=\cfrac{8}{ \sqrt{14}}$$
Question 10
1 / -0

For $$O$$ being the origin and $$3$$ points $$P,Q$$ and $$R$$ lie on a plane. If $$\displaystyle \vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}$$, then $$P, Q, R$$ are

A
the vertices of an equilateral triangle

B
the vertices of an isoceles triangle

C
collinear

D
none of these

Solution

$$\displaystyle \vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}$$
$$\Rightarrow \displaystyle -\vec{OP}+\vec{OQ}=-\vec{OQ}+\vec{OR}$$
$$\Rightarrow \displaystyle \vec{OQ}=\dfrac{\vec{OP}+\vec{OR}}{2}$$
Clearly $$Q$$ is the mid point of $$P$$ and $$R$$
Hence $$P,Q,R$$ are collinear.

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Vector Algebra Test - 15

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Vector Algebra Test - 15

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Question 1

$$2\widehat{i}+\widehat{j}+\widehat{k}$$

$$3\widehat{i} + 2\widehat{j} + \widehat{k}$$

$$-3\widehat{i} - \widehat{j} + \widehat{k}$$

None

Solution

Question 2

$$\displaystyle \frac{1}{\sqrt{14}}$$

$$\displaystyle \sqrt{14}$$

$$\displaystyle \frac{2}{\sqrt{14}}$$

none of these

Solution

Question 3

$$90^{0}$$

$$60^{0}$$

$$45^{0}$$

$$30^{0}$$

Solution

Question 4

$$2\widehat{i}- \widehat{k}$$

$$\widehat{i}- 2\widehat{k}$$

$$2\widehat{i} + \widehat{k}$$

None of these

Solution

Question 5

F

$$\frac{NF}{2}$$

NF

Zero

Solution

Question 6

Zero

Towards AB

Towards BC

Towards AD

Solution

Question 7

$$3\hat{i}-4\hat{j}$$

$$3\hat{i}+4\hat{j}$$

$$4\hat{i}-4\hat{j}$$

$$4\hat{i}+4\hat{j}$$

Solution

Question 8

10

24

28

38

Solution

Question 9

$$\displaystyle 8/\sqrt{\left ( 35 \right )}$$

$$\displaystyle 8/\sqrt{\left ( 39 \right )}$$

$$\displaystyle 8/\sqrt{\left ( 14 \right )}$$

$$\displaystyle \sqrt{\left ( 14 \right )}$$

Solution

Question 10

the vertices of an equilateral triangle

the vertices of an isoceles triangle

collinear

none of these

Solution

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