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Notes, Exercises, Videos, Tests and Things to Remember on Vector

Please scroll down to get to the study materials. ## Note on Vector

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#### Scalar and Vector

The Physical quantity which has the only magnitude but no direction is called a scalar quantity. Examples of scalars are time, length, area, temperature, mass, speed etc.

A physical quantity which has both magnitude and direction is called a vector quantity or simply vector. Examples of vectors are velocity, displacement, force, acceleration, weight etc.

#### Directed line Segment

If A and B are two different points, A as initial point and B as the terminal point. Then the line segment from A to B is called a directed line segment.It is denoted by $$\overrightarrow {AB}$$ (from initial point A to terminal point B) or $$\overrightarrow {a}$$ and the length of line segment is denoted by$$|\overrightarrow {AB}|$$ or$$|\overrightarrow {a}|$$ which is known as magnitude of$$|\overrightarrow {AB}|$$.

A directed line segment has both direction and magnitude. So it represents a vector. While denying a vector, we write first the initial point, then the terminal point.

### Representing a Vector in terms of co-ordinates

#### Position Vector

Let P(x,y) be a point in a plane. Then, OP is a directed line segment representing a vector with origin as initial point and P as a terminal point. Hence $$\overrightarrow {OP}$$ is a vector. Draw PA perpendicular to OX. Then OA = x and AP = y. OA =x - component and AP = y is called y-component of the vector $$\overrightarrow {OP}$$. Here, $$\overrightarrow {OP}$$ is called the position vector of P. Then $$\overrightarrow {OP}$$ = $$\begin{pmatrix}x\\y\\\end{pmatrix}$$

A vector $$\overrightarrow {OP}$$ whose initial point is origin is called the position vector of the given point P.

#### Triangle law of vectors

If two sides of a triangle represent two vectors as shown in the figure, then their sum is represented by the third side in magnitude and direction whose initial point is the initial point of 1st vector and terminal point is terminal point of 2nd vector.This is called the triangle law of vector.

• If  $$\overrightarrow {OP}$$ =  $$\begin{pmatrix}x\\y\\\end{pmatrix}$$ then the magnitude of $$\overrightarrow {OP}$$ i.e |$$\overrightarrow {OP}$$| = $$\sqrt{x^2+y^2}$$ and Direction : tan$$\theta$$ = $$\frac{y}{x}$$
• If A(x1,y1) and B (x2,y2) then column vector ( $$\overrightarrow {AB}$$) =  $$\begin{pmatrix}x_2-x_1\\y_2-y_1\\\end{pmatrix}$$
.

### Very Short Questions

Given points A (1, -1) , B(2 , 4)

$$\therefore$$  $$\overrightarrow{AB}$$ =$$\begin{pmatrix} 2-1 \\ 4+1 \\ \end{pmatrix}$$

= $$\begin{pmatrix} 1 \\ 5 \\ \end{pmatrix}$$

Magnitude of vector | $$\overrightarrow{AB}$$ | = $$\sqrt{1^{2} + 5^{2}}$$
= $$\sqrt{1 + 25}$$
= $$\sqrt{26}$$ Ans.

Direction of $$\overrightarrow{AB}$$ is tan $$\theta$$ = $$\frac{5}{1}$$= tan(tan-15)
$$\therefore$$ $$\theta$$ = tan-15

Given points A(3 , -4) , B(-2 , 1)

$$\therefore$$ $$\overrightarrow {AB}$$ = $$\begin {pmatrix} -5 \\ 5 \end{pmatrix}$$ Ans.

Magnitude of $$\overrightarrow {AB}$$ = $$\sqrt { (-5) ^{2} + 5^{2}}$$
= $$\sqrt{25 + 25}$$
= 5$$\sqrt{2}$$ units.
Direction of $$\overrightarrow {AB}$$ ,
tan $$\theta$$ = $$\frac{5}{-5}$$ = -1 = tan135o. Ans.

Given points A(1 , 1) , B(2 , 4)

$$\therefore$$ Vector $$\overrightarrow {AB}$$ = $$\begin{pmatrix}1 \\ 3 \end{pmatrix}$$
Magnitude of $$\overrightarrow {AB}$$ = $$\sqrt{1 + 3^{2}}$$
= $$\sqrt{ 1+9}$$
= $$\sqrt{10}$$ units. Ans.

Direction of $$\overrightarrow {AB}$$ is tan $$\theta$$ = $$\frac{3}{1}$$ = tan(tan-13)
$$\theta$$ = tan-13 Ans.

Given points A( -1 , 3) , B(1 , -3)

$$\therefore$$ Vector $$\overrightarrow {AB}$$ = $$\begin{pmatrix} 2 \\ -6 \end {pmatrix}$$ Ans.

$$\therefore$$ Magnitude of $$\overrightarrow {AB}$$ = $$\sqrt{2 ^{2} + 6 ^{2}}$$
= $$\sqrt{4 + 36}$$
= $$\sqrt{4 + 36}$$
= $$\sqrt{40}$$
= 2$$\sqrt{10}$$units. Ans.

Direction of $$\overrightarrow {AB}$$ , tan$$\theta$$ = $$\frac{-6}{2}$$
= -3
= tan(tan-1(-3)

$$\therefore$$ $$\theta$$ =tan-1(-3) .

Given points A(-2 , 0) , B(-3 , -2)

$$\therefore$$ Vector $$\overrightarrow {AB}$$ = $$\begin {pmatrix} - 1 \\ -2 \end{pmatrix}$$ Ans.

Magnitude of $$\overrightarrow {AB}$$ = $$\sqrt { -1^{2} + -2 ^{2}}$$
= $$\sqrt{1 +4}$$
= $$\sqrt{5}$$ Ans.

Direction of $$\overrightarrow {AB}$$ , tan$$\theta$$ = $$\frac{-2}{-1}$$
= tan(tan-12)
$$\theta$$ = (tan-12)

Here , A(x1 y1) = (0 , 0) , B = (1 , 4)
Now , x -component of $$\overrightarrow {AB}$$ = x2 - x1 = 1 - 0 = 1
y- component of $$\overrightarrow {AB}$$ = y2 - y1 = 4 - 0 = 4

$$\therefore$$ $$\overrightarrow {AB}$$ = $$\begin {pmatrix} 1 \\ 4 \end {pmatrix}$$
$$\therefore$$ - $$\overrightarrow {AB}$$ = $$\overrightarrow {BA}$$ = -1 $$\begin {pmatrix} 1 \\ 4 \end {pmatrix}$$ = $$\begin {pmatrix} -1 \\ -4 \end {pmatrix}$$ Ans.

Here , A(x1 , y1) = (1 , -1) , B(x2 , y2) = (2 , 4)
Now , x - component of $$\overrightarrow {AB}$$ = x2 - x1 = 2 - 1 = 1
y - component of $$\overrightarrow {AB}$$ = y2 - y1 = 4 - (-1) = 4 + 1 = 5.
$$\therefore$$ $$\overrightarrow {AB}$$ = $$\begin {pmatrix} 1 \\ 5 \end{pmatrix}$$
$$\therefore$$ - $$\overrightarrow {AB}$$ = $$\overrightarrow {BA}$$ = -$$\begin {pmatrix} 1 \\ 5 \end{pmatrix}$$ = $$\begin {pmatrix} -1 \\ -5 \end{pmatrix}$$ Ans.

Here , A(x1 , y1) = (-1 , 3) , B(x2 , y2) = (1 , -3)
Now , x - component of $$\overrightarrow {AB}$$ = x2 - x1 = 1 -(- 1) = 2
$$\therefore$$ y - component of $$\overrightarrow {AB}$$ = y2 - y1 = -3 - 3 = -6
$$\therefore$$ $$\overrightarrow {AB}$$ = $$\begin {pmatrix} 2 \\ -6 \end{pmatrix}$$
$$\therefore$$ - $$\overrightarrow {AB}$$ =$$\overrightarrow {BA}$$ = - $$\begin {pmatrix} 2 \\ -6 \end{pmatrix}$$ = $$\begin {pmatrix} -2 \\6 \end{pmatrix}$$ Ans.

Here , A(x1 , y1) = (2 , 0) , B(x2 - x1) = (-3 , -2)

Now , x - component of $$\overrightarrow {AB}$$ = x2 - x1 = -3 + 2 = -1
y - component of $$\overrightarrow {AB}$$ =y2- y1 = -2 ,-0 -2
$$\therefore$$ $$\overrightarrow {AB}$$ = $$\begin {pmatrix} -1 \\ -2 \end{pmatrix}$$
$$\therefore$$ -$$\overrightarrow {AB}$$ = $$\overrightarrow {BA}$$ = $$\begin{pmatrix} -1 \\ -2 \end{pmatrix}$$ = $$\begin {pmatrix} 1 \\ 2 \end {pmatrix}$$ Ans.

Here given ,
Position vector of A is $$\overrightarrow {OA}$$ = $$\begin {pmatrix} 3 \\ 4 \end{pmatrix}$$
$$\therefore$$ A(x1 , y1) = (3 , 4)
Position vector of B is $$\overrightarrow {OA}$$ = $$\begin {pmatrix} 1 \\ 5 \end{pmatrix}$$
$$\therefore$$ B (x2 , y2) = (1 , 5)
Now , x - component of $$\overrightarrow {AB}$$ = x2 - x1 = (1 -3) = -2
y - component of $$\overrightarrow {AB}$$ = y2 - y1 = 5 - 4 = 1
$$\therefore$$ $$\overrightarrow {AB}$$ = $$\begin {pmatrix} -2 \\ 1 \end{pmatrix}$$
$$\therefore$$ Magnitude of $$\overrightarrow {AB}$$ = $$\sqrt {x -component ^{2} + y - component ^{2}}$$
= $$\sqrt{(-2) ^{2} + 1^{2}}$$ = $$\sqrt{(4 + 1)}$$ = $$\sqrt{5}$$ unit Ans.
and Direction of $$\overrightarrow {AB}$$ is tan$$\theta$$ = $$\frac{y - component}{x - component}$$ = $$\frac{1}{-2}$$ = $$tan(tan^{-1})\frac{-1}{2}$$.
$$\therefore$$ $$\theta$$ = $$tan^{-1}\frac{-1}{2}$$ Ans.

Here , given point ,
P9x1 , y1) = (1 , 1) and Q(x2 , y2) = (2 , 5)
Now , x - component of $$\overrightarrow {PQ}$$ = x2 - x1 = 2 - 1 = 1
y - component of $$\overrightarrow {PQ}$$ = y2 - y1 = 5 - 1 = 4

$$\therefore$$ = $$\begin{pmatrix} 1 \\ 4 \end{pmatrix}$$ Ans.
Direction of $$\overrightarrow {PQ}$$ is tan $$\theta$$ = $$\frac{4}{1}$$ = tan(tan-14)
$$\therefore$$ $$\theta$$ = tan-14

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• ### Find the column vector (overrightarrow {AB}) in the following conditions:A(4,6),B(12,16)

(egin{pmatrix} 8\ 10\ end{pmatrix})

(egin{pmatrix} 8\ 8\ end{pmatrix})

(egin{pmatrix} 9\ 9\ end{pmatrix})

(egin{pmatrix} 7\ 11\ end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(8,10),B(14,18)

(egin{pmatrix}7\
7\
end{pmatrix})

(egin{pmatrix}6\
1\
end{pmatrix})

(egin{pmatrix}6\
8\
end{pmatrix})

(egin{pmatrix}4\
5\
end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(-6,4),B(12,14)

(egin{pmatrix}12\ 1\end{pmatrix})
(egin{pmatrix}18\ 10\end{pmatrix})
(egin{pmatrix}10\ 5\end{pmatrix})
(egin{pmatrix}20\ 6\end{pmatrix})
• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(-8,10),B(10,12)

(egin{pmatrix}18\

2\ end{pmatrix})

(egin{pmatrix}1\

2\ end{pmatrix})

(egin{pmatrix}14\

7\ end{pmatrix})

(egin{pmatrix}20\

9\ end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(-2,-4),B(12,8)

(egin{pmatrix}16\ 18\
end{pmatrix})

(egin{pmatrix}13\ 11\
end{pmatrix})

(egin{pmatrix}14\ 12\
end{pmatrix})

(egin{pmatrix}15\ 10\
end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(12,6), B(8,-10)

(egin{pmatrix}-4\

-16end{pmatrix})

(egin{pmatrix}12\

9end{pmatrix})

(egin{pmatrix}4\

16end{pmatrix})

(egin{pmatrix}2\

11end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(8,6), B(-16,-10)

(egin{pmatrix}42\
-16
end{pmatrix})

(egin{pmatrix}-24\
-16
end{pmatrix})

(egin{pmatrix}2\
-11
end{pmatrix})

(egin{pmatrix}-14\

end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(16,18), B(-20,-24)

(egin{pmatrix}-36\
-42
end{pmatrix})

(egin{pmatrix}-36\
-2
end{pmatrix})

(egin{pmatrix}16\
2
end{pmatrix})

(egin{pmatrix}-46\
-32
end{pmatrix})

• ### Find the column vector (overrightarrow {AB}) in the following conditions: A(24,-30), B(-36,26)

(egin{pmatrix}-66\  54
end{pmatrix})

(egin{pmatrix}-30\  26
end{pmatrix})

(egin{pmatrix}-60\  56
end{pmatrix})

(egin{pmatrix}-20\  16
end{pmatrix})