Notes, Exercises, Videos, Tests and Things to Remember on Magnitude and Direction of Vector

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## Note on Magnitude and Direction of Vector

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#### Magnitude of a vector

The magnitude of the vector is a positive number which represents the length of the vector or directed line segment. It is also known as modulus of the vector. If $$\overrightarrow {OP}$$ is the position vector of P, then its magnitude is denoted by $$|\overrightarrow {OP}|$$.If $$|\overrightarrow {AB}|$$is a vector, then magnitude is denoted by $$|\overrightarrow {AB}|$$. If the vector is written in the single letter from $$\overrightarrow {a}$$, then its magnitude is denoted by $$\overrightarrow {a}$$.

If the initial point is at the origin and the terminal point at P (x,y).

$$\overrightarrow {OP}$$ = $$\frac{x}{y}$$ is given by $$|\overrightarrow {OP}|$$ = $$\sqrt{(x)^2+(y)^2}$$

If A (x1,y1) and B(x2,y2) are two points,

We know, $$\overrightarrow {AB}$$ =$$\begin{pmatrix} x_2-x_1\\y_2-y_1\end{pmatrix}$$. Then $$|\overrightarrow {AB}|$$ = $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Thus$$|\overrightarrow {AB}|$$ = $$\sqrt{(x-component)^2+ (y-component)^2}$$

If the initial point is at A(x1,y1) and terminal point at B(x2,y2).

Here, For $$\overrightarrow{AB}$$

x component of $$\overrightarrow{AB}$$= AE = x2 - x1

y component of $$\overrightarrow{AB}$$ = BE = y2 - y1

Using pythagoras theorem,

(AB)2 = (AE)2 + (BE)2

or, AB2 = (x2 - x1)2 + (y2 - y1)2

or, AB = $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

or, AB = $$\sqrt{(x\;component)^2 + (y\;component)^2}$$

$$\therefore$$ $$|\overrightarrow {AB}|$$ = AB =$$\sqrt{(x\; comp)^2 + (y\; comp)^2}$$

#### Direction of Vector

Direction of a vector is the value of angle made by the vector with x-axis in positive direction. It is generally measured in degree which ranges from 0º to 360º. Direction of a vector is denoted by θ. In the figure, $$\overrightarrow {OP}$$ makes an angle θwith x-axis in anticlockwise direction. Draw PM perpendicular to x-axis.

When the initial point is at the origin and terminal point at P(x,y).

Then, OM = x and MP =y.

In ΔPOM, tan θ = $$\frac{Perpendicular}{base}$$

or, tan θ =$$\frac{MP}{OM}$$

or, tan θ = $$\frac{y}{x}$$

∴ θ = tan-1 $$(\frac{y}{x}$$)

Which gives the direction of the vector.

Hence, direction of vector (θ) =tan-1 $$(\frac{y-component}{x-component}$$)

When the initial point is at A(x1,y1) and the terminal point is at B(x2,y2)

If A(x1,y1) and B(x2,y2) are two points, then direction of $$\overrightarrow {AB}$$ is given by θ = tan-1 $$\begin{pmatrix}y_2-y_1\\x_2-x_1\\\end{pmatrix}$$.

• If a vector is parallel to x-axis from left to right, $$\theta$$ = 0o
• If a vector is parallel to x-axis from right to left, $$\theta$$ = 180o
• If a vector is parallel to y-axis down to up, then $$\theta$$ = 90o
• If a vector is parallel to y-axis up to down, then $$\theta$$ = 270o
.

### Very Short Questions

Here, $$\overrightarrow{OA}$$ = {3,4}

$$\therefore$$ x component = 3 and y component = 4

$$\therefore$$ |$$\overrightarrow{OA}$$| = OA = $$\sqrt{(x comp)^2+(y comp)^2}$$

= $$\sqrt{(3)^2 + (4)^2}$$

= $$\sqrt{9+16}$$ = $$\sqrt(25)$$ = 5

$$\therefore$$ $$\overrightarrow{OA}$$ = 5 units

Here, P = ($$\sqrt{3}$$,$$\sqrt{3}$$)

$$\therefore$$ x-component = $$\sqrt{3}$$

y - component = $$\sqrt{3}$$

Let $$\theta$$ be the angle made by $$\overrightarrow{OP}$$ with the positive direction of x-axis.

tan$$\theta$$ = $$\frac{y - component}{x - component}$$ = $$\frac{\sqrt{3}}{\sqrt{3}}$$ = 1

$$\therefore$$ tan$$\theta$$ = tan45

$$\theta$$ = 45 [$$\therefore$$ x-component and y-component are +ve the value must lie in 1st quadrant.]

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• ### What is the value of column vector if (overrightarrow {AB}) = (egin {pmatrix} 3 \ 4 end {pmatrix}) , (overrightarrow {CD}) = (egin {pmatrix} -4 \ -3 end {pmatrix})  then express (overrightarrow {AB}) + (overrightarrow {CD})?

(egin {pmatrix} -1 \ 1 end {pmatrix})
(egin {pmatrix} 2 \ 2 end {pmatrix})
(egin {pmatrix} -2 \ 2 end {pmatrix})
(egin {pmatrix} 1 \ 1 end {pmatrix})
• ### What is the magnitude of direction if (overrightarrow {AB}) = (egin {pmatrix} 3 \ 4 end {pmatrix}) , (overrightarrow {CD}) = (egin {pmatrix} -4 \ -3 end {pmatrix})  then express (overrightarrow {AB}) + (overrightarrow {CD}) ?

( heta) = 100o
( heta) = 120o
( heta) = 110o
( heta) = 135o
• ### What is the value of  column vector if (overrightarrow {PQ}) = (egin {pmatrix} -2 \ 7 end {pmatrix}) and (overrightarrow {RS}) = (egin {pmatrix} 3 \ -2 end {pmatrix})  , then express (overrightarrow {PQ}) + (overrightarrow {RS}) ?

(egin {pmatrix} 2 \ 3 end {pmatrix})
(egin {pmatrix} 2 \ 5 end {pmatrix})
(egin {pmatrix} 1 \ 3 end {pmatrix})
(egin {pmatrix} 1 \ 5 end {pmatrix})
• ### What is the magnitude if (overrightarrow{OA}) = {3,4} ?

(overrightarrow{OA}) = 5 units
(overrightarrow{OA}) = 8 units
(overrightarrow{OA}) =2 units
(overrightarrow{OA}) = 4 units
• ### What is the direction of (overrightarrow{OP}) where P = ((sqrt{3}),(sqrt{3}))?

( heta) = 60°
( heta) = 90°
( heta) = 30°
( heta) = 45°
• ### What is the column vector of (overrightarrow{AB}) if the vector (overrightarrow{AB}) displaces a point A(2,4) to B(5,7)?

(egin {pmatrix} 2 \ 2 end {pmatrix})
(egin {pmatrix} -3 \ 3 end {pmatrix})
(egin {pmatrix} 3 \ 3 end {pmatrix})
(egin {pmatrix} -2\ 2 end {pmatrix})
• ### What is the direction of (overrightarrow{AB}) if the vector (overrightarrow{AB}) displaces a point A(2,4) to B(5,7)?

( heta) = 90°
( heta) = 45°
( heta) = 30°
( heta) = 60°

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##### parallel vector

what happen when two vector are parallel to each other

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