Magnitude and Direction of a Vector

Magnitude and Direction of a Vector

Magnitude and Direction of a Vector

Magnitude of a Vector

Magnitude of a vector is the length of its directed line segment from the initial point to the terminal point. So, the magnitude of a vector is a positive number which is the measure of the line segment representing that vector. Magnitude of a vector is also known as the modulus or the absolute value of the vector.

The magnitude of a vector (OP) ⃗ is the length of OP and it is written as OP or |(OP) ⃗|. Similarly, the magnitude of vector a ⃗ is denoted by |a ⃗| or a. Let (OP) ⃗ = position vector of P = (■(x_1@y_1 )). Draw PM⊥OX, then x-component of (OP) ⃗ = OM = x1 y-component of (OP) ⃗ = MP = y1 Then, OP2 = OM2 + MP2 or, OP2 = x12 + y12 or, OP = √(x_1^2+y_1^2 ) Hence, OP = |(OP) ⃗|= √(x_1^2+y_1^2 ) In general, if a vector (AB) ⃗ displaces a point A(x1, y1) to the point B(x2, y2) then vector (AB) ⃗ is represented by (AB) ⃗ = (■(〖x_2-x〗_1@y_2-y_1 )). ∴ x-component (x) = 〖x_2-x〗_1 y-component (y) = y_2-y_1 ∴ Magnitude of (AB) ⃗ is |(AB) ⃗ |= √((〖x_2-x〗_1 )^2-(y_2-y_1 )^2 ) Hence the magnitude of a vector is given by, √((x-component)^2-(y-component)^2 )

Direction of a Vector

Direction of a vector is the direction of the directed line segment representing that vector. Direction of a vector is measured as the angle made by the vector with positive (i.e anticlockwise) direction of X-axis.

Let (OP) ⃗ = position vector of P = (■(x@y)) and (OP) ⃗ makes angle θ with positive direction of X-axis as shown in the figure. Draw PM⊥OX. Then OM = x, PM = y. Now, from ΔOPM, tanθ = y/x or, θ = tan-1(y/x)

Therefore, for the direction of a vector, we find the tangent of the angle (θ) made by the given vector with the X-axis in positive (anticlockwise) direction.

For example: Let vector (OA) ⃗ = (■(3@2)) and θ be the angle made by the vector (OA) ⃗ with X-axis in positive (anticlockwise) direction. Then, tanθ = AN/ON = 2/3 = 0.666 or, tanθ = tan34° (approximately) ∴ θ = 34° ∴ direction of (OA) ⃗ (θ) = 34°.

If A(x1, y1) and B(x2, y2) are the initial and terminal points of a vector (AB) ⃗ then, (AB) ⃗ = (■(x_2-x_1@y_2-y_1 )). Suppose (AB) ⃗ makes an angle θ with positive (anticlockwise) direction of X-axis as shown in the figure. Draw AM⊥OX, BN⊥OX and AR⊥BN. Then, BR = y2 – y1 and AR = x2 – x1 Now, from ΔABR, tanθ = BR/AR or, tanθ = (y_2-y_1)/(x_2-x_1 ) ∴ θ = tan-1((y_2-y_1)/(x_2-x_1 )) Therefore, if A(x1, y1) and B(x2, y2) are the initial and terminal points of a vector (AB) ⃗ and makes angle θ with positive (anticlockwise) direction of X-axis, then the direction θ of vector (AB) ⃗ is given by tanθ = (y_2-y_1)/(x_2-x_1 ) = (y-component)/(x-component)

Note: a. If x-component = +ve and y-component = +ve then the directin of vector is θ. Figure b. If x-component = -ve and y-component = +ve then the direction of vector is (180° - θ). Figure c. If x-component = -ve and y-component = -ve then the direction of vector is (180° + θ). Figure d. If x-component = +ve and y-component = -ve then the direction of vector is (360° - θ). Figure e. If x-component = +ve and y-component = 0 then the direction of vector is 0°. f. If x-component = 0 and y-component = +ve then the direction of vector is 90°. g. If x-component = -ve and y-component = 0 then the direction of vector is 180°. h. If x-component = 0 and y-component = -ve then the direction of vector is 270°.

Worked Out Examples

Example 1: If PQ displaces from point P(10, 13) into Q(7, 8), (a) Find the components of (PQ) ⃗ (b) Find the magnitude of (PQ) ⃗ and (c) Find the direction of (PQ) ⃗. Solution: (a) x-component of (PQ) ⃗ = x2 – x1 = 7 – 10 = -3 y-component of (PQ) ⃗ = y2 – y1 = 8 – 13 = -5 Hence, (PQ) ⃗ = (■(-3@-5)) (b) Magnitude (PQ) ⃗ = |(PQ) ⃗| = √(〖(-3)〗^2+〖(-5)〗^2 ) = √(9+25) = √34 units (c) If θ be the angle made by (PQ) ⃗ with positive X-axis, then tanθ = (y-component)/(x-component) = (-5)/(-3) or, tanθ = 1.666 or, tanθ = tan59° ∴ θ = 59° Since, both the components of (PQ) ⃗ are negetive, the direction of (PQ) ⃗ is 180° + 59° = 239°.

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