Question

If a=4i-3j then find the magnitude and direction of the vector is

Answer 1

Answer:

a = 4i - 3j

Magnitude of a = |a| = √[(4^2) + {(-3)^2}] = 5 units

Direction of a = arc tan {(-3) / 4} = (360° - 36.87°) = 323.13° with positive X-axis

b = 6i + 8j

Magnitude of b = |b| = √{(6^2) + (8^2)} = 10 units

Direction of b = arc tan (8 / 6) = 53.13° with positive X-axis

(a - b) = (4 - 6)i + (-3 - 8)j = -2i - 11j

Magnitude of (a - b) = |a - b| = √[{(-2)^2} + {(-11)^2}] = 11.18 units

Direction of (a - b) = arc tan {(-11) / (-2)} = 180° + 79.695° = 259.695° with positive X-axis

(b - a) = - (a - b) = 2i + 11j

Magnitude of (b - a) = |b - a| = 11.18 units

Direction of (b - a) = arc tan (11 / 2) = 79.695° with positive X-axis

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