Question

Q26) Vector A is equal to 8 unit and vector B is equal to 6 unit, angle between them is 30° then find angle between (A+B) and (A-B). ​

Answer 1

Given:

Vector A = 8 unit

Vector B = 6 unit

Angle = 30°

To find:

Angle between (A+B) and (A-B)

Solution:

(A-B) = $\sqrt{A^{2} +B^2-2ABcos$∅

(A-B) = $\sqrt{6^{2} + 8^{2} -2(6)(8)cos30}$°

(A-B) = $\sqrt{136-96(0.86)}$

(A-B) =$\sqrt{53.44}$

(A-B) = 7.31

Therefore, the angle between (A-B) is 7.31 unit.