Question

If vector $\overset{\rightarrow}{P},\overset{\rightarrow}{Q}$ and $\overset{\rightarrow}{R}$ have magnitudes 5, 12 and 13 units

and $\overset{\rightarrow}{P} + \overset{\rightarrow}{Q} = \overset{\rightarrow}{R}$, the angle between $\overset{\rightarrow}{Q}$ and $\overset{\rightarrow}{R}$ is-

• A. cos^–1 (5/12)
• B. cos^–1 (5/13)
• C. cos^–1 (12/13)
• D. cos^–1 (2/13)

Answer 1

Answer: (C)

cos^–1 (12/13)

Answer 2

Here R² = P² + Q² + 2PQ cos q

q = angle between $\overset{\rightarrow}{P}$and $\overset{\rightarrow}{Q}$ \ q = 90º

From tan a = $\frac{Q\sin\theta}{P + Q\cos\theta}$

(a = angle between $\overset{\rightarrow}{P}$and$\overset{\rightarrow}{R}$ )

= $\frac{12.\sin 90^{0}}{5 + 12\cos 90^{0}}$ = $\frac{12}{5}$

cos b = $\frac{12}{13}$ \ b = cos^–1 $\left( \frac{12}{13} \right)$