ISRO 2016- Vectors [Mech]

Question

The sine of the angle between the two vectors a = 3i + j + k and b = 2i -2j + k is
(a) √ (74/99)

(b) √ (25/99)

(c) √ (37/99)

(d) √ (5/99)

Answer 1

$$\begin{align*} &a = 3 {\hat{\textbf{i}}} + {\hat{\textbf{j}}} + {\hat{\textbf{k}}} \\ &b = 2 {\hat{\textbf{i}}} -2 {\hat{\textbf{j}}} + {\hat{\textbf{k}}} \\ &\Rightarrow |a| = \sqrt{3^2+1^2+1^2} = \sqrt{11} \\ &\Rightarrow |b| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{9} = 3 \\ &\Rightarrow a\cdot b = 3*2 + 1*(-2) + 1*1 = 5 \\ &\Rightarrow \cos \theta =\frac{a\cdot b}{|a||b|} =\frac{5}{3*\sqrt{11}} \\ &\Rightarrow \sin ^2 \theta = 1 - \left ( \frac{5}{3*\sqrt{11}} \right )^2 \\ &\Rightarrow \sin \theta = \sqrt{\frac{74}{99}} \\ \end{align*}$$