Vectors 1 Question 12
13. If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}} _1$ are two unit vectors such that $\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}$ and $5 \overrightarrow{\mathbf{a}}-4 \overrightarrow{\mathbf{b}}$, are perpendicular to each other, then the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is
$(2002,1 M)$
(a) $45^{\circ}$
(b) $60^{\circ}$
(c) $\cos ^{-1}\left(\frac{1}{3}\right)$
(d) $\cos ^{-1}\left(\frac{2}{7}\right)$
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Answer:
Correct Answer: 13. $I=\frac{\alpha \overrightarrow{\mathbf{a}}+\beta \overrightarrow{\mathbf{b}}+\gamma \overrightarrow{\mathbf{c}}}{\alpha+\beta+\gamma}$
Solution:
- Since,
$$ (\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}) \cdot(5 \overrightarrow{\mathbf{a}}-4 \overrightarrow{\mathbf{a}})=0 $$
$$ \begin{aligned} & \Rightarrow \quad 5|\overrightarrow{\mathbf{a}}|^{2}+6 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}-8|\overrightarrow{\mathbf{b}}|^{2}=0 \\ & \Rightarrow \quad 6 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=3 \quad[\because|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|=1] \\ & \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \end{aligned} $$
