Q75 / 160 · TS EAMCET 2009 · Mathematics
Suppose $\overrightarrow{\mathbf{a}}=\lambda \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}$. If the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality
- Correct answerA$-7 < \lambda < 1$
- B$\lambda>1$
- C$1 < \lambda < 7$
- D$-5 < \lambda < 1$
Solution
Given, $\overrightarrow{\mathbf{a}}=\lambda \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}$
$\begin{array}{rlrl}
\therefore & \cos \theta= \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|} \\
= \frac{\lambda^2-7+6 \lambda}{\sqrt{\lambda^2+49+9} \sqrt{\lambda^2+1+4 \lambda^2}} < 0 \\
\Rightarrow (\lambda+7)(\lambda-1) < 0 \\
\Rightarrow \quad-7 < \lambda < 1
\end{array}$
Hence, the correct answer is option: 'A'.