Suppose $\overrightarrow{\mathbf{a}}=\lambda \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},…

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
  1. $-7 < \lambda < 1$
  2. $\lambda>1$
  3. $1 < \lambda < 7$
  4. $-5 < \lambda < 1$

Solution

Given, $\overrightarrow{\mathbf{a}}=\lambda \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}$ $\begin{array}{rlrl} \therefore & \cos \theta= \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|} \\ = \frac{\lambda^2-7+6 \lambda}{\sqrt{\lambda^2+49+9} \sqrt{\lambda^2+1+4 \lambda^2}} < 0 \\ \Rightarrow (\lambda+7)(\lambda-1) < 0 \\ \Rightarrow \quad-7 < \lambda < 1 \end{array}$

Asked in: AP EAMCET 2009

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