The angle between the lines $\hat{\mathbf{r}}=(2 \hat{\mathbf{i}}-3…

The angle between the lines $\hat{\mathbf{r}}=(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \quad$ and $\hat{\mathbf{r}}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ is
  1. $\frac{\pi}{2}$
  2. $\cos ^{-1}\left(\frac{9}{\sqrt{91}}\right)$
  3. $\cos ^{-1}\left(\frac{7}{\sqrt{84}}\right)$
  4. $\frac{\pi}{2}$

Solution

Given lines are and $ \begin{aligned} & \mathrm{r}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & \mathrm{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \end{aligned} $ Here DR's of given lines are $(1,4,3)$ and $(1,2,-3)$. $\therefore$ Angle between these lines is $ \begin{aligned} \cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\ & =\frac{1 \times 1+4 \times 2+3 \times(-3)}{\sqrt{1^2+4^2+3^2} \sqrt{1^2+2^2+(-3)^2}} \\ & =\frac{1+8-9}{\sqrt{1+16+9} \sqrt{1+4+9}}=0 \\ \Rightarrow \quad \theta & =\frac{\pi}{2} \end{aligned} $

Asked in: AP EAMCET 2014

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