Question
Let a line L be perpendicular to both the lines
$L_{1}:\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and
$L_{2}:\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}$.
If $\theta$ is the acute angle between the lines L and $L_{3}:\frac{x - \frac{8}{7}}{2} = \frac{y - \frac{4}{7}}{1} = \frac{z}{2}$, then $tan\theta$ is equal to:
Options
Solution
Sol. [Key Concept: The direction of $L$ is the cross product of the direction vectors of $L_{1}$ and $L_{2}$; then $cos\theta = \frac{{|d}_{L} \cdot \ d_{3}|}{{|d}_{L}{||d}_{3}|}$ between $L$ and $L_{3}$ gives $tan\theta$.]
$${\overrightarrow{r} = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| = 7\widehat{i} - 14\widehat{j} + 7\widehat{k} }{\equiv 7(\widehat{i} - 2\widehat{j} + \widehat{k}) }$$Now,
$${\because cos\theta = \frac{2 - 2 + 2}{\sqrt{1 + 4 + 1}\sqrt{4 + 1 + 4}} }{\Rightarrow cos\theta = \frac{2}{3\sqrt{6}} \Rightarrow tan\theta = \frac{5\sqrt{2}}{2} }$$
Create a free account to view solution
View Solution Free