Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,
$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$
be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then, the point which lies on $L_3$ is
Solution
<p>$$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& 3-1+2 \mathrm{P}=0 \\
& \mathrm{P}=-1 \\
& \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -1 & 2 \\
3 & 1 & -1
\end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\
& \therefore(-\delta, 7 \delta, 4 \delta) \text { will lie on } \mathrm{L}_3
\end{aligned}$$</p>
<p>For $\delta=1$ the point will be $(-1,7,4)$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
This question is part of PrepWiser's free JEE Main question bank. 182 more solved questions on Three Dimensional Geometry are available — start with the harder ones if your accuracy is >70%.