JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 11)
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
Explanation
$$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$$
$$\begin{aligned} & 3-1+2 \mathrm{P}=0 \\ & \mathrm{P}=-1 \\ & \left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \therefore(-\delta, 7 \delta, 4 \delta) \text { will lie on } \mathrm{L}_3 \end{aligned}$$
For $$\delta=1$$ the point will be $$(-1,7,4)$$
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