Three Dimensional Geometry Question 18
Question 18 - 2024 (30 Jan Shift 2)
Let $L_{1}: \overrightarrow{\mathbf{r}}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R L_{2}: \overrightarrow{\mathbf{r}}=(\hat{\mathbf{j}}-\hat{\mathbf{k}})+\mu(3 \hat{\mathrm{i}}+\hat{\mathbf{j}}+\mathrm{p} \hat{\mathbf{k}}), \mu \in R$ and $\mathrm{L}_{3}: \overrightarrow{\mathbf{r}}=\delta(\ell \hat{\mathbf{i}}+\mathrm{m} \hat{\mathbf{j}}+\mathrm{n} \hat{\mathbf{k}}) \delta \in \mathrm{R}$
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
(1) $(-1,7,4)$
(2) $(-1,-7,4)$
(3) $(1,7,-4)$
(4) $(1,-7,4)$
Show Answer
Answer (1)
Solution
$\mathrm{L}{1} \perp \mathrm{L}{2}$
$3-1+2 \mathrm{P}=0$
$\mathrm{P}=-1$
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & -1\end{array}\right|=-\hat{i}+7 \hat{j}+4 \hat{k}$
$\therefore(-\delta, 7 \delta, 4 \delta)$ will lie on $\mathrm{L}_{3}$
For $\delta=1$ the point will be $(-1,7,4)$
