Three Dimensional Geometry Question 18
Question 18 - 2024 (30 Jan Shift 2)
Let $L _1: \overrightarrow{\mathbf{r}}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R L _2: \overrightarrow{\mathbf{r}}=(\hat{\mathbf{j}}-\hat{\mathbf{k}})+\mu(3 \hat{i}+\hat{\mathbf{j}}+p \hat{\mathbf{k}}), \mu \in R$ and $L _3: \overrightarrow{\mathbf{r}}=\delta(\ell \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}}) \delta \in R$
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
$L _1 \perp L _2$
$3-1+2 P=0$
$P=-1$
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1\end{array}\right|=-\hat{i}+7 \hat{j}+4 \hat{k}$
$\therefore(-\delta, 7 \delta, 4 \delta)$ will lie on $L _3$
For $\delta=1$ the point will be $(-1,7,4)$
