Three Dimensional Geometry Question 20
Question 20 - 2024 (31 Jan Shift 1)
The distance of the point $Q(0,2,-2)$ form the line passing through the point $\mathrm{P}(5,-4,3)$ and perpendicular to the lines $\overrightarrow{\mathrm{r}}=(-3 \hat{\mathrm{i}}+2 \hat{\mathrm{k}})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k}), \quad \lambda \in \mathbb{R} \quad$ and $\quad \vec{r}=(\hat{i}-2 \hat{j}+\hat{k})+$ $\mu(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}), \mu \in \mathbb{R}$ is
(1) $\sqrt{86}$
(2) $\sqrt{20}$
(3) $\sqrt{54}$
(4) $\sqrt{74}$
Show Answer
Answer (4)
Solution
A vector in the direction of the required line can be obtained by cross product of
$$ \begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ -1 & 3 & 2 \end{array}\right| \ & =-9 \hat{i}-9 \hat{j}+9 \hat{k} \end{aligned} $$
Required line,
$\overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})$
$\overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
Now distance of $(0,2,-2)$
P.V. of $\mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{j}+(3-\lambda) \hat{k}$
$\overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{j}+(5-\lambda) \hat{k}$
$\overrightarrow{\mathrm{AP}} \cdot(\hat{i}+\hat{j}-\hat{k})=0$
$5+\lambda+\lambda-6-5+\lambda=0$
$\lambda=2$
$|\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9}$
$\overrightarrow{\mathrm{AP}}=\sqrt{74}$
