Question

Find the distance of the point (2, 12, 5) from the point of intersection of the line $\overrightarrow{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda \left(3\hat{i}+4\hat{j}+2\hat{k}\right)$ and $\overrightarrow{r}.\left(\hat{i}-2\hat{j}+\hat{k}\right)=0$. [CBSE 2014]

Answer 1

The equation of the given line is $\overrightarrow{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda \left(3\hat{i}+4\hat{j}+2\hat{k}\right)$.

The position vector of any point on the given line is

$\overrightarrow{r}=\left(2+3\lambda \right)\hat{i}+\left(-4+4\lambda \right)\hat{j}+\left(2+2\lambda \right)\hat{k}$ .....(1)

If this lies on the plane $\overrightarrow{r}.\left(\hat{i}-2\hat{j}+\hat{k}\right)=0$, then

$$\begin{gathered} \left[\left(2+3\lambda \right)\hat{i}+\left(-4+4\lambda \right)\hat{j}+\left(2+2\lambda \right)\hat{k}\right].\left(\hat{i}-2\hat{j}+\hat{k}\right)=0 \\ \Rightarrow \left(2+3\lambda \right)-2\left(-4+4\lambda \right)+\left(2+2\lambda \right)=0 \\ \Rightarrow 2+3\lambda +8-8\lambda +2+2\lambda =0 \\ \Rightarrow 3\lambda =12 \\ \Rightarrow \lambda =4 \end{gathered}$$

Putting $\lambda =4$ in (1), we get $\left(2+3\times 4\right)\hat{i}+\left(-4+4\times 4\right)\hat{j}+\left(2+2\times 4\right)\hat{k}$ or $14\hat{i}+12\hat{j}+10\hat{k}$ as the coordinate of the point of intersection of the given line and the plane.

The position vector of the given point is $2\hat{i}+12\hat{j}+5\hat{k}$.

∴ Required distance = Distance between $14\hat{i}+12\hat{j}+10\hat{k}$ and $2\hat{i}+12\hat{j}+5\hat{k}$

$$\begin{gathered} =\left|\left(14\hat{i}+12\hat{j}+10\hat{k}\right)-\left(2\hat{i}+12\hat{j}+5\hat{k}\right)\right| \\ =\left|12\hat{i}+5\hat{k}\right| \\ =\sqrt{{12}^2+0^2+5^2} \\ =\sqrt{169} \\ =13\mathrm{units} \end{gathered}$$