The distance of the point (-1,2,3) from the plane r (i-2j+3k)=10 parallel to the line of t…

Question

The distance of the point $(-1,2,3)$ from the plane $\vec{r}\cdot (\hat{i}-2\hat{j}+3\hat{k})=10$ parallel to the line of the shortest distance between the lines $\vec{r}=(\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j})+\mu (\hat{i}-\hat{j}+\hat{k})$ is :

Accepted Answer

1. Determine the line of shortest distance between the given two lines:

Direction vector of line 1: $\vec{{d}_{1}}=2\hat{i}+\hat{k}$

Direction vector of line 2: $\vec{{d}_{2}}=\hat{i}-\hat{j}+\hat{k}$

Now, let's find the cross product $\vec{N}=\vec{{d}_{1}}	imes \vec{{d}_{2}}$

$\vec{N}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 2 \quad  0 \quad  1 \ 1 \quad  -1 \quad  1 \end{vmatrix}=(\hat{i}(0+1)-\hat{j}(2-1)+\hat{k}(-2-0))=\hat{i}-\hat{j}-2\hat{k}$

So, the direction vector of the line of shortest distance is $\vec{N}=\hat{i}-\hat{j}-2\hat{k}$.

2. Find the equation of a line passing through point $A(-1,2,3)$ and having the direction vector $\vec{N}$:

$\frac{x+1}{1}=\frac{y-2}{-1}=\frac{z-3}{-2}=\lambda $

3. Find the point $P$ where the line intersects the given plane:

Let the coordinates of point $P$ be in terms of $\lambda $:

$P(\lambda -1,-\lambda +2,-2\lambda +3)$

Since $P$ lies on the plane $x-2y+3z=10$, we can substitute the coordinates of $P$ in terms of $\lambda $ into the equation of the plane:

$(\lambda -1)-2(-\lambda +2)+3(-2\lambda +3)=10$

$\lambda =-2$

$P(-3,4,7)$

4. Calculate the distance between points $A$ and $P$:

$AP=\sqrt{(-3-(-1){)}^{2}+(4-2{)}^{2}+(7-3{)}^{2}}=2\sqrt{6}$

Solutions · 1