Question

Find the distance of the point $(-1,-5,-10)$ from the point of the intersection of the line $\vec{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda \left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right)$ and the plane $\overrightarrow{r}.\left( \widehat{i}-\widehat{j}+\widehat{k} \right)=5$ .

Answer 1

Hint:In the question we have given the equation of line and the plane. So, in order to find the intersection of the line and plane substitute $\vec{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda \left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right)$ in place of $\overrightarrow{r}$ in the equation of plane. After solving the equation value of $\lambda $ can be found, and with this $\lambda $ we can get the point of intersection. After finding the intersection point, using distance formula of two points which states that if $A({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and $B({{x}_{2}},{{y}_{2}},{{z}_{2}})$ then the distance $AB$ will be equal to $\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}$, we can calculate the distance between intersection point and $(-1,-5,-10)$ and that will be our answer.

Complete step by step answer:
It is given that the equation of line and plane is $\vec{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda \left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right)$ and $\overrightarrow{r}.\left( \widehat{i}-\widehat{j}+\widehat{k} \right)=5$ respectively.
Now, at the point of intersection a common $\overrightarrow{r}$ will exist, so substituting $\vec{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda \left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right)$ from line in place of $\overrightarrow{r}$ in the equation of plane, we get
\[\therefore \left( 2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda \left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right) \right).\left( \widehat{i}-\widehat{j}+\widehat{k} \right)=5\]
\[\Rightarrow \left( 2\widehat{i}-\widehat{j}+2\widehat{k} \right).\left( \widehat{i}-\widehat{j}+\widehat{k} \right)+\lambda \left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right).\left( \widehat{i}-\widehat{j}+\widehat{k} \right)=5\]
Now, simplifying further, we get
$\Rightarrow 2+1+2+\lambda \left( 3-4+2 \right)=5$
$\Rightarrow 5+\lambda =5$
Subtracting 5 from both sides, we get
$\Rightarrow \lambda =0$
Now, substituting the value of $\lambda =0$ in the equation of line to get the point of intersection, so
$\Rightarrow \vec{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+0\left( 3\widehat{i}+\widehat{4j}+2\widehat{k} \right)=2\widehat{i}-\widehat{j}+2\widehat{k}$
As $\overrightarrow{r}$ can be represented as \[x\widehat{i}+\widehat{yj}+z\widehat{k}\], hence
$\Rightarrow x\widehat{i}+\widehat{yj}+z\widehat{k}=2\widehat{i}-\widehat{j}+2\widehat{k}$
Comparing LHS with RHS, we get
$x=2,y=-1$ and $z=2$
Hence, the point of intersection of the line and plane is $\left( 2,-1,2 \right)$.
According to distance formula if $A({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and $B({{x}_{2}},{{y}_{2}},{{z}_{2}})$ are two points then the distance between them i.e. $AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}$.
Now, using distance formula distance between $\left( 2,-1,2 \right)$ and $(-1,-5,-10)$ is,
$=\sqrt{{{(-1-2)}^{2}}+{{(-5+1)}^{2}}+{{(-10-2)}^{2}}}$
$=\sqrt{9+16+144}$
$=\sqrt{169}=13$
Hence, the distance between the intersection point and $(-1,-5,-10)$ is 13.

Note:
In this question the one noticeable thing was we used the fact that at point of intersection a common $\overrightarrow{r}$ will exist, and so we substituted the $\overrightarrow{r}$ of line into the plane and the find the unknown
$\lambda $ and hence the intersection point. After finding the intersection point it was very easy to find the distance between them using the distance formula. The distance formula also is easy to recollect, we just have to extend the 2D formula and add z variables inside the root.