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

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Last updated date: 17th Jul 2024
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Answer
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Hint: We can find a general coordinate of a point on the line and we can substitute these coordinates in an algebraic form of plane to get the point of intersection.Now use distance between two points formula.

To easily solve such types of questions, it is a good practice to convert the line and the plane from vector form to algebraic form.
Let us consider a line having its equation in vector form,
\[\overrightarrow{r}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)+\lambda \left(
d\widehat{i}+e\widehat{j}+f\widehat{k} \right)\]
The algebraic coordinate of a general point lying on this line is given by,
$\left( a+\lambda d,b+\lambda e,c+\lambda f \right)............\left( 1 \right)$
Let us consider a plane having its equation in vector form,
\[\overrightarrow{r}.\left(a'\widehat{i}+b'\widehat{j}+c'\widehat{k} \right)=d'\]
The equation of this plane in algebraic form is given by,
$a'x+b'y+c'z=d'............\left( 2 \right)$
In the question, the equation of the line is \[\overrightarrow{r}=\left( 2\widehat{i}-
\widehat{j}+2\widehat{k} \right)+\lambda \left( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right)\].
Then substituting $a=2,b=-1,c=2,d=3,e=4,f=2$ in equation $\left( 1 \right)$, the general coordinate of a point lying on the line is given by,
$\left( 2+3\lambda ,-1+4\lambda ,2+2\lambda \right)...........\left( 3 \right)$
Also, in the question, the equation of plane is \[\overrightarrow{r}.\left( \widehat{i}-
\widehat{j}+\widehat{k} \right)=5\].
 Substituting $a'=1,b'=-1,c'=1,d'=5$ in equation $\left( 2 \right)$,
we get the algebraic equation of the plane,
$x-y+z=5.........\left( 4 \right)$
Since we have to find the intersection point of the line and the plane, we will substitute the general coordinates of line from equation $\left( 3 \right)$ in the algebraic equation of the plane in equation
$\left( 4 \right)$.
 Substituting $x=2+3\lambda ,y=-1+4\lambda ,z=2+2\lambda $ in the plane $x-y+z=5$,
we get,
$\begin{align}
& \left( 2+3\lambda \right)-\left( -1+4\lambda \right)+\left( 2+2\lambda \right)=5 \\
& \Rightarrow 5+\lambda =5 \\
& \Rightarrow \lambda =0........\left( 5 \right) \\
\end{align}$
Substituting $\lambda =0$ from equation $\left( 5 \right)$ in equation $\left( 3 \right)$, the
coordinate of the intersection point of the line and the plane is,
$\begin{align}
& \left( 2+3\left( 0 \right),-1+4\left( 0 \right),2+2\left( 0 \right) \right) \\
& \Rightarrow \left( 2,-1,2 \right) \\
\end{align}$
In the question, we have to find the distance between the intersection point of line and the plane
i.e. $\left( 2,-1,2 \right)$ and $\left( -1,-5,-10 \right)$ .
From distance formula, the distance between any two coordinates $\left( x,y,z \right)$ and $\left(x',y',z' \right)$ is given by,
$d=\sqrt{{{\left( x-{x}' \right)}^{2}}+{{\left( y-{y}' \right)}^{2}}+{{\left( z-z' \right)}^{2}}}$
Substituting \[~x=2,y=-1,z=2,x'=-1,y'=-5,z'=-10\] in the distance formula, we get,
$\begin{align}
& d=\sqrt{{{\left( 2-\left( -1 \right) \right)}^{2}}+{{\left( -1-\left( -5 \right) \right)}^{2}}+{{\left( 2-\left( -10 \right) \right)}^{2}}} \\
& \Rightarrow d=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( -1+5 \right)}^{2}}+{{\left( 2+10 \right)}^{2}}}\\
& \Rightarrow d=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}+{{\left( 12 \right)}^{2}}} \\
& \Rightarrow d=\sqrt{9+16+144} \\
& \Rightarrow d=\sqrt{169} \\
& \Rightarrow d=13 \\
\end{align}$

Note: To remember how to find the general coordinate of line in algebraic form from vector form of line, just find the coefficients of $\widehat{i},\widehat{j},\widehat{k}$. The coefficients of $\widehat{i},\widehat{j},\widehat{k}$ are the coordinate of the general point which is lying on the line. To remember how to convert the plane from vector to algebraic form, substitute $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ and then apply dot product to get the algebraic form of the plane.