Question

The length of the perpendicular drawn from the point $\left( 2,1,4 \right)$ to the plane containing the lines $\overrightarrow{r}=\left( \widehat{i}+\widehat{j} \right)+\lambda \left( \widehat{i}+2\widehat{j}-\widehat{k} \right)$ and $\overrightarrow{r}=\left( \widehat{i}+\widehat{j} \right)+\mu \left( \widehat{-i}+\widehat{j}-2\widehat{k} \right)$ is

(a)\[\sqrt{3}\]
(b)$\dfrac{1}{\sqrt{3}}$
(c)$\dfrac{1}{3}$
(d)3

Answer 1

Hint: We will first find the equation of the plane that contains two lines, $\overrightarrow{r}=\left( \widehat{i}+\widehat{j} \right)+\lambda \left( \widehat{i}+2\widehat{j}-\widehat{k} \right)$ and $\overrightarrow{r}=\left( \widehat{i}+\widehat{j} \right)+\mu \left( \widehat{-i}+\widehat{j}-2\widehat{k} \right)$. In the final step, we will find the perpendicular distance from the point $\left( 2,1,4 \right)$ and the obtained plane using the formula $\text{distance}=\left| \dfrac{\left( ap+bq+cr+d \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$ here a, b, c, d are the constants of the plane and p, q, r are the coordinates of the point.

Complete step-by-step answer:

We have been given that the plane contains these two lines, $\overrightarrow{r}=\left( \widehat{i}+\widehat{j} \right)+\lambda \left( \widehat{i}+2\widehat{j}-\widehat{k} \right)$ and $\overrightarrow{r}=\left( \widehat{i}+\widehat{j} \right)+\mu \left( \widehat{-i}+\widehat{j}-2\widehat{k} \right)$. We have to find the perpendicular distance between this plane and the point $\left( 2,1,4 \right)$.

Now first, we will find the equation of the plane containing these two lines. Since the plane contains the lines, the normal to plane would be perpendicular to the vectors parallel to the line. So, we can find the normal vector to the plane by taking the cross product of the parallel vectors of the lines. We have the parallel vectors as $\left( \widehat{i}+2\widehat{j}-\widehat{k} \right)$ and $\left( -\widehat{i}+\widehat{j}-2\widehat{k} \right)$. So, the normal to plane can be found as

$\begin{align}

  & \left| \begin{matrix}

   \widehat{i} & \widehat{j} & \widehat{k} \\

   1 & 2 & -1 \\

   -1 & 1 & -2 \\

\end{matrix} \right| \\

 & =\widehat{i}\left[ \left( -2\times 2 \right)-\left( -1\times 1 \right) \right]-\widehat{j}\left[ \left( -2\times 1 \right)-\left( -1\times -1 \right) \right]+\widehat{k}\left[ \left( 1\times 1 \right)-\left( -1\times 2 \right) \right] \\

 & =\widehat{i}\left[ -4+1 \right]-\widehat{j}\left[ -2-1 \right]+\widehat{k}\left[ 1+2 \right] \\
 & =-3\widehat{i}+3\widehat{j}+3\widehat{k} \\

\end{align}$

Now, we can write the equation of plane in vector form as $\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\left( -3\widehat{i}+3\widehat{j}+3\widehat{k} \right)=d$. Applying dot product, we get $-3x+3y+3z=d$. Cancelling 3, we get the equation of the plane as $-x+y+z=d$. Now, to find the value of d, we know that point $\left( \widehat{i}+\widehat{j} \right)\Rightarrow \left( 1,1,0 \right)$ lies on this plane, so we get $-1+1+0=d\Rightarrow d=0$. Therefore, the equation of the plane is $-x+y+z=0$.

Now we will find the perpendicular distance of the point $\left( 2,1,4 \right)$ from the plane $-x+y+z=0$. We know that perpendicular distance of a point $\left( p,q,r \right)$ from a given plane $ax+by+cz+d=0$ can be found using the following formula $\text{distance}=\left| \dfrac{\left( ap+bq+cr+d \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$. Therefore, putting the required values as $a=-1,b=1,c=1,d=0,p=2,q=1,r=4$ in the formula, we get,

$d=\left| \dfrac{\left( -1\times 2 \right)+\left( 1\times 1 \right)+\left( 1\times 4 \right)+0}{\sqrt{{{\left( -1 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right|$
$d=\left| \dfrac{-2+1+4}{\sqrt{1+1+1}} \right|$

$d=\dfrac{3}{\sqrt{3}}$

$d=\sqrt{3}$

Thus, the distance between the point and plane will be $d=\sqrt{3}$ and therefore option d) is the correct answer.


Note: Students may do wrong in the calculation while finding the cross product, especially in the sign of the terms. So it is recommended to do the calculation step by step rather than solving the whole determinant in one go.