Question

The plane containing the line $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}$ and also containing its projection on the plane $2x+3y-z=5$, contains which one of the following point?
(a) $\left( 2,0,-2 \right)$,
(b) $\left( -2,2,2 \right)$,
(c) $\left( 0,-2,2 \right)$,
(d) $\left( 2,2,0 \right)$.

Answer 1

We start solving the problem by finding the vector parallel to the given lines using the fact that the vector equation of the line passing through point $\left( a,b,c \right)$ and parallel to the vector $d\hat{i}+e\hat{j}+f\hat{k}$ is $x\hat{i}+y\hat{j}+z\hat{k}=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)+\alpha \left( d\hat{i}+e\hat{j}+f\hat{k} \right)$. We then assume the normal vector to the required line and take the dot product with the parallel vector obtained. We also find the normal vector from the given plane using the fact that the equation of the plane passing through the point $\left( a,b,c \right)$ with normal vector $p\hat{i}+q\hat{j}+r\hat{k}$ is $\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\bullet \left( p\hat{i}+q\hat{j}+r\hat{k} \right)=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( p\hat{i}+q\hat{j}+r\hat{k} \right)$. We take dot product with the normal vector obtained to get another equation and we solve for the normal vector of the plane. We then find the equation of the plane and substitute points to verify.

Complete step by step answer:
We have given that the plane containing the line $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}$ and also containing its containing projection on the plane $2x+3y-z=5$. We need to find which of the points given in options lies in the plane.
Let us consider $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}=\lambda$(say).
$\Rightarrow \dfrac{x-3}{2}=\lambda$, $\dfrac{y+2}{-1}=\lambda$, $\dfrac{z-1}{3}=\lambda$.
$\Rightarrow x=2\lambda +3$, $y=-\lambda -2$, $z=3\lambda +1$.
Let us consider $x\hat{i}+y\hat{j}+z\hat{k}$.
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left( 3\hat{i}-2\hat{j}+\hat{k} \right)+\lambda \left( 2\hat{i}-\hat{j}+3\hat{k} \right)$ ---(1).
We know that the vector equation of the line passing through point $\left( a,b,c \right)$ and parallel to the vector $d\hat{i}+e\hat{j}+f\hat{k}$ is $x\hat{i}+y\hat{j}+z\hat{k}=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)+\alpha \left( d\hat{i}+e\hat{j}+f\hat{k} \right)$, where $\alpha$is an arbitrary constant. Using this we can say that the line in equation (1) is parallel to $2\hat{i}-\hat{j}+3\hat{k}$ and passing through the point $\left( 3,-2,1 \right)$.
Let us assume the normal vector to the plane containing the line $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}$ be $a\hat{i}+b\hat{j}+c\hat{k}$.
We know that the normal vector is perpendicular to the plane and it makes us to know that the normal vector is perpendicular to the vector that is parallel to the plane. We know that the dot product of perpendicular vectors is 0.
$\Rightarrow \left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( 2\hat{i}-\hat{j}+3\hat{k} \right)=0$.
$\Rightarrow 2a-b+3c=0$ ---(1).
We have the equation of the plane as $2x+3y-z=5$. Let us assume the point in the plane is $x\hat{i}+y\hat{j}+z\hat{k}$. So, we get the equation of the plane as $\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\bullet \left( 2\hat{i}+3\hat{j}-\hat{k} \right)=5$.
We know that the equation of the plane passing through the point $\left( a,b,c \right)$ with normal vector $p\hat{i}+q\hat{j}+r\hat{k}$ is $\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\bullet \left( p\hat{i}+q\hat{j}+r\hat{k} \right)=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( p\hat{i}+q\hat{j}+r\hat{k} \right)$.
So, the normal vector for the plane $2x+3y-z=5$ is $2\hat{i}+3\hat{j}-\hat{k}$.
According to the problem, the required plane contains the projection of line $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}$ on the plane $2x+3y-z=5$. We know that the projection vector of $\overrightarrow{a}$ on $\overrightarrow{b}$ is $\dfrac{\left| \overrightarrow{a}.\overrightarrow{b} \right|}{{{\left| \overrightarrow{b} \right|}^{2}}}\overrightarrow{b}$, here $\overrightarrow{b}$ will be a normal vector. So, the normal vector of the required plane will be perpendicular to $\overrightarrow{b}$.
So, we have $\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( 2\hat{i}+3\hat{j}-\hat{k} \right)=0$.
$\Rightarrow 2a+3b-c=0$---(2).
Let us solve equations (1) and (2).
$\Rightarrow 2a-b+3c=0$.
$\Rightarrow 2a+3b-c=0$.
So, we have three variables, two equations to find those and the equations are equal to zero. So, there will be solutions other than the trivial solution $\left( 0,0,0 \right)$. So, we solve it as follows:
$\Rightarrow \dfrac{a}{\left( -1\times -1 \right)-\left( 3\times 3 \right)}=\dfrac{b}{\left( 3\times 2 \right)-\left( -1\times 2 \right)}=\dfrac{c}{\left( 2\times 3 \right)-\left( 2\times -1 \right)}$.
$\Rightarrow \dfrac{a}{\left( 1 \right)-\left( 9 \right)}=\dfrac{b}{\left( 6 \right)-\left( -2 \right)}=\dfrac{c}{\left( 6 \right)-\left( -2 \right)}$.
$\Rightarrow \dfrac{a}{1-9}=\dfrac{b}{6+2}=\dfrac{c}{6+2}$.
$\Rightarrow \dfrac{a}{-8}=\dfrac{b}{8}=\dfrac{c}{8}$.
$\Rightarrow \dfrac{a}{1}=\dfrac{b}{-1}=\dfrac{c}{-1}$.
So, the value of a, b and c is 1, –1, –1.
So, the vector normal to the plane is $\hat{i}-\hat{j}-\hat{k}$. We know that the line $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}$ is passing through the point $\left( 3,-2,1 \right)$ and this makes the required plane contains this line so, the plane also contains the point $\left( 3,-2,1 \right)$.
We know that the equation of the plane passing through the point $\left( a,b,c \right)$ with normal vector $p\hat{i}+q\hat{j}+r\hat{k}$ is $\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\bullet \left( p\hat{i}+q\hat{j}+r\hat{k} \right)=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( p\hat{i}+q\hat{j}+r\hat{k} \right)$.
The required equation of the plane is $\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\bullet \left( \hat{i}-\hat{j}-\hat{k} \right)=\left( 3\hat{i}-2\hat{j}+\hat{k} \right)\bullet \left( \hat{i}-\hat{j}-\hat{k} \right)$.
$\Rightarrow \left( x\times 1 \right)+\left( y\times -1 \right)+\left( z\times -1 \right)=\left( 3\times 1 \right)+\left( -2\times -1 \right)+\left( 1\times -1 \right)$.
$\Rightarrow \left( x \right)+\left( -y \right)+\left( -z \right)=3+2-1$.
$\Rightarrow x-y-z=4$.
$\Rightarrow x-y-z-4=0$.
So, we have found the equation of the plane as $x-y-z-4=0.$
Let us substitute each point in the equation of the plane and check.
We have option (a) as $\left( 2,0,-2 \right)$.
$\Rightarrow 2-0-\left( -2 \right)-4=0$.
$\Rightarrow 2-2=0$.
$\Rightarrow 0=0$. This point lies in the plane as we get the same value for L.H.S and R.H.S.
We have option (b) as $\left( -2,2,2 \right)$.
$\Rightarrow -2-2-2-4=0$.
$\Rightarrow -10=0$, which is a contradiction. So, this point doesn’t lie on the plane as we get different values for L.H.S and R.H.S.
We have option (c) as $\left( 0,-2,2 \right)$.
$\Rightarrow 0-\left( -2 \right)-2-4=0$.
$\Rightarrow -6+2=0$.
$\Rightarrow -4=0$. which is a contradiction. So, this point doesn’t lie on the plane as we get different values for L.H.S and R.H.S.
We have option (d) as $\left( 2,2,0 \right)$.
$\Rightarrow 2-2-0-4=0$.
$\Rightarrow -4=0$. which is a contradiction. So, this point doesn’t lie on the plane as we get different values for L.H.S and R.H.S.
We have found that the point $\left( 2,0,-2 \right)$ lies in the plane.
∴ The plane containing the line $\dfrac{x-3}{2}=\dfrac{y+2}{-1}=\dfrac{z-1}{3}$ and also containing its projection on the plane $2x+3y-z=5$ has point $\left( 2,0,-2 \right)$.

So, the correct answer is “Option A”.

Note: We can also solve this problem in cartesian form. But in cartesian form calculating the normal vector requires a huge amount of time and needs thorough calculation. We can use the fact that the vector perpendicular to vectors is the cross product of those two vectors to get the normal vector of the plane. We should not confuse the equation of line and equation of line in vector or cartesian form. We represent lines with direction ratios that represent all points on it and it doesn’t contain any involvement of addition of coordinates, which is not the case in the equation of plane.