Question

Find the equation of the plane containing the straight line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$ and perpendicular to the plane containing the lines $\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}$ and $\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}$?
(a) $x+2y-2z=0$,
(b) $x-2y+z=0$,
(c) $5x+2y-4z=0$,
(d) $3x+2y-3z=0$.

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 two parallel vectors that we have just obtained. We use the fact that if two planes are perpendicular, then their normal vectors are perpendicular to solve for the normal vector and using the normal vector, we find the equation of the required plane.

Complete step by step answer:
According to the problem, we need to find the equation of plane containing the straight line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$ and perpendicular to the plane containing the lines $\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}$ and $\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}$.
Let us consider $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\lambda$(say).
$\Rightarrow \dfrac{x}{2}=\lambda$, $\dfrac{y}{3}=\lambda$, $\dfrac{z}{4}=\lambda$.
$\Rightarrow x=2\lambda$, $y=3\lambda$, $z=4\lambda$.
Let us consider $x\hat{i}+y\hat{j}+z\hat{k}=\lambda \left( 2\hat{i}+3\hat{j}+4\hat{k} \right)$.
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left( 0\hat{i}+0\hat{j}+0\hat{k} \right)+\lambda \left( 2\hat{i}+3\hat{j}+4\hat{k} \right)$ ---(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 $2\hat{i}+3\hat{j}+4\hat{k}$ and passing through the point $\left( 0,0,0 \right)$.
Let us assume the normal vector to the plane containing the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$ 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 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.
$\Rightarrow \left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( 2\hat{i}+3\hat{j}+4\hat{k} \right)=0$.
$\Rightarrow 2a+3b+4c=0$---(1).
We have another plane which contains the lines $\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}$ and $\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}$. So, we know that the normal vector perpendicular to every line present in the line. So, let us find the vectors that these lines are parallel.
Let us consider $\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}=k$(say),
$\Rightarrow \dfrac{x}{3}=k$, $\dfrac{y}{4}=k$, $\dfrac{z}{2}=k$.
$\Rightarrow x=3k$, $y=4k$, $z=2k$.
Let us consider $x\hat{i}+y\hat{j}+z\hat{k}=k\left( 3\hat{i}+4\hat{j}+2\hat{k} \right)$.
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left( 0\hat{i}+0\hat{j}+0\hat{k} \right)+k\left( 3\hat{i}+4\hat{j}+2\hat{k} \right)$ ---(2).
Let us consider $\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}=l$(say),
$\Rightarrow \dfrac{x}{4}=l$, $\dfrac{y}{2}=l$, $\dfrac{z}{3}=l$.
$\Rightarrow x=4l$, $y=2l$, $z=3l$.
Let us consider $x\hat{i}+y\hat{j}+z\hat{k}=l\left( 4\hat{i}+2\hat{j}+3\hat{k} \right)$.
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left( 0\hat{i}+0\hat{j}+0\hat{k} \right)+l\left( 4\hat{i}+2\hat{j}+3\hat{k} \right)$ ---(3).
From equations (2) and (3), we can see that the lines $\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}$ and $\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}$ are parallel to the vectors $3\hat{i}+4\hat{j}+2\hat{k}$ and $4\hat{i}+2\hat{j}+3\hat{k}$.
We know that the normal vector is perpendicular to both vectors $3\hat{i}+4\hat{j}+2\hat{k}$ and $4\hat{i}+2\hat{j}+3\hat{k}$. We know that the vector perpendicular to the vectors $\overrightarrow{x}$ and $\overrightarrow{y}$ is $\overrightarrow{x}\times \overrightarrow{y}$. We know that the cross product of two vectors $a\hat{i}+b\hat{j}+c\hat{k}$ and $d\hat{i}+e\hat{j}+f\hat{k}$ is defined as $\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ a & b & c \\\ d & e & f \\\ \end{matrix} \right|$.
So, the vector perpendicular to the vectors $3\hat{i}+4\hat{j}+2\hat{k}$ and $4\hat{i}+2\hat{j}+3\hat{k}$ is $\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{matrix} \right|=\hat{i}\times \left| \begin{matrix} 4 & 2 \\\ 2 & 3 \\\ \end{matrix} \right|-\hat{j}\times \left| \begin{matrix} 3 & 2 \\\ 4 & 3 \\\ \end{matrix} \right|+\hat{k}\left| \begin{matrix} 3 & 4 \\\ 4 & 2 \\\ \end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{matrix} \right|=\hat{i}\times \left( \left( 4\times 3 \right)-\left( 2\times 2 \right) \right)-\hat{j}\times \left( \left( 3\times 3 \right)-\left( 4\times 2 \right) \right)+\hat{k}\left( \left( 3\times 2 \right)-\left( 4\times 4 \right) \right)$.
$\Rightarrow \left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{matrix} \right|=\hat{i}\times \left( 12-4 \right)-\hat{j}\times \left( 9-8 \right)+\hat{k}\left( 6-16 \right)$.
$\Rightarrow \left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{matrix} \right|=\hat{i}\times \left( 8 \right)-\hat{j}\times \left( 1 \right)+\hat{k}\left( -10 \right)$.
$\Rightarrow \left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{matrix} \right|=8\hat{i}-\hat{j}-10\hat{k}$ ---(4).
We know that if two planes are perpendicular, then their normal vectors are perpendicular to each other. We know that the normal vector for the plane containing the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$ is $a\hat{i}+b\hat{j}+c\hat{k}$.
So, we get $\left( a\hat{i}+b\hat{j}+c\hat{k} \right).\left( 8\hat{i}-\hat{j}-10\hat{k} \right)=0$.
$\Rightarrow 8a-b-10c=0$ ---(5).
Let us solve the equations (1) and (5).
$\Rightarrow 2a+3b+4c=0$,
$\Rightarrow 8a-b-10c=0$.
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:
$\Rightarrow \dfrac{a}{\left( 3\times -10 \right)-\left( -1\times 4 \right)}=\dfrac{b}{\left( 4\times 8 \right)-\left( -10\times 2 \right)}=\dfrac{c}{\left( 2\times -1 \right)-\left( 8\times 3 \right)}$.
$\Rightarrow \dfrac{a}{\left( -30 \right)-\left( -4 \right)}=\dfrac{b}{\left( 32 \right)-\left( -20 \right)}=\dfrac{c}{\left( -2 \right)-\left( 24 \right)}$.
$\Rightarrow \dfrac{a}{-30+4}=\dfrac{b}{32+20}=\dfrac{c}{-2-24}$.
$\Rightarrow \dfrac{a}{-26}=\dfrac{b}{52}=\dfrac{c}{-26}$.
$\Rightarrow \dfrac{a}{1}=\dfrac{b}{-2}=\dfrac{c}{1}$.
So, the value of a, b and c is 1, –2, 1.
So, the vector normal to the plane is $\hat{i}-2\hat{j}+\hat{k}$. We know that the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$ is passing through the point $\left( 0,0,0 \right)$ and this makes the required plane contains this line so, the plane also contains the point $\left( 0,0,0 \right)$.
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 $\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\bullet \left( \hat{i}-2\hat{j}+\hat{k} \right)=\left( 0\hat{i}+0\hat{j}+0\hat{k} \right)\bullet \left( \hat{i}-2\hat{j}+\hat{k} \right)$.
$\Rightarrow \left( x\times 1 \right)+\left( y\times -2 \right)+\left( z\times 1 \right)=\left( 0\times 1 \right)+\left( 0\times -2 \right)+\left( 0\times 1 \right)$.
$\Rightarrow \left( x \right)+\left( -2y \right)+\left( z \right)=0+0+0$.
$\Rightarrow x-2y+z=0$.

So, the correct answer is “Option B”.

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 should not get confused with 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.