Question

The plane passing the points $A\left( 1,0,1 \right),$ $B\left( 3,2,-1 \right)$ and parallel to the $y-$axis meets the line $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ at the point
$\left( a \right) \left( 0,0,2 \right)$
$\left( b \right) \left( 1,2,3 \right)$
$\left( c \right)\left( 0,0,0 \right)$
$\left( d \right)\left( 2,4,8 \right)$

Answer 1

Hint: We will find the equation of the required plane. Then, we will use the direction cosines of the $y-$axis. At last, we apply the given points in the obtained equation.

Complete step by step answer:
Consider the given points through which the plane parallel to $y-$axis meets the line $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ at a certain point.
The points are $A\left( 1,0,1 \right)$ and $B\left( 3,2,-1 \right).$
The equation of the plane that passes through a point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is given by $\Rightarrow a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0.$
If this plane passes through another point $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right),$ then the equation of the plane is $\Rightarrow a\left( {{x}_{1}}-{{x}_{2}} \right)+b\left( {{y}_{1}}-{{y}_{2}} \right)+c\left( {{z}_{1}}-{{z}_{2}} \right)=0.$
Therefore, equation of the plane that passes through the point $A\left( 1,0,1 \right)$ is given by
$\Rightarrow a\left( x-1 \right)+b\left( y-0 \right)+c\left( z-1 \right)=0.......\left( 1 \right)$
Since the plane passes through another point $B\left( 3,2,-1 \right),$ then the equation is
$\Rightarrow a\left( 1-3 \right)+b\left( 0-2 \right)+c\left( 1+1 \right)=0.$
That is,
\[\Rightarrow -2a-2b+2c=0.\]
This becomes,
\[\Rightarrow 2a+2b-2c=0.\]
And,
\[\Rightarrow a+b-c=0.......\left( 2 \right)\]
It is given that the plane is parallel to the $y-$axis.
And we know that the direction cosines of $y-$axis are $0,1,0.$
Since $y-$axis is parallel to the required plane whose equation is written above, equation $\left( 1 \right),$ we get
$\Rightarrow a\cdot 0+b\cdot 1-c\cdot 0=0+b-0=0$
So, we will get the value of $b$ as $b=0.$
Thus, the equation $\left( 2 \right)$ becomes
$\Rightarrow a-c=0$
This will lead us to,
$\Rightarrow a=c$
Let us suppose that $a=c=k$
Now consider the equation $\left( 1 \right).$
After applying $b=0,$ the equation $\left( 1 \right)$ becomes
$\Rightarrow a\left( x-1 \right)+c\left( z-1 \right)=0.$
Now, let us substitute the value of $a$ and $c,$ which is $k,$ in the above equation.
We get,
$\Rightarrow k\left( x-1 \right)+k\left( z-1 \right)=0.$
In this equation, $k$ is the common factor of both the summands. Let us take it out and transpose it into the RHS to get
$\Rightarrow \left( x-1 \right)+\left( z-1 \right)=0.$
We get,
$\Rightarrow x+z-2=0$
Apply the points $A\left( 1,0,1 \right)$ and $B\left( 3,2,-1 \right)$ in the above equation,
$\begin{align}
  & \Rightarrow 1+1-2=0 \\
 & \Rightarrow 3-1-2=0 \\
\end{align}$
This proves the plane $x+z-2=0$ passes through the points $A\left( 1,0,1 \right),B\left( 3,2,-1 \right)$ and is parallel to the $y-$axis.
We need to find the point at which the plane $x+z-2=0$ meets the line $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}.$
We are, now, going to apply the points given in the options one by one to check which one of them satisfies the equation of the plane.
Let us begin with $\left( 0,0,2 \right).$
We get,
$\Rightarrow x+z-2=0+2-2=2-2=0.$
Therefore, this point satisfies the equation of the plane given.
We can apply the rest of the points to confirm that none of them satisfies the equation of the plane. However, it is optional or necessary if you are given with a multiple selection question.
$\begin{align}
  & \Rightarrow 1+3-2\ne 0 \\
 & \Rightarrow 0+0-2\ne 0 \\
 & \Rightarrow 2+8-2\ne 0 \\
\end{align}$

Hence, the point at which the plane that passes through the points $A\left( 1,0,1 \right),B\left( 3,2,-1 \right)$ and parallel to the $y-$axis meets the line $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ is $\left( 0,0,2 \right).$

Note: The direction cosines of the $x-$axis are $1,0,0.$ The direction cosines of the $y-$axis are $0,1,0.$ And the direction cosines of the $z-$axis are $0,0,1.$