Question

The image of the point (1, 3, 4 ) in the plane x + 2y – z + 3 = 0:
(a) (1, 1, -6 )
(b) (-1, -1, 6 )
(c) (-1, 1, -6 )
(d) None of these

Answer 1

Hint: First we are going to draw the given plane and plot the point and then we will find the direction ratio of normal to the plane and then using the point and direction ratio we will find the equation of line and then we will check which points satisfy that equation of line and that will be our answer.

Complete step-by-step answer:

Let’s look at the figure,

In this figure we have drawn the given plane, point A and the line which is perpendicular to the plane and passing through the point.

The equation of plane is x + 2y – z + 3 = 0, then the direction ratio of normal will be coefficient of x,y,z which is,

1:2:-1

Now we will find the equation of line with the direction ratio 1:2:-1 and point (1, 3, 4 ) is,

The equation of line will be:

$\dfrac{x-1}{1}=\dfrac{y-3}{2}=\dfrac{z-4}{-1}$

Now we will check all the options by putting the values,

For (a):

$\dfrac{1-1}{1}=\dfrac{1-3}{2}=\dfrac{-6-4}{-1}$

It’s incorrect hence option (a) is wrong.

For (b):

$\begin{align}

  & \dfrac{-1-1}{1}=\dfrac{-1-3}{2}=\dfrac{6-4}{-1} \\

 & -2=-2=-2 \\

\end{align}$

Hence (b) may be correct.

For (c):

$\begin{align}

  & \dfrac{-1-1}{1}=\dfrac{1-3}{2}=\dfrac{6-4}{-1} \\

 & -2=-1=-2 \\

\end{align}$

It is incorrect, hence option (c) is wrong.

Now the perpendicular distance from the two points to the plane should also be the same.

The formula is if the plane is Ax + By + Cz + D = 0 and the point is (a, b, c),

$\dfrac{Aa+Bb+Cc+D}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$

Now using this let’s check if it is equal or not,

$\begin{align}

  & \left| \dfrac{1+2\times 3-4+3}{\sqrt{1+{{2}^{2}}+1}} \right|=\left| \dfrac{-1+2\left( -1 \right)-6+3}{\sqrt{1+{{2}^{2}}+1}} \right| \\

 & \left| \dfrac{6}{\sqrt{1+{{2}^{2}}+1}} \right|=\left| \dfrac{-6}{\sqrt{1+{{2}^{2}}+1}} \right| \\

\end{align}$

Hence both are equal.

Therefore, option (b) is the correct answer.

Note: We could have also solved this question by taking a variable point on the line and we can equate the perpendicular distance of both those points the same and find the value of the image point.