Question

The image of the point $\left( { - 1,3,4} \right)$ in the plane $x - 2y = 0$ is
A) $\left( {15,11,4} \right)$
B) $\left( {\dfrac{9}{5}, - \dfrac{{13}}{5},4} \right)$
C) $\left( {8,4,4} \right)$
D) None of these.

Answer 1

Here we need to find the image of the given point in the plane. For that, we will first assume the image of the point as variable and we will find the midpoint of the given point and its image. The midpoint will lie on the plane, so it will satisfy the equation of the plane. From there, we will get the first equation including the variables. The line joining the given point and its image will be perpendicular to the plane. We will use the condition for a line to be perpendicular to the plane. From there, we will get a second equation including the variables. After solving these two equations, we will get the value of variables and hence, the image of the given point in the plane.

Formula used:
The midpoint of any points $\left( {e,f,g} \right)$ and $\left( {x,y,z} \right)$ can by calculated using the formula
$\left( {\dfrac{{e + x}}{2},\dfrac{{f + y}}{2},\dfrac{{g + z}}{2}} \right)$.

Complete step by step solution:
Let the image of the point $\left( { - 1,3,4} \right)$ in the plane $x - 2y = 0$ be $\left( {a,b,c} \right)$.
We know the midpoint of the point $\left( { - 1,3,4} \right)$ and the point $\left( {a,b,c} \right)$ will lie on the plane.
Now, we will find the midpoint of these two points.
We will calculate the midpoint of the point $\left( { - 1,3,4} \right)$ and the point $\left( {a,b,c} \right)$ using the formula $\left( {\dfrac{{e + x}}{2},\dfrac{{f + y}}{2},\dfrac{{g + z}}{2}} \right)$.
Therefore, we get
$\Rightarrow {\rm{midpoint}} = \left( {\dfrac{{ - 1 + a}}{2},\dfrac{{3 + b}}{2},\dfrac{{4 + c}}{2}} \right)$
We know the midpoint of the point $\left( { - 1,3,4} \right)$ and the point $\left( {a,b,c} \right)$ will lie on the plane. So, it will satisfy the equation of the plane.
Therefore,
$\Rightarrow \dfrac{{ - 1 + a}}{2} - 2 \times \dfrac{{3 + b}}{2} = 0$
On further simplification, we get
$\Rightarrow - 1 + a - 6 - 2b = 0$
Adding and subtracting the like terms, we get
$\Rightarrow a - 2b = 7$ ……… $\left( 1 \right)$
We know that the line joining the given point and its image will be perpendicular to the plane.
Then the ratio of the difference of the coordinates of the two points to the coefficients of the equation of the plane will be equal.
$\Rightarrow \dfrac{{a + 1}}{1} = \dfrac{{b - 3}}{{ - 2}} = \dfrac{{c - 4}}{0}$
Solving these ratios, we get
$c = 4$ and $\dfrac{{a + 1}}{1} = \dfrac{{b - 3}}{{ - 2}}$
We will solve these two ratios, $\dfrac{{a + 1}}{1} = \dfrac{{b - 3}}{{ - 2}}$.
On cross multiplication, we get
$\Rightarrow - 2a + - 2 = b - 3$
On further simplification, we get
$\Rightarrow 2a + b = 1$ …… $\left( 2 \right)$
Multiplying equation 2 by 2 and then adding it with equation 1, we get
\begin{array} \underline \begin{array} 4a + 2b = 2 \\\ a - 2b = 7 \end{array} \\\ 5a = 9\end{array}
Now we got the equation as $5a = 9$.
Dividing both sides by 5, we get
$\Rightarrow a = \dfrac{9}{5}$
Now, substituting $a = \dfrac{9}{5}$ in equation $\left( 1 \right)$, we get
$\Rightarrow \dfrac{9}{5} - 2b = 7$
Adding $2b$ on both the sides, we get
$\Rightarrow 2b = \dfrac{9}{5} - 7$
On subtracting the numbers, we get
$\Rightarrow 2b = - \dfrac{{26}}{5}$
Dividing both sides by 2, we get
$\Rightarrow b = - \dfrac{{13}}{5}$
Hence, the required image of the point $\left( { - 1,3,4} \right)$ in the plane $x - 2y = 0$ is equal to $\left( {\dfrac{9}{5}, - \dfrac{{13}}{5},4} \right)$

Hence, the correct option is option B.

Note:
Here, it is important to note that if a point lies on the plane, then it will satisfy the equation of that plane. In other words we can say that when we put the point in the equation of the plane, then get the value as 0. Also, if a line is perpendicular to a plane, it will be parallel to the plane’s normal vector.