Question

Find the reflection of the point \[(0,3, - 2)\] in the line $\dfrac{{1 - x}}{2} = 2 - y = z + 1$.
(A). $\left( {1,2, - 1} \right)$
(B). $\left( {2,1,0} \right)$
(C). $\left( {2,1,4} \right)$
(D). $\left( {0,0,1} \right)$

Answer 1

Hint: First we have to express the given line in the standard form and find the direction ratios. Then we can find the direction ratios of the line perpendicular to this line and passing through the given point. Then establishing the relation between direction ratios of perpendicular lines we get the required solution.

Formula used: If a line $L$ is expressed in its Cartesian form, $L:\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}} = k$ (say), then ${a_1},{b_1},{c_1}$ are the direction ratios and $({x_1},{y_1},{z_1})$ is a point on the line.
If ${a_1},{b_1},{c_1}$ are the direction ratios of a line and ${a_2},{b_2},{c_2}$ are direction ratios of a line perpendicular to it, then ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$.
Distance between two points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$ is $\sqrt {({x_1} - {x_2})_{}^2 + ({y_1} - {y_2})_{}^2 + ({z_1} - {z_2})_{}^2} $.
Direction ratios of a line joining two points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$ is ${x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}$

Complete step-by-step answer:
The line given is $L:\dfrac{{1 - x}}{2} = 2 - y = z + 1$ and we have to find the reflection of the point $(0,3, - 2)$ in the line.
$L$ can be written as $\dfrac{{x - 1}}{{ - 2}} = \dfrac{{y - 2}}{{ - 1}} = \dfrac{{z - ( - 1)}}{1}$
Now it is in the standard form $\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$
With ${x_1} = 1,{y_1} = 2,{z_1} = - 1,{a_1} = - 2,{b_1} = - 1,{c_1} = 1$

Let the point $(0,3, - 2)$ be $P$ and its image be $P'({u_1},{u_2},{u_3})$
Consider a perpendicular from $P$ to the given line $L$.
Let $M$ be the foot of the perpendicular on $L$.
Then since $ - 2, - 1,1$ are the direction ratios of $L$ and $(1,2, - 1)$ is a point on $L$(obtained from the Cartesian equation of $L$), $M$ is $( - 2k + 1, - k + 2,k - 1)$ for some $k$.
Now the direction ratios of the line $PM$ can be calculated using two points $P(0,3, - 2)$ and $M( - 2k + 1, - k + 2,k - 1)$.
Direction ratios of a line joining two points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$ is ${x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}$.
Therefore, direction ratios of $PM$ is $ - 2k + 1 - 0, - k + 2 - 3,k - 1 - ( - 2)$ , that is $ - 2k + 1, - k - 1,k + 1$

Since the given line $L$ and the line $PM$ are perpendicular to each other, the sum of the product of the direction ratios is zero.
$ \Rightarrow - 2( - 2k + 1) + - 1( - k - 1) + 1(k + 1) = 0$
On simplification we get,
$ \Rightarrow 4k - 2 + k + 1 + k + 1 = 0$
$ \Rightarrow 6k = 0$
$6 \ne 0 \Rightarrow k = 0$
Therefore, the point $M( - 2k + 1, - k + 2,k - 1) = M(1,2, - 1)$ and the direction ratios of $PM$ is $ - 2k + 1, - k - 1,k + 1$ which is equal to $1, - 1,1$

So, the image $P'$ belongs to the line with direction ratios $1, - 1,1$ and a point in the line is $M(1,2, - 1)$.
Then $P'({u_1},{u_2},{u_3})$ is $P'(1 - k',2 - ( - k'), - 1 - k')$.
Since $M$ is the midpoint of $PP'$, $PM = MP'$.
Distance between two points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$ is$\sqrt {({x_1} - {x_2})_{}^2 + ({y_1} - {y_2})_{}^2 + ({z_1} - {z_2})_{}^2} $.
Using this equation and substituting $P(0,3, - 2),M(1,2, - 1)$ and $P'(1 - k',2 + k', - 1 - k')$
$PM = MP' \Rightarrow $
$\sqrt {(0 - 1)_{}^2 + (3 - 2)_{}^2 + ( - 2 + 1)_{}^2} = \sqrt {(1 - (1 - k')_{}^2 + (2 - (2 + k'))_{}^2 + ( - 1 - ( - 1 - k'))_{}^2} $
Squaring both sides,
${( - 1)^2} + {1^2} + {( - 1)^2} = {(1 - 1 + k')^2} + {(2 - 2 - k')^2} + {( - 1 + 1 + k')^2}$
$ \Rightarrow 1 + 1 + 1 = k{'^2} + k{'^2} + k{'^2}$
On simplification we get,
$ \Rightarrow 3 = 3k{'^2}$
Dividing both sides by $3$,
$ \Rightarrow \dfrac{3}{3} = \dfrac{{3k{'^2}}}{3}$
$ \Rightarrow 1 = k{'^2}$
This implies $k'$ is either $1$ or $ - 1$.
$k' = 1 \Rightarrow P'(1 - k',2 + k', - 1 - k') = P'(1 - 1,2 + 1, - 1 - 1) = P'(0,3, - 2) = P$, which is not possible.
Therefore, $k' \ne 1 \Rightarrow k' = - 1$.
Then the point $P'(1 - k',2 + k', - 1 - k') = P'(1 - - 1,2 + - 1, - 1 - ( - 1)) = P'(2,1,0)$
Therefore, the image of the point $P(0,3, - 2)$ is $P'(2,1,0)$.
               

So, the correct answer is “Option B”.

Note: The reflection or image of a point and the point are equidistant from the line of reflection. If in case the reflection is taken about $x$ or $y$ axes calculation is much simpler. In all other cases it is better to first write the equation in standard form and then proceed.