Question

If Q (0,-1,-3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, -1, -2), then the area (in square units) of $\Delta PQR$ is:
(a) $\dfrac{\sqrt{64}}{2}$
(b) $\dfrac{\sqrt{91}}{4}$
(c) $2\sqrt{13}$
(d) $\dfrac{\sqrt{91}}{2}$

Answer 1

Hint: First of all, we would use plane equation and coordinates of Q to obtain the coordinates of P. After that, we have three points and by applying the area formula we would evaluate the area of the triangle.

Complete step-by-step answer:

For an equation of plane ax + by + cz = d, the corresponding direction normal vector is (a, b, c).
Hence, the given plane 3x – y + 4z = 2, the direction normal can be given as (3, -1, 4).
The equation of a line passing through a point $\left( {{a}_{1}},{{a}_{2}},{{a}_{3}} \right)$ and parallel to another line whose direction vector is $\left( {{b}_{1}},{{b}_{2}},{{b}_{3}} \right)$ could be given as:
$\dfrac{x-{{a}_{1}}}{{{b}_{1}}}=\dfrac{y-{{a}_{2}}}{{{b}_{2}}}=\dfrac{z-{{a}_{3}}}{{{b}_{3}}}$
So, the equation of a line passing through Q and parallel to direction normal of plane:
$\dfrac{x}{3}=\dfrac{y+1}{-1}=\dfrac{z+3}{4}$
Now let the point A be on the plane which is the mid-point of the line joining PQ. Hence, the coordinates of A in the form of $\lambda $.
$\begin{align}
  & \dfrac{x}{3}=\dfrac{y+1}{-1}=\dfrac{z+3}{4}=\lambda \\
 & x=3\lambda ,y=-\lambda -1,z=4\lambda -3 \\
 & A=\left( 3\lambda ,-\lambda -1,4\lambda -3 \right) \\
\end{align}$
Satisfying the coordinates of A in the equation of plane:
$\begin{align}
  & 3\times 3\lambda -\left( -\lambda -1 \right)+4\cdot \left( 4\lambda -3 \right)=2 \\
 & 9\lambda +\lambda +1+16\lambda -12=2 \\
 & 26\lambda =13 \\
 & \lambda =\dfrac{1}{2} \\
\end{align}$
Therefore, the coordinates of A $=\left( \dfrac{3}{2},-\dfrac{3}{2},-1 \right)$.
So, the coordinate of P can be evaluated by using the mid-point formula:
$x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2},z=\dfrac{{{z}_{1}}+{{z}_{2}}}{2}$
So, we have A$=\left( \dfrac{3}{2},-\dfrac{3}{2},-1 \right)$ as the mid-point of P (x, y, z) and Q (0, -1, -3).
Now, to evaluate coordinates of P:
$\begin{align}
  & \dfrac{3}{2}=\dfrac{x}{2},\dfrac{-3}{2}=\dfrac{y-1}{2},-1=\dfrac{z-3}{2} \\
 & \therefore x=3,y=-2,z=1 \\
\end{align}$
Therefore, P (3, -2, 1).
So, now we are required to calculate the area of $\Delta PQR$ whose coordinates P (3, -2, 1)
 Q (0, -1, -3) and R (3, -1, -2).
The vector formed from two position vectors is:
$\overrightarrow{AB}=position\text{ vector of B }-\text{ }position\text{ vector of A}$
Let two vectors formed from P, Q, R would be:
$\begin{align}
  & \overrightarrow{PQ}=\left( 0-3 \right)i+\left( -1+2 \right)j+\left( -3-1 \right)k \\
 & \overrightarrow{PQ}=-3i+j-4k \\
 & \overrightarrow{QR}=\left( 3-0 \right)i+\left( -1+1 \right)j+\left( -2+3 \right)k \\
 & \overrightarrow{QR}=3i+k \\
\end{align}$
As we know that Area of triangle in determinant form for a vector:
$A=\dfrac{1}{2}\times \left| \begin{matrix}
   i & j & k \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$
Area of $\Delta PQR$:
$\begin{align}
  & A=\dfrac{1}{2}\times\left| \begin{matrix}
   i & j & k \\
   -3 & 1 & -4 \\
   3 & 0 & 1 \\
\end{matrix} \right| \\
 & A=\dfrac{1}{2}\times i\left( 1-0 \right)-j\left( -3+12 \right)+k\left( 0-3 \right) \\
 & A=\dfrac{1}{2}\times i-9j-3k \\
 & \left| A \right|=\dfrac{\sqrt{91}}{2} \\
\end{align}$
So, the correct answer is option (d).

Note: Alternate solution of this question is by using the distance formula of plane to a point and calculate the distance of point Q and hence calculate the total length of PQ which is twice of QM.
Then, calculate the perpendicular distance RM and finally evaluate the area.