JEE Mathematics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: The point is $P(1,2,3)$, the plane is $2x-y+z=9$, and $R=(6,10,7)$.

Find: The square of the area of triangle $PQR$, where $Q$ is the image of $P$ in the given plane.

From the extracted solution, the reflected point is obtained using the image formula and comes out to be

$$\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{3}=-2\frac{(2\times 1-2+3\times 3-2)}{1^{2}+2^{2}+3^{2}}$$

Hence,

$$Q(-1,3,0)$$

Using the vector area formula for triangle,

$$\text{Area}=\frac{1}{2}\begin{Vmatrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 5 & 7 & 12\\ 2 & -1 & 3 \end{Vmatrix}$$

Evaluating the determinant,

$$=\frac{1}{2}\left|33\widehat{i}+9\widehat{j}-19\widehat{k}\right|$$

Therefore,

$$\text{Area}=\frac{1}{2}\sqrt{(33)^{2}+9^{2}+(19)^{2}}$$

So,

$$\text{Area}=\frac{1}{2}\sqrt{1089+81+361}=\frac{1}{2}\sqrt{1531}$$

Hence, the square of the area is

$$\left(\frac{1}{2}\sqrt{1531}\right)^2=\frac{1531}{4}$$

Therefore, the square of the area of triangle $PQR$ is $\frac{1531}{4}$. The provided answer field on the solution's states $594$, but the extracted working leads to $\frac{1531}{4}$.

Common mistakes

• Using the wrong reflection formula for a point in a plane. This gives an incorrect image point $Q$. First find the reflected point carefully from the plane equation, then use that point in the area formula.

• Using $\overrightarrow{PR} \times \overrightarrow{PQ}$ correctly but forgetting the factor $\frac{1}{2}$ for the area of a triangle. The cross product gives the area of the parallelogram, so divide by $2$.

• Calculating the area correctly and then not squaring it when the question asks for the square of the area. After finding the area, square the final expression.