JEE Mathematics 2023 Question with Solution

Given: $\mathbf{a}$ is parallel to the line of intersection of the two planes with normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ corresponding to $i + j + k$ and $-i - j - k$. Also, $|\mathbf{a}| = 6$ and $\mathbf{b} = -2i - 2j + 2k$.

Find: $\mathbf{a} \cdot \mathbf{b}$.

From the solution, take

$$\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \qquad \mathbf{n}_2 = \begin{pmatrix} -1 \\ -1 \\ -1 \end{pmatrix}$$

and compute the direction vector using the cross product:

$$\mathbf{a} = \mathbf{n}_1 \times \mathbf{n}_2$$

The extracted working gives

$$\mathbf{a} = \begin{pmatrix} -2 \\ 0 \\ -2 \end{pmatrix}$$

Now scale it to magnitude $6$ as shown:

$$\mathbf{a} = \frac{6}{\sqrt{8}} \begin{pmatrix} -2 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ -3 \end{pmatrix}$$

Using $\mathbf{b} = \begin{pmatrix} -2 \\ -2 \\ 2 \end{pmatrix}$,

$$\mathbf{a} \cdot \mathbf{b} = (-3)(-2) + (0)(-2) + (-3)(2)$$ $$\mathbf{a} \cdot \mathbf{b} = 6 + 0 - 6 = 0$$

The worked steps in the solution therefore give $\mathbf{a} \cdot \mathbf{b} = 0$, but the same the solution concludes that the correct option is D, i.e. $\left( \frac{2}{5} \sqrt{6} \right)$. This is an internal discrepancy in the source solution, and by the page conclusion the correct option is D.

The solution is self-contradictory. Its algebra leads to

$$\mathbf{a} \cdot \mathbf{b} = 0$$

but the solution explicitly states The Correct Option is D and ends with $\frac{2}{5}\sqrt{6}$. Since the