JEE Mathematics 2023 Question with Solution

Given: $\mathbf{P} = \hat{i} + 2\hat{j} + m\hat{k}$ and $\mathbf{Q} = 4\hat{i} - 2\hat{j} + \hat{k}$ are perpendicular.

Find: The value of $m$.

For perpendicular vectors, the dot product must be zero.

$$\vec{P} \cdot \vec{Q} = 0$$

Using the vectors exactly as stated in the question,

$$(\hat{i} + 2\hat{j} + m\hat{k}) \cdot (4\hat{i} - 2\hat{j} + \hat{k}) = 0$$

Now expand the dot product:

$$1\cdot 4 + 2\cdot(-2) + m\cdot 1 = 0$$ $$4 - 4 + m = 0$$ $$m = 0$$

So, from the question text, the vectors are perpendicular only when $m = 0$.

However, the solution works with different vectors, namely $\mathbf{P} = \hat{i} + 2m\hat{j} + m\hat{k}$ and $\mathbf{Q} = 4\hat{i} - 2\hat{j} + m\hat{k}$, and concludes:

$$4 - 4m + m^2 = 0$$ $$(m-2)^2 = 0$$ $$m = 2$$

Thus, the solution concludes that the correct option is B, but this conflicts with the question text and the listed option values. Since the solution is the primary source here, the answer is recorded as B.

Given: the solution treats the vectors as $\mathbf{P} = \hat{i} + 2m\hat{j} + m\hat{k}$ and $\mathbf{Q} = 4\hat{i} - 2\hat{j} + m\hat{k}$.

Find: The value of $m$ according to the provided working.

Since the vectors are perpendicular,

$$\vec{P} \cdot \vec{Q} = PQ\cos 90^\circ = 0$$

Now expand term by term:

$$(\hat{i} + 2m\hat{j} + m\hat{k}) \cdot (4\hat{i} - 2\hat{j} + m\hat{k}) = 0$$

Using the fact that like unit vectors have dot product $1$ and unlike unit vectors have dot product $0$,

$$(\hat{i}\cdot 4\hat{i}) + (2m\hat{j}\cdot -2\hat{j}) + (m\hat{k}\cdot m\hat{k}) = 0$$ $$4 - 4m + m^2 = 0$$

Rearranging,

$$m^2 - 4m + 4 = 0$$

Factorizing,

$$(m-2)^2 = 0$$ $$m = 2$$

Therefore, the numerical value obtained in the solution is $2$. This corresponds to option D in the listed options, even though the solution labels the correct option as B.