Q29 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\overrightarrow{\mathrm{r}}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$.
Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6 \hat{i}+2 \hat{j}+3 \hat{k}$ is :
Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6 \hat{i}+2 \hat{j}+3 \hat{k}$ is :
- A$\frac{15}{7}$
- B4
- Correct answerC$\frac{18}{7}$
- D3
Solution
Line: $\vec{r} = (-1, 3, 1) + \lambda(2, 3, -1)$, Point P = (5, 4, 2).
Any point on line: $Q = (-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda)$.
$\vec{PQ} = (-6 + 2\lambda, -1 + 3\lambda, -1 - \lambda)$.
For perpendicular: $\vec{PQ} \cdot (2, 3, -1) = 0$ gives $14\lambda - 14 = 0$, so $\lambda = 1$.
Foot $(\alpha, \beta, \gamma) = (1, 6, 0)$.
Projection of $(1, 6, 0)$ on $(6, 2, 3)$: $\frac{|6 + 12 + 0|}{\sqrt{49}} = \frac{18}{7}$.
Any point on line: $Q = (-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda)$.
$\vec{PQ} = (-6 + 2\lambda, -1 + 3\lambda, -1 - \lambda)$.
For perpendicular: $\vec{PQ} \cdot (2, 3, -1) = 0$ gives $14\lambda - 14 = 0$, so $\lambda = 1$.
Foot $(\alpha, \beta, \gamma) = (1, 6, 0)$.
Projection of $(1, 6, 0)$ on $(6, 2, 3)$: $\frac{|6 + 12 + 0|}{\sqrt{49}} = \frac{18}{7}$.
Hence, the correct answer is option: 'C'.