Vector Algebra – 3D – JEE Main 21 Jan 2026 Shift 1

Solution:

Let $A = (5, 4, 2)$.
Any general point $P$ on the line is given by:
$$P = (-1+2\lambda, 3+3\lambda, 1-\lambda)$$
The vector $\vec{AP}$ is:
$$\vec{AP} = (2\lambda – 1 – 5)\hat{i} + (3\lambda + 3 – 4)\hat{j} + (-\lambda + 1 – 2)\hat{k}$$
$$\vec{AP} = (2\lambda – 6)\hat{i} + (3\lambda – 1)\hat{j} + (-\lambda – 1)\hat{k}$$

Since $P$ is the foot of the perpendicular, $\vec{AP}$ is perpendicular to the direction vector of the line $\vec{b} = 2\hat{i} + 3\hat{j} – \hat{k}$.
Dot product $\vec{AP} \cdot \vec{b} = 0$:
$$2(2\lambda – 6) + 3(3\lambda – 1) – 1(-\lambda – 1) = 0$$
$$4\lambda – 12 + 9\lambda – 3 + \lambda + 1 = 0$$
$$14\lambda – 14 = 0 \Rightarrow \lambda = 1$$

Substitute $\lambda = 1$ into the coordinates of $P$ to find $(\alpha, \beta, \gamma)$:
$$\alpha = -1 + 2(1) = 1$$
$$\beta = 3 + 3(1) = 6$$
$$\gamma = 1 – 1 = 0$$
So, the vector $\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} = \hat{i} + 6\hat{j}$.

We need the projection of $\vec{u}$ on $\vec{v} = 6\hat{i} + 2\hat{j} + 3\hat{k}$.
Magnitude $|\vec{v}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.

$$\text{Projection} = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|}$$
$$= \frac{|1(6) + 6(2) + 0(3)|}{7}$$
$$= \frac{|6 + 12|}{7} = \frac{18}{7}$$

Ans. (3)