Q.The vector equation of the line passing through the point $(-1, 5, 4)$ and perpendicular to the plane $z = 0$ is
(A) $\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda(\hat{i} + \hat{j})$
(B) $\vec{r} = -\hat{i} + 5\hat{j} + (4 + \lambda)\hat{k}$
(C) $\vec{r} = \hat{i} - 5\hat{j} - 4\hat{k} + \lambda\hat{k}$
(D) $\vec{r} = \lambda\hat{k}$

The key idea is that a line perpendicular to the plane $z=0$ must be parallel to the $z$-axis, so its direction vector is along $\hat{k}$.

Step 1: The plane $z=0$ is the $xy$-plane. A line perpendicular to it is vertical, parallel to the $z$-axis.

Step 2: The direction vector of such a line is $\vec{d} = \hat{k}$ (or any scalar multiple). …

The plane $z = 0$ is the $xy$-plane — a flat horizontal surface. Any line perpendicular to it must point straight up or down, i.e., parallel to the $z$-axis. That’s the core geometric insight.

A line’s vector equation is $\vec{r} = \vec{a} + \lambda \vec{d}$, where $\vec{a}$ is a point on the line and $\vec{d}$ is the direction vector. Here, the direction vector must be along $\hat{k}$ (or any scalar multiple of it). The given point is $(-1, 5, 4)$, so $\vec{a} = -\hat{i} + 5\hat{j} + 4\hat{k}$.

Now check each option:

1. Option (A): $\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda(\hat{i} + \hat{j})$

   Direction is $\hat{i} + \hat{j}$, which lies in the $xy$-plane — parallel to $z=0$, not perpendicular. So this is wrong.

2. Option (B): $\vec{r} = -\hat{i} + 5\hat{j} + (4 + \lambda)\hat{k}$

   Rewrite as $-\hat{i} + 5\hat{j} + 4\hat{k} + \lambda\hat{k}$. Direction is $\hat{k}$ — exactly what we need. This is correct.

3. Option (C): $\vec{r} = \hat{i} - 5\hat{j} - 4\hat{k} + \lambda\hat{k}$ …