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(C) $1$. The equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{

Vedclass · Accepted Answer

$\vec{r} = (4\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{j})$

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(C) $1$. The equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.$2$. The given point is $(4, -2, 3)$, so $\vec{a} = 4\hat{i} - 2\hat{j} + 3\hat{k}$.$3$. $A$ line perpendicular to the $XZ$-plane is parallel to the $Y$-axis.$4$. The direction vector of the $Y$-axis is $\hat{j}$. Thus, $\vec{b} = \hat{j}$.$5$. Substituting these into the formula, we get $\vec{r} = (4\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{j})$.

The equation of a line passing through a point $(4, -2, 3)$ and perpendicular to the $XZ$-plane is:

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