Let a straight line $ L $ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines \[ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \quad \text{and} \quad \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. \] If the line $L$ intersects the yz-plane at the point Q, then the distance between the points P and Q is:

Question

Let $P$ be a point in the plane of the vectors \[ \vec{AB}=3\hat{i}+\hat{j}-\hat{k} \quad\text{and}\quad \vec{AC}=\hat{i}-\hat{j}+3\hat{k} \] such that $P$ is equidistant from the lines $AB$ and $AC$. If $|\vec{AP}|=\frac{\sqrt5}{2}$, then the area of triangle $ABP$ is:

Answer 1

Step 1: Determine the direction vector of the line.
The direction vector $\mathbf{v}$ is found using the cross product of direction vectors: \[ \mathbf{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = 10\hat{i} - 10\hat{j} + 5\hat{k} = 5(2\hat{i} - 2\hat{j} + \hat{k}) \]

Step 2: Express the line's equation.
The parametric form of the line is: \[ \frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 3}{1} = \lambda \]

Step 3: Locate the intersection with the yz-plane.
The yz-plane is defined by $x = 0$. Setting $x = 0$ in the line equation yields: \[ \frac{0 - 2}{2} = \lambda \implies \lambda = -1 \] Substituting $\lambda = -1$ into the parametric equations gives the intersection point: \[ Q(0, 0, 2) \]

Step 4: Compute the distance.
The distance between points P and Q is calculated as: \[ d(P, Q) = \sqrt{4 + 3 + 2} = \sqrt{9} = 3 \]

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The Correct Option is C

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