Question

The straight line $\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t(2\hat{i} + 5\hat{j} + 3\hat{k})$ is parallel to the plane $\vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 5$. Then the distance between the straight line and the plane is:

• A. $\frac{9}{\sqrt{14}}$
• B. $\frac{8}{\sqrt{14}}$
• C. $\frac{7}{\sqrt{14}}$
• D. $\frac{6}{\sqrt{14}}$
• E. $\frac{5}{\sqrt{14}}$

Hint:

Verify parallelism first by checking $\vec{b} \cdot \vec{n} = 0$. Here, $(2, 5, 3) \cdot (2, 1, -3) = 4 + 5 - 9 = 0$. This confirms the line is parallel and the distance is constant.

Answer 1

The Correct Option is B

Concept: If a line is parallel to a plane, every point on the line is equidistant from the plane. To find the distance, we simply calculate the perpendicular distance from any point on the line (usually the position vector $\vec{a}$) to the plane.

Step 1: Convert the plane equation to Cartesian form.
Plane: $\vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 5 \Rightarrow 2x + y - 3z - 5 = 0$.

Step 2: Identify a point on the line.
From $\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t(2\hat{i} + 5\hat{j} + 3\hat{k})$, the point $A$ is $(1, 1, 2)$.

Step 3: Apply the point-to-plane distance formula.
$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$ $d = \frac{|2(1) + 1(1) - 3(2) - 5|}{\sqrt{2^2 + 1^2 + (-3)^2}}$ $d = \frac{|2 + 1 - 6 - 5|}{\sqrt{4 + 1 + 9}} = \frac{|-8|}{\sqrt{14}} = \frac{8}{\sqrt{14}}$