Question

Let A be a point having position vector \( \vec{i}-3\vec{j} \) and \( \bar{r} = (\vec{i}-3\vec{j}) + t(\vec{j}-2\vec{k}) \) be a line. If P is a point on this line and is at a minimum distance from the plane \( \bar{r} \cdot (2\vec{i}+3\vec{j}+5\vec{k}) = 0 \), then the equation of the plane through P and perpendicular to AP is

• A. \( \bar{r} \cdot (-\vec{j}+2\vec{k}) = 8 \)
• B. \( \bar{r} \cdot (\vec{j}+\vec{k}) = 4 \)
• C. \( \bar{r} \cdot (\vec{i}+\vec{j}+\vec{k}) = 8 \)
• D. \( \bar{r} \cdot (\vec{i}-\vec{j}) = 12 \)

Hint:

Minimum distance from a line to a plane is zero if they intersect.

Answer 1

The Correct Option is A

Step 1: Intersection Point P:

Line: \( x=1, y=-3+t, z=-2t \). Plane: \( 2x+3y+5z=0 \). Substitute: \( 2(1) + 3(-3+t) + 5(-2t) = 0 \implies 2 - 9 + 3t - 10t = 0 \implies -7t = 7 \implies t = -1 \). P is \( (1, -4, 2) \).
Step 2: Plane Equation:

Normal \( \vec{n} = \vec{AP} = P - A = (1-1, -4-(-3), 2-0) = (0, -1, 2) = -\vec{j} + 2\vec{k} \). Plane: \( \vec{r} \cdot \vec{n} = P \cdot \vec{n} \) \( \vec{r} \cdot (-\vec{j} + 2\vec{k}) = (1)(0) + (-4)(-1) + (2)(2) = 8 \).
Step 3: Final Answer:

\( \bar{r} \cdot (-\vec{j}+2\vec{k}) = 8 \).