Question

If M is the foot of the perpendicular drawn from \( P(1,2,-1) \) to the plane passing through the point \( A(3,-2,1) \) and perpendicular to the vector \( 4\overline{i}+7\overline{j}-4\overline{k} \) then the length of PM is

• A. \( \frac{32}{9} \)
• B. \( \frac{28}{9} \)
• C. \( \frac{26}{5} \)
• D. \( \frac{22}{5} \)

Hint:

You don't need to waste time finding the equation of the plane first. The shortest distance from a point to a plane can always be found directly by projecting the relative position vector onto the normal vector.

Answer 1

The Correct Option is B

Concept: The length of the perpendicular \( PM \) from point \( P(x_1, y_1, z_1) \) to a plane is equal to the absolute projection of the vector \( \overline{AP} \) along the normal vector \( \overline{n} \) of the plane: \[ \text{Length } PM = \frac{|\overline{AP} \cdot \overline{n}|}{|\overline{n}|} \]

Step 1: Finding the components of vector \( \overline{AP} \).
Given points \( A(3, -2, 1) \) and \( P(1, 2, -1) \): \[ \overline{AP} = (1 - 3)\overline{i} + (2 - (-2))\overline{j} + (-1 - 1)\overline{k} = -2\overline{i} + 4\overline{j} - 2\overline{k} \]

Step 2: Calculating the scalar dot product with normal vector \( \overline{n} \).
The normal vector is given as \( \overline{n} = 4\overline{i} + 7\overline{j} - 4\overline{k} \): \[ \overline{AP} \cdot \overline{n} = (-2)(4) + (4)(7) + (-2)(-4) = -8 + 28 + 8 = 28 \]

Step 3: Calculating the magnitude of the normal vector and final length.
\[ |\overline{n}| = \sqrt{4^2 + 7^2 + (-4)^2} = \sqrt{16 + 49 + 16} = \sqrt{81} = 9 \] Thus, the perpendicular distance is: \[ PM = \frac{|\overline{AP} \cdot \overline{n}|}{|\overline{n}|} = \frac{28}{9} \]