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

Step 1: Choose the method.
The length $PM$ equals the projection of $\overline{AP}$ along the plane's normal $\overline{n}$: $PM = \dfrac{|\overline{AP}\cdot\overline{n}|}{|\overline{n}|}$.
Step 2: Find vector AP.
From $A(3,-2,1)$ to $P(1,2,-1)$: $\overline{AP} = -2\overline{i} + 4\overline{j} - 2\overline{k}$.
Step 3: Write the normal.
The normal is $\overline{n} = 4\overline{i}+7\overline{j}-4\overline{k}$.
Step 4: Take the dot product.
$\overline{AP}\cdot\overline{n} = (-2)(4) + (4)(7) + (-2)(-4) = -8 + 28 + 8 = 28$.
Step 5: Find the length of the normal.
$|\overline{n}| = \sqrt{16+49+16} = \sqrt{81} = 9$.
Step 6: Divide for the distance.
\[ PM = \frac{28}{9} \] \[ \boxed{\tfrac{28}{9}} \]