Question

Let d be the distance between the foot of perpendiculars of the point P(1, 2, –1) and Q(2, –1, 3) on the plane – x + y + z = 1. Then d 2 is equal to ____.

Answer 1

The correct answer is: 26

To find the distance between a point and a plane, we can use the formula for the distance from a point $(x_0​,y_0​,z_0​)$ to a plane $A_x+B_y+C_z+D=0:$

$d = \frac{\left|Ax_0 + By_0 + Cz_0 + D\right|}{\sqrt{A^2 + B^2 + C^2}}$

In this case, the equation of the plane is $−x+y+z−1=0$, so $A=−1, B=1, C=1,$ and $D=−1.$

Let's use this formula for both points $P(1,2,−1)$ and $Q(2,−1,3)$, and then find the square of the distance $d^2$:

For point P(1,2,−1): $d_P = \frac{\left|(-1)(1) + (1)(2) + (1)(-1) - 1\right|}{\sqrt{(-1)^2 + (1)^2 + (1)^2}}$

For point Q(2,−1,3): $d_Q = \frac{\left|(-1)(2) + (1)(-1) + (1)(3) - 1\right|}{\sqrt{(-1)^2 + (1)^2 + (1)^2}}$

Now, calculate $d_P​$ and $d_Q$​, and then find $d^2 = d_P^2 + d_Q^2$​. The squared distance $d^2$ is equal to 26.