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 d² is equal to ____.

Answer 1

Correct Answer: 26

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.