A plane $E$ is perpendicular to the two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, and passes through the point $P (1,-1,1)$ If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3 \sqrt{2}$, then $( PQ )^2$ is equal to
To solve this problem, we must find a plane \(E\) that is perpendicular to the given planes and passes through a specified point, and then calculate the distance of this plane from another point. From there, we determine the square of the distance between two points.
The equations of the given planes are:
\(2x - 2y + z = 0\)
\(x - y + 2z = 4\)
For \(2x - 2y + z = 0\), the normal vector is \(\vec{n_1} = (2, -2, 1)\)
For \(x - y + 2z = 4\), the normal vector is \(\vec{n_2} = (1, -1, 2)\)
Thus, the equation of the plane \(E\) is of the form: \(-3x - 3y + c = 0\). Since the plane passes through point \(P(1, -1, 1)\), substitute these values into the plane equation to solve for \(c\): \(-3(1) - 3(-1) + c = 0 \Rightarrow c = 0\). Thus, the plane \(E\) is \(-x - y = 0\) or equivalently \(x + y = 0\).
Find the distance from point \(Q(a, a, 2)\) to this plane. The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is \(d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\). Substituting into the distance formula: \(d = \frac{|a + a|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|2a|}{\sqrt{2}} = \sqrt{2}|a|\). We are given that this distance is \(3\sqrt{2}\), so set: \(\sqrt{2}|a| = 3\sqrt{2}\), which simplifies to \(|a| = 3\).
Find \((PQ)^2\) where \(P\) is \((1, -1, 1)\) and \(Q\) is \((3, 3, 2)\) or \((-3, -3, 2)\) based on value of \(a\) being \(3\) or \(-3\). Calculate \((PQ)^2\):
