Question

Let Q be the foot of the perpendicular from the point P(7, -2, 13) on the plane containing the lines \(\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}\) and \(\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}\). Then (PQ)\(^2\), is equal to ___________.

Hint:

To find the equation of a plane containing two lines, find the cross product of their direction vectors to get the normal vector. Use any point from either line to find the constant term 'd' in the plane equation \(ax+by+cz+d=0\). The distance from a point to a plane formula can also be used to find the length PQ directly.

Answer 1

Correct Answer: 96

Step 1: Find the Equation of the Plane
The plane contains two lines. The direction vector of the first line is \(\vec{d_1} = (6, 7, 8)\) and it passes through \(A(-1, 1, 3)\). The direction vector of the second line is \(\vec{d_2} = (3, 5, 7)\) and it passes through \(B(1, 2, 3)\). A normal vector to the plane, \(\vec{n}\), can be found by taking the cross product of the direction vectors of the lines. \[ \vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
6 & 7 & 8
3 & 5 & 7 \end{vmatrix} \] \[ \vec{n} = \hat{i}(49 - 40) - \hat{j}(42 - 24) + \hat{k}(30 - 21) = 9\hat{i} - 18\hat{j} + 9\hat{k} \] We can use a simpler normal vector by dividing by 9: \(\vec{n}' = (1, -2, 1)\). The equation of the plane is of the form \(x - 2y + z + d = 0\). The plane passes through point A(-1, 1, 3) (from the first line). We use this to find d. \[ (-1) - 2(1) + (3) + d = 0 \implies -1 - 2 + 3 + d = 0 \implies d = 0 \] So the equation of the plane is \(x - 2y + z = 0\). (Let's check with point B(1,2,3): 1 - 2(2) + 3 = 1 - 4 + 3 = 0. It works).
Step 2: Find the Foot of the Perpendicular Q
The line passing through P(7, -2, 13) and perpendicular to the plane has the direction of the normal vector \(\vec{n}' = (1, -2, 1)\). The parametric equation of this line (let's call it L) is: \[ x = 7 + t, \quad y = -2 - 2t, \quad z = 13 + t \] The point Q is the intersection of line L and the plane. We substitute these coordinates into the plane's equation to find the value of the parameter \(t\) for point Q. \[ (7+t) - 2(-2 - 2t) + (13+t) = 0 \] \[ 7 + t + 4 + 4t + 13 + t = 0 \] \[ 6t + 24 = 0 \implies 6t = -24 \implies t = -4 \] Now find the coordinates of Q by substituting \(t=-4\) back into the line's equations: \(x_Q = 7 + (-4) = 3\) \(y_Q = -2 - 2(-4) = -2 + 8 = 6\) \(z_Q = 13 + (-4) = 9\) So, Q = (3, 6, 9).
Step 3: Calculate (PQ)\(^2\)
P = (7, -2, 13) and Q = (3, 6, 9). We use the distance formula squared: \[ (PQ)^2 = (x_P - x_Q)^2 + (y_P - y_Q)^2 + (z_P - z_Q)^2 \] \[ (PQ)^2 = (7-3)^2 + (-2-6)^2 + (13-9)^2 \] \[ (PQ)^2 = (4)^2 + (-8)^2 + (4)^2 = 16 + 64 + 16 = 96 \] Step 4: Final Answer
The value of (PQ)\(^2\) is 96.