Question

Let P be a plane containing the line (x-1)/3 = (y+6)/4 = (z+5)/2 and parallel to the line (x-3)/4 = (y-2)/(-3) = (z+5)/7. If the point (1, -1, α) lies on the plane P, then the value of |α| is equal to ________.

Hint:

The normal to a plane containing one line and parallel to another is the cross product of the direction vectors of the two lines.

Answer 1

Correct Answer: 38

Step 1: Normal to the plane \[ \vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix}. \] \[ \vec{n} = \hat{i}(28 + 6) - \hat{j}(21 - 8) + \hat{k}(-9 - 16) = 34\hat{i} - 13\hat{j} - 25\hat{k}. \]
Step 2: Equation of the plane passing through \( (1, -6, -5) \): \[ 34(x - 1) - 13(y + 6) - 25(z + 5) = 0. \]
Step 3: Since the point \( (1, -1, \alpha) \) lies on the plane, \[ 34(1 - 1) - 13(-1 + 6) - 25(\alpha + 5) = 0. \] \[ -65 - 25\alpha - 125 = 0 \Rightarrow 25\alpha = -190 \Rightarrow \alpha = -\frac{38}{5}. \]