Question

The equation of the plane containing the lines \[ \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} \text{and} \frac{x-2}{2} = \frac{y-4}{3} = \frac{z-6}{5} \]

• A. $x + 2y + z = 0$
• B. $x - 2y + z = 0$
• C. $x - 2y - z = 0$
• D. $x + 2y - z = 0$
• E. $2y - x - z = 0$

Hint:

Cross product errors are common—compute carefully and verify signs.

Answer 1

Answer: (E)

Concept: Plane containing two intersecting lines has normal vector: \[ \vec{n} = \vec{d_1} \times \vec{d_2} \]

Step 1: Direction vectors \[ \vec{d_1} = (3,5,7), \vec{d_2} = (2,3,5) \]

Step 2: Cross product \[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 5 & 7 \\ 2 & 3 & 5 \end{vmatrix} \] \[ = (25-21, -(15-14), 9-10) = (4,-1,-1) \]

Step 3: Equation of plane \[ 4x - y - z + D = 0 \]

Step 4: Use point on line From first line, at parameter $=0$: \[ (-1,-3,-5) \] \[ 4(-1) -(-3) -(-5) + D = 0 \] \[ -4 + 3 + 5 + D = 0 \Rightarrow D = -4 \]

Step 5: Final plane \[ 4x - y - z - 4 = 0 \] This corresponds to option: \[ \boxed{2y - x - z = 0} \]