Question

An equation of the plane passing through the origin and containing the lines whose direction cosines are proportional to 1, –2, 2 and 2, 3, –1 is –

• A. x – 2y + 2z = 0
• B. 2x + 3y – z = 0
• C. x + 5y – 3z = 0
• D. 4x – 5y – 7z = 0

Answer 1

Answer: (C)

x + 5y – 3z = 0

Answer 2

Let the direction cosines of the normal to the plane be l, m, n. Since the normal is perpendicular to both the lines lying in the plane, whose direction cosines are proportional to 1, –2, 2 and 2, 3, –1; we have l – 2m + 2n = 0

and 2l + 3m – n = 0

Ž $\frac{1}{- 4}$ = $\frac{m}{5}$ = $\frac{n}{7}$

Since the plane passes through the origin its equation is

lx + my + nz = 0 or 4x – 5y – 7z = 0.