Question

The equation to the plane through the points (2, -1, 0), (3, -4, 5) parallel to a line with direction cosines proportional to (2, 3, 4 ) is 9x – 2y – 3z = k, where k is

• A. 20
• B. -20
• C. 10
• D. -10

Answer 1

The correct option is A 20

Equation of plane through (2, -1, 0) is
a(x – 2) + b (y + 1) + c(z - 0) = 0 …..(i)
It also passes through (3, -4, 5), then
a – 3b + 5c = 0 ……(ii)
Given plane (i) is parallel to a line with direction cosines proportional to 2, 3, 4
2a + 3b + 4c = 0 ….. (iii)
From Eqs. (ii) and (iii), we get
$\frac{a}{9}=\frac{b}{-2}=\frac{c}{-3}=\lambda$ (say)
$\therefore a=9\lambda ,b=-2\lambda ,C=-3\lambda$
Now, from Eq. (i)
$9\lambda (x-2)-2\lambda (y+1)-3\lambda (z-0)=0$
$\Rightarrow 9x-2\lambda -3z=20$
$\therefore k=20$