Equation of the plane passing through (-1, 1, 4) and containing the line $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z}{5}$, is :
Answer & explanation
Correct answer: option 4
Equation of any plane containing the line $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z}{5}$ will be
a(x − 1) + b(y − z) + cz = 0
where, 3a + b + 4c = 0 …. (i)
It is given that plane passes through (–1, 1, 4).
∴ –2a – b + 4c = 0 …. (ii)
From (i) and (ii), we get
$\frac{a}{-9}=\frac{b}{22}=\frac{c}{1}$
Thus the equation of required plane is,
–9(x – 1) + 22(y – 2) + z = 0
i.e., 22y – 9x + z = 35