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  4. The equation of the plane passing through the origin and containing the lines whose d.c.'s are proportional to 1, -2, 2 and 2, 3, -1 is

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Q. The equation of the plane passing through the origin and containing the lines whose d.c.'s are proportional to 1, -2, 2 and 2, 3, -1 is

Introduction to Three Dimensional Geometry

Solution:

Any plane through the origin is $a(x - 0) + b(y - 0) + c(z - 0)$ = 0.
The plane contains the lines whose d.c. are proportional to 1, - 2, 2 and 2, 3, -1 $\therefore $ normal to the plane are to $\bot$ to these lines.
$\therefore $ $a - 2b + 2c$ = 0
and $2a + 3b + c$ = 0
$\therefore $ $\frac{a}{2-6}=\frac{b}{4+1}=\frac{c}{3+4}\Rightarrow \frac{a}{-4} =\frac{b}{5}=\frac{c}{7}$
$\therefore $ reqd . plane is
$-4\left(x-0\right)+5\left(y -0\right) +7 \left(z -0\right)=0$
$ \Rightarrow -4x + 5y + 7z = 0$
$ i.e., 4x - 5y - 7z = 0 $