Question

The plane passing through the origin and containing the lines whose direction cosines are proportional to 1,  −2,  21,  −2,  2 and 2,  3,  −12,  3,  −1 passes through the point.

Answer 1

Coordinate Geometry (3D)

To find: We have to find the plane passing through the origin and containing the lines whose direction cosines are proportional to $(1,\:-2,\:2)$ and $(2,\:3,\:-1)$.

Solution:

The equation of the plane passing through the origin $(0,\:0,\:0)$ is given by

$\quad A(x-0)+B(y-0)+C(z-0)=0$

$\Rightarrow Ax+By+Cz=0\quad.....(i)$

Since the plane $(i)$ contains the straight lines with direction cosines $(1,\:-2,\:2)$ and $(2,\:3,\:-1)$, we get

$\quad A-2B+2C=0\quad.....(ii)$

$\quad 2A+3B-C=0\quad.....(iii)$

Now eliminating $A,\:B,\:C$ from $(i),\:(ii),\:(iii)$, we have

$\quad$$\left|\begin{array}{ccc}x&y&z\\1&-2&2\\2&3&-1\end{array}\right|=0$

$\Rightarrow x(2-6)-y(-1-4)+z(3+4)=0$

$\quad$( Expanding along $R_{1}$)

$\Rightarrow -4x+5y+7z=0$

$\Rightarrow \boxed{4x-5y-7z=0}$

Answer: The equation of the required plane is

$\quad\quad 4x-5y-7z=0$.