Question

Find the equation of the plane which contains the two parallel lines x+1/5=y-2/3 =z+6/4 &x/10=y-3/6=z+5/8

Answer 1

Therefore the equation of plane is

$x+y-2z-13=0$

Step-by-step explanation:

Given straight lines

$\frac{x+1}{5} =\frac{y-2}{3} =\frac{z+6}{4}$

and

$\frac{x}{10} =\frac{y-3}{6} =\frac{z+5}{8}$

The direction ratio of the above lines are (5,3,4)

The $1^{st}$ line passes through the point (-1,2,-6)

and the $2^{nd}$ line passes through the point (0,3,-5)

since the required plane contains the above two straight line then the plane also contains the points  (-1,2,-6) and (0,3,-5)

Let the equation of the plane be

a(x+1)+b(y-2)+c(z+6)=0 ........(1)   [ since (-1,2,-6)point lies on the plane]

Again the plane contain the plane .

Therefore,

5a+3b+4c=0........(2)

Again the plane contains the point(0,3,-5). It will be satisfy the equation of plane.

∴a(0+1)+b(3-2)+c(-5+6)=0

⇔a+b+c=0........(3)

Eliminating a, b,and c from equation (1),(2) and (3)

$\left|\begin{array}{ccc}x+1&y-2&z+6\\5&3&4\\1&1&1\end{array}\right|=0$

$\Leftrightarrow -x-y+2z+13=0$

$\Leftrightarrow x+y-2z-13=0$

Therefore the equation of plane is

$x+y-2z-13=0$