Question

Find the equation of the planes through (6,-4,3),(0,4,3) and cutting of intercepts whose sum is zero

Answer 1

Therefore the equation of the plane is

$\frac{x}{3} -\frac{y}{2} -\frac{z}{1} =1$

and

$\frac{x}{3} +\frac{y}{6} -\frac{z}{9} =1$

Step-by-step explanation:

Plane: A plane is a flat surface area joining by minimum two lines.

The direction ratio of plane is equal to the direction ratio of a line which is perpendicular to the plane.

Intercept: The intercept of a plane is the points of coordinate axes where the plane cuts the x-axis, y-axis and z-axis.

The intercept form of a plane is

$\frac{x}{a} +\frac{y}{b} +\frac{z}{c} =1$

Thus the plane intersects x-axis, y-axis and z-axis at (a,00) , (0,b,0) and (0,0,c)

respectively.

Let the equation of the required plane is

$\frac{x}{a} +\frac{y}{b} +\frac{z}{c} =1$

Since the sum cutting intercepts are zero.

∴a+b+c=0........(1)

Since the plane passes through the points (6,-4,3) and (0,4,-3).This two points will be satisfy the equation of plane.

$\frac{6}{a} +\frac{-4}{b} +\frac{3}{c} =1$.........(2)

and

$\frac{4}{b} +\frac{-3}{c} =1$........(3)

Adding (2) and (3)

$\frac{6}{a} +\frac{-4}{b} +\frac{3}{c}+\frac{4}{b} +\frac{-3}{c} =1+1$

$\Rightarrow \frac{6}{a} =2$

⇒a=3

Putting a=3 in equation (1)

∴3+b+c=0

⇒c= -(b+3)

Putting c= -(b+3) in equation (3)

$\frac{4}{b} +\frac{-3}{-(b+3)} =1$

$\Rightarrow \frac{4b+12+3b}{b(b+3)} =1$

$\Rightarrow 7b+12=b^2+3b$

$\Rightarrow b^2+3b-7b-12=0$

$\Rightarrow b^2-4b-12=0$

$\Rightarrow b^2-6b+2b-12=0$

$\Rightarrow b(b-6)+2(b-6)=0$

$\Rightarrow (b+2)(b-6)=0$

⇒b= -2,6

When b = -2, c=-(-2+3)=-1

And when b=6 , c=-(6+3)= -9

Therefore the equation of the plane is

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

$\Rightarrow\frac{x}{3} -\frac{y}{2} -\frac{z}{1} =1$

And$\frac{x}{3} +\frac{y}{6} +\frac{z}{-9} =1$

$\Rightarrow \frac{x}{3} +\frac{y}{6} -\frac{z}{9} =1$