Question

Find the direction ratios of the normal to the plane, which passes through the points (1, 0, 0) and (0, 1, 0) and makes angle π/4 with the plane x + y = 3. Also find the equation of the plane

Answer 1

Let the equation of the plane passing through (1, 0, 0) be

a(x−1)+b(y−0)+c(z−0)=0       .....(1)

Here, a, b and c are the direction ratios of the normal to the plane.

If this plane passes through (0, 1, 0), then

a(0−1)+b(1−0)+c(0−0)=0

⇒−a+b=0

⇒a=b          .....(2)

It is given that plane (1) makes an angle of π/4 with the plane x + y = 3.

`therefore cos(pi/4)=(axx1+bxx1+cxx0)/(sqrt(a^2+b^2+c^2)sqrt(1^2+1^1+0^2))`

`=>1/sqrt2=a+b/(sqrt(a^2+b^2+c^2)sqrt2)`

`⇒sqrt(a^2+b^2+c^2)=a+b`

Squaring on both sides, we get

a²+b²+c²=a²+b²+2ab

⇒c²=2ab                        .....(3)

From (2) and (3), we get

c=±sqrt2a

So, the equation of the plane is

`a(x−1)+a(y−0)±sqrt2a(z−0)=0`

`⇒x−1+y±sqrt2z=0`

`⇒x+y±sqrt2z=`

Hence, the direction ratios of the normal to the plane are proportional to 1, 1,`±sqrt2`