Question

Find equation of plane passing through the point (-1,3,2) and perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z=0.

Answer 1

Let the equation of the plane passing through the point (–1, 3, 2) be a(x + 1) + b(y -3) + c(z - 2) = 0  ......(i)

The given equation of the planes are: x + 2 y + 3 z = 5 and 3 x + 3 y + z = 0

a+2b+3c=0........(ii)

3a+3b+c=0.......(iii)

 

Solving for a , b and c . we get,

( left steps in the picture above )

a=-7k,b=8k,c=-3k

 

Substituting for a ,b, and c in (i) we get,

⇒-7k(x + 1) + 8k(y + 1) -3k(z - 2) = 0

⇒-7x-7+8y+8-3z+6=0

⇒-7x+8y-3z+7=0

 

which is the required equation.

hope helpful for you

 

Answer 2

Answer:

= 7x - 8y + 3z + 25 = 0

Step-by-step explanation:

The equation of plane through ( - 1 , 3 , 2 ) can be expressed as

             A (x + 1) + B (y - 3) + C (z - 2) = 0

As the required plane is perpendicular to x + 2y + 3z = 5 and 3x + 3y + 3z = 0

               A + 2B + 3C = 0   and    3A + 3B + C = 0

=> $\frac{A}{2-9} = \frac{B}{9-1} = \frac{C}{3-6}$

=> $\frac{A}{-1} = \frac{B}{8} = \frac{C}{-3}$

Direction ratios of normal to the required plane are -7 , 8 , -3

Hence, equation of the plane will be

             -7(x + 1) + 8 (y - 3) - 3 (z - 2) = 0

=> -7x - 7 + 8y - 24 - 3z + 6 = 0

FINAL RESULT = 7x - 8y + 3z + 25 = 0  

GOOD LUCK !!