Question

hcf of 6x^2y^3z^4,9x^4y^5z and 12xy^2z^3​

Answer 1

Answer:

Let the DR of the normal of the plane be <a,b,c>

∴ The plane is perpendicular to the plane x+2y+2z=1

∴ Sum of product of the DRs of the normal be zero

∴a+2b+2c=0 ...... (1)

It is also perpendicular to the plane

x−2y+3z=4

∴ Sum of the product of the DRs of the normal be zero

∴a−2b+3c=0 ...... (2)

a+2b+2c=0

a−2b+3c=0

6+4

a

=

3−2

−b

=

−2−2

c

a=10,b=−1,c=−4

The plane passes through point (2,5,−3)

∴10(x−2)+(−1)(y−5)−4(z+3)=0

⇒10x−20−y+5−4z−12=0

⇒10x−y−4z−27=0

10x−y−4z=27