Question

the equation of the plane which is a distance of 12 units from the origin and perpendicular to the vector 6i+2j-3k is​

Answer 1

Answer:

When equation of a plane is ax+by+cz+d=0, its normal vector is a

i

^

+b

j

^

+c

k

^

and its distance from origin is

a

2

+b

2

+c

2

∣d∣

Similarly, a=2,b=−3,c=6 here.

Since 5=

4+9+36

∣d∣

⇒∣d∣=35

The equation of the plane can thus be 2x−3y+6z±3