Question

the equation of the plane passing through the point (2,-1,3)and perpendicular to the vector 3i+2j-k​

Answer 1

Step-by-step explanation:

Equation of plane is given by

r

.(2

i

+3

j

−

k

)=2

Distance of plane from origin =2 units.

Now, to find the equation of the plane passing through the point a=2

i

+3

j

−

k

and perpendicular to a vectors 3

i

−2

j

−2

k

and the distance of this plane from the origin.

We know that equation of a plane in vector form is given by

r

.

n

=d

here plane is passing through

a=2

i

+3

j

−

k

n

=3

i

−2

j

−2

k

So, the plane passes through vector a and perpendicular to vector n is given by equation.

(

r

−

a

).

n

=0

⇒

r

n

=

a

n

⇒

r

.(2

i

+3

j

−

k

)=(2

i

+3

j

−

k

).(3

i

−2

j

−2

k

)

⇒

r

.(2

i

+3

j

−

k

)=6−6+2

⇒

r

.(2

i

+3

j

−

k

)=2

is the required equation of plane.

Distance of plane from origin =2 units