Question

Find the equation of the plane passing through the point a = 2i + 3j - k and perpendicular to a vector 3i - 2j - 2k and the distance of this plane from the origin.

Answer 1

Answer:

Equation of plane is given by

$\vec r.(2\hat i+3\hat j-\hat k)=2$

Distance of plane from origin = 2 units

Step-by-step explanation:

To find the equation of the plane passing through the point a = 2i + 3j - k and perpendicular to a vector 3i - 2j - 2k and the distance of this plane from the origin.

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

$\vec r.\vec n =d$

here plane is passing through

$\vec a =2\hat i+3\hat j-\hat k\\\\and\\\\\vec n=3\hat i-2\hat j-2\hat k$

so, the plane passes throught vector a and perpendicular to vector n is given by equation

$(\vec r-\vec a).\vec n=0\\\\\vec r.\vec n =\vec a.\vec n\\\\\vec r.(2\hat i+3\hat j-\hat k)=(2\hat i+3\hat j-\hat k).(3\hat i-2\hat j-2\hat k)$

$\vec r.(2\hat i+3\hat j-\hat k)=6-6+2\\\\\vec r.(2\hat i+3\hat j-\hat k)=2$

is the required equation of plane.

Distance of plane from origin = 2 units