The equation of the plane is
6x – 2y + 3z – 7 = 0
∴ its vector equation is
∴ \(\bar{n} = 6\hat{i} - 2\hat{j} + 3\hat{k}\) is normal to the plane
Unit vector along \(\bar{n}\) is
Dividing both sides of (1) by 7, we get
Comparing with normal form of equation of the plane \(\bar{n}.\hat{n} = p,\) it follows that length of perpendicular from origin is 1 unit.
Alternative Method:
The equation of the plane is 6x – 2y + 3z – 7 = 0 i.e. 6x – 2y + 3z = 7
This is the normal form of the equation of plane.
∴ perpendicular distance of the origin from the plane is p = 1 unit.
