# Plane | Part-6 | Ex-5A

##### In the following article, we are going to discuss/solve few Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Plane. In the previous article , we have solved few Short Answer type Questions.

10. If the points and are coplanar then prove that

Solution.

If the points and are coplanar then

11. Let be a vector of magnitude such that it makes equal acute angles with the coordinate axes. Find the vector and cartesian forms of the equation of a plane passing through and normal to .

Solution.

Suppose that the vector makes an angle with the coordinate axes.

So, the equation of the plane passing through and normal to is given by

So, the equation represents the vector equation of the plane. Now, putting in (1) we get,

represents the cartesian equation of the plane.

12. Through the point a plane is drawn perpendicular to where is the origin. Let this plane meet the coordinate axes at and Show that the area of the triangle where

Solution.

The direction ratios of are or,

Again, is normal to the plane . The equation of the plane through the point and perpendicular to the plane is

So, the direction ratios of and are respectively.

The Volume of

Again, the volume of

13. A variable plane which is at a constant distance from the origin cuts the axes at and Show that the locus of the point of intersection of the planes through drawn parallel to the coordinate axes parallel to the coordinate planes is

Solution.

Let be the point of intersection of three planes passing through the points and

The equation of planes passing through and parallel to and planes are and respectively and these planes meet z-axis, x-axis and y-axis at and respectively.

The equation of the plane through the points can be written as

Now, the distance of the plane (1) from the origin is unit.

Hence, the locus of the point is

14. A variable plane passes through the point and meets the axes at and If the planes through and parallel to the axes meet at , then prove that the locus of is

Solution.

Let the equation of the plane be If this plane passes through then

The equation of the planes through are respectively.

So, using (1) we get,

So, from (2) we get the locus of P and hence the result follows. (proved)

15. A point moves on the plane The plane, drawn through perpendicular to meets the axes in If the planes through parallel to the coordinate planes meet in a point then show that the locus of is given by the equation :

Solution.

Let So, and

The equation of any plane through the points and is given by

Let So, the direction ratios of are i.e.,

Now, is normal to the plane

The point lies on the plane

Again, the point lies on the plane

Hence, from and we get,

Hence, the locus of is