##### In the previous article , we have solved few VSA type questions of Plane Chapter (Ex-2B) of S N De Mathematics(Chhaya). In the following article, we are going to discuss/solve Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Plane (Ex-5B).

1. Find the equation of the plane passing through the intersection of the planes and and parallel to the line with direction ratios proportional to Find also the perpendicular distance of from this plane. [CBSE ’05]

Solution.

The given equations of the planes are

The equation of the plane passing through the intersection of planes (1) and (2) is

This plane (3) is parallel to the line with direction ratios proportional to

The required equation of the plane is

2. Find the equation of the plane parallel to the plane and situated at a distance of units from it.

Solution.

The equation of any plane parallel to the plane is

This plane (1) is situated at a distance units from the given plane .

So, the equation of the plane is and

3. Find the equation of the plane passing through the intersection of the planes and and parallel to the line

Solution.

Equation of plane passing through the intersection of planes (1) and (2) is

The plane (3) is parallel to the straight line

Now, we calculate the following values (from (3)).

So, the required equation of the plane is given by

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4. Find the distance of the point from the plane measured along a line parallel to

Solution.

The equation of any straight line passing through the point and parallel to the straight line is

Any point on this straight line (1) can be written as where is any non-zero real number.

If this point lies on the plane then

The point is

So, the required distance = The distance between and

5. Find the length and foot of the perpendicular from the point to the plane

Solution.

The equation of the straight line passing through the point and perpendicular to the plane is

The coordinates of any point on the straight line (1) can be written as where is any non-zero real number.

If this point lies on the given plane , then

Now, the distance between the points and is

The coordinates of the foot of the perpendicular is

6. Find the equation of the plane that contains the line of intersection of the planes and and which is perpendicular to the plane .

Solution.

The equation of the plane that contains the line of intersection of planes (1) and (2) can be written as

The plane (3) is perpendicular to the plane

Now, we calculate the following rules (from (3)).

After substituting the aforesaid values in (3) we get the required equation of the plane as follows .

Putting we get from (4),

7. Find the distance between the point and the plane passing through and

Solution.

The equation of the plane passing through the points and is

Distance of the point from the plane (1) is

8. Find the equation of the plane through the points and and find the distance of this plane from the point \

Hints : Follow Question No. 7

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9. Find the equation of the plane through the points and parallel to the line Convert to vector form.

Solution.

The equation of the plane passing through the point is

The plane (1) is passing through the point

Since the plane is parallel to the straight line

Eliminating from (1), (2), (3) we get,

The vector equation of (4) is

where

10. Show that the lines and are coplanar.

Solution.

The straight line (1) passes through the point and is parallel to

Again, the straight line (2) passes through the point and is parallel to

Clearly, given two straight lines are coplanar whenever and are coplanar.

are coplanar and hence the result follows.