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

1. Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point from the plane Find also the image of the point in the plane.

Solution.

Let be the image of the point in the plane

PO is perpendicular to the plane and S is the midpoint of PO and the foot of the perpendicular.

Direction ratios of PS are

the equation of PS are

The general point on line is

If this point lies on plane, then

So, the coordinates of S are

As S is the midpoint of PO

By comparing both sides , we get

So, the image of the point

Perpendicular distance between two points is

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2. Show that, is the circum-centre of the triangle formed by the points and

Solution.

Let

is a right angled triangle with hypotenuse

So, circum centre of the triangle is the midpoint of AB.

the circum-centre of the triangle is

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3. Find the equations of the three planes which are parallel to the coordinate axes and which pass through the points and

Solution.

For Planes parallel to x-axis :

The direction ratios of x-axis are So, the equation of the plane parallel to the x-axis and passing through the points and is

For Planes parallel to y-axis :

The direction ratios of y-axis are So, the equation of the plane parallel to the y-axis and passing through the points and is

For Planes parallel to z-axis :

The direction ratios of z-axis are So, the equation of the plane parallel to the z-axis and passing through the points and is

4. Let be the coordinates of the foot of the perpendicular drawn from the origin to a plane. Find equation of that plane.

Solution.

Since is the coordinates of the foot of the perpendicular drawn from the origin to the plane, the direction ratios of the normal to the plane are or,

Let be the equation of the plane which passes through the point

So, the equation of the plane is given by

5. Show that the equation of the plane passing through the point (1, 2, 3) and parallel to the plane 3x+4y-5z=3 is given by 3x+4y-5z=-4.

Solution.

The equation of the plane parallel to can be written as

Since the plane (1) passes through the point ,

So, the equation of the plane is given by

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6. Find the equation of the plane passing through the points and and perpendicular to the plane convert the equation to vector form.

Solution.

The equation of any plane passing through the point is

where are constants.

Since the plane (1) passes through the point

Since the plane (1) is perpendicular to the plane so

From and , we get by cross-multiplication,

Hence, by (1) we get the cartesian equation of plane as follows :

Hence, the vector equation of the plane is

7. Prove that the equation of the plane which passes through the points and and which is perpendicular to the plane is

Solution.

Equation of any plane passing through the point is where are constants.

If the plane (1) passes through the point

Again, the plane (1) is perpendicular to the given plane

From (2) and (3), we get by cross-multiplication,

Hence, the equation of the plane is

8. Prove that the equation of the plane which passes through the point and is perpendicular to the planes and is

Solution.

The equation of any plane passing through the point is

where are constants.

Since the plane (1) is perpendicular to the planes and so

Now, from (2) and (3) we get by cross-multiplication,

So, the equation of the plane is

9. Show that the planes and are perpendicular to and planes respectively.

Solution.

For the plane , the direction ratios of the normal to the plane are

For xy-plane, the direction ratios of the normal to the plane are

Now,

So, the plane is perpendicular to the xy-plane.

For the plane , the direction ratios of the normal to the plane are

For yz-plane, the direction ratios of the normal to the plane are

Now,

So, the plane is perpendicular to the yz-plane.

For the plane , the direction ratios of the normal to the plane are

For zx-plane, the direction ratios of the normal to the plane are

Now,

So, the plane is perpendicular to the zx-plane.