# NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 12 - Three Dimensional Geometry

Three Dimensional Geometry is the 11th chapter of NCERT Class 12 Mathematics. The chapter covers a few concepts from lines and planes in three dimensions. Concepts like equations of a line, plane and distance between them etc. A the end of the chapter the miscellaneous exercise comes. Here the NCERT solutions for Class 12 Maths chapter 11 miscellaneous exercise is given. The NCERT book class 12 maths chapter 11 miscellaneous exercise solutions covers questions related to all the concepts of the chapter. These Class 12 Maths chapter 11 miscellaneous solutions are designed by mathematics expert faculties. Along with the miscellaneous exercise, the following exercise is also present.

Three Dimensional Geometry 11.1

Three Dimensional Geometry 11.2

Three Dimensional Geometry 11.3

**Three Dimensional Geometry Class 12th Chapter 11-Miscellaneous Exercise **

** Question:1 ** Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

** Answer: **

We can assume the line joining the origin, be OA where and the point and PQ be the line joining the points and .

Then the direction ratios of the line OA will be and that of line PQ will be

So to check whether line OA is perpendicular to line PQ then,

Applying the relation we know,

** Therefore OA is perpendicular to line PQ. **

** Question:2 ** If l _{ 1 } , m _{ 1 } , n _{ 1 } and l _{ 2 } , m _{ 2 } , n _{ 2 } are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are .

** Answer: **

Given that are the direction cosines of two mutually perpendicular lines.

Therefore, we have the relation:

** .........................(1) **

** .............(2) **

Now, let us assume be the new direction cosines of the lines which are perpendicular to the line with direction cosines.

Therefore we have,

Or,

......(3)

So, l,m,n are the direction cosines of the line.

where, ** ........................(4) **

Then we know that,

So, from the equation (1) and (2) we have,

Therefore, ** ..(5) **

Now, we will substitute the values from the equation (4) and (5) in equation (3), to get

Therefore we have the direction cosines of the required line as;

** Question:3 ** Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

** Answer: **

Given direction ratios and .

Thus the angle between the lines A is given by;

a

** Thus, the angle between the lines is **

** Question:4 ** Find the equation of a line parallel to x-axis and passing through the origin.

** Answer: **

Equation of a line parallel to the x-axis and passing through the origin is itself ** x-axis ** .

So, let A be a point on the x-axis.

Therefore, the coordinates of A are given by , where .

Now, the direction ratios of OA are

So, the equation of OA is given by,

or

Thus, the equation of the line parallel to the x-axis and passing through origin is

** Question:5 ** If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

** Answer: **

Direction ratios of AB are

and Direction ratios of CD are

So, it can be noticed that,

Therefore, AB is parallel to CD.

** Thus, we can easily say the angle between AB and CD which is either . **

** Question:6 ** If the lines and are perpendicular, find the value of k.

** Answer: **

Given both lines are perpendicular so we have the relation;

For the two lines whose direction ratios are known,

We have the direction ratios of the lines, and are and respectively.

Therefore applying the formula,

or

** For, the lines are perpendicular. **

** Question:7 ** Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane

** Answer: **

Given that the plane is passing through the point so, the position vector of the point A is and perpendicular to the plane whose direction ratios are and the normal vector is

So, the equation of a line passing through a point and perpendicular to the given plane is given by,

, where

** Question:8 ** Find the equation of the plane passing through (a, b, c) and parallel to the plane .

** Answer: **

Given that the plane is passing through and is parallel to the plane

So, we have

The position vector of the point is,

and any plane which is parallel to the plane, is of the form,

. ** .......................(1) **

Therefore the equation we get,

Or,

So, now substituting the value of in equation (1), we get

** .................(2) **

** So, this is the required equation of the plane . **

Now, substituting in equation (2), we get

Or,

** Question:9 ** Find the shortest distance between lines and .

** Answer: **

Given lines are;

and

So, we can find the shortest distance between two lines and by the formula,

** ...........................(1) **

Now, we have from the comparisons of the given equations of lines.

** **

** **

So,

and

Now, substituting all values in equation (3) we get,

** Hence the shortest distance between the two given lines is 9 units. **

** Question:10 ** Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

** Answer: **

We know that the equation of the line that passes through the points and is given by the relation;

and the line passing through the points,

And any point on the line is of the form .

So, the equation of the YZ plane is

Since the line passes through YZ- plane,

we have then,

or and

So, therefore the required point is

** Question ** : ** 11 ** Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

** Answer: **

We know that the equation of the line that passes through the points and is given by the relation;

and the line passing through the points,

And any point on the line is of the form .

So, the equation of ZX plane is

Since the line passes through YZ- plane,

we have then,

or and

So, therefore the required point is

** Question:12 ** Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

** Answer: **

We know that the equation of the line that passes through the points and is given by the relation;

and the line passing through the points, .

And any point on the line is of the form.

This point lies on the plane,

or .

Hence, the coordinates of the required point are or .

** Question:13 ** Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

** Answer: **

Given

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors of these plane are

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

Now, as we know

the equation of a plane in vector form is :

Now Since this plane passes through the point (-1,3,2)

Hence the equation of the plane is

** Question:14 ** If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane then find the value of p.

** Answer: **

Given that the points and are equidistant from the plane

So we can write the position vector through the point is

Similarly, the position vector through the point is

The equation of the given plane is

and We know that the perpendicular distance between a point whose position vector is and the plane, and

Therefore, the distance between the point and the given plane is

** nbsp; .........................(1) **

Similarly, the distance between the point , and the given plane is

** .........................(2) **

** And it is given that the distance between the required plane and the points, and is equal. **

** therefore we have, **

** or or **

** Question:15 ** Find the equation of the plane passing through the line of intersection of the planes and and parallel to x-axis.

** Answer: **

So, the given planes are:

and

The equation of any plane passing through the line of intersection of these planes is

** ..............(1) **

Its direction ratios are and = 0

The required plane is parallel to the x-axis.

Therefore, its normal is perpendicular to the x-axis.

The direction ratios of the x-axis are 1,0, and 0.

Substituting in equation (1), we obtain

So, the Cartesian equation is

** Question:16 ** If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

** Answer: **

We have the coordinates of the points and respectively.

Therefore, the direction ratios of OP are

And we know that the equation of the plane passing through the point is

where a,b,c are the direction ratios of normal.

Here, the direction ratios of normal are and and the point P is .

Thus, the equation of the required plane is

** Question:17 ** Find the equation of the plane which contains the line of intersection of the planes and which is perpendicular to the plane

** Answer: **

The equation of the plane passing through the line of intersection of the given plane in

,,,,,,,,,,,,,(1)

The plane in equation (1) is perpendicular to the plane, Therefore

Substituting in equation (1), we obtain

.......................(4)

So, this is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting in equation (1).

Therefore we get the answer

** Question:18 ** Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line and the plane .

** Answer: **

Given,

Equation of a line :

Equation of the plane

Let's first find out the point of intersection of line and plane.

putting the value of into the equation of a plane from the equation from line

Now, from the equation, any point p in line is

So the point of intersection is

SO, Now,

The distance between the points (-1,-5,-10) and (2,-1,2) is

Hence the required distance is 13.

** Question:19 ** Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes and .

** Answer: **

Given

A point through which line passes

two plane

And

it can be seen that normals of the planes are

since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

So, a vector perpendicular to both these normal vector is

Now a line which passes through and parallels to is

So the required line is

** Questio ** n: ** 20 ** Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

and

** Answer: **

Given

Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5)

Now the two vectors which are parallel to the two lines are

and

As we know, a vector perpendicular to both vectors and is , so

A vector parallel to this vector is

Now as we know the vector equation of the line which passes through point p and parallel to vector d is

Here in our question, give point p = (1,2,-4) which means position vector of this point is

So, the required line is

** Question:21 ** Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then .

** Answer: **

The equation of plane having a, b and c intercepts with x, y and z-axis respectively is given by

The distance p of the plane from the origin is given by

Hence proved

** Question:22 ** Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units (C) 8 units (D)

** Answer: **

Given equations are

and

Now, it is clear from equation (i) and (ii) that given planes are parallel

We know that the distance between two parallel planes is given by

Put the values in this equation

we will get,

Therefore, the correct answer is (D)

** Question:23 ** The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through

** Answer: **

Given equations of planes are

and

Now, from equation (i) and (ii) it is clear that given planes are parallel to each other

Therefore, the correct answer is (B)

**More About NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise**

Twenty-three questions are there in the Class 12 Maths chapter 11 miscellaneous solutions. Miscellaneous exercise chapter 11 Class 12 are important to understand the concepts explained in the chapter. The NCERT syllabus for Class 12 Maths chapter 11 miscellaneous exercise covers all the concepts of the chapter.

**Also Read| **Three Dimensional Geometry Class 12th Notes

**Benefits of NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise**

To practise the complete chapter the miscellaneous exercise chapter 11 Class 12 is useful

To practice for the board exam the NCERT solutions for Class 12 Maths chapter 11 miscellaneous exercise is helpful.

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