NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry: In class 11 ^{ th } you have learnt about 2D geometry. The 3D geometry is an extension of 2D geometry with taking consideration of 3 orthogonal axes. It requires more of your imagination power to visualize 3D geometry concepts. In NCERT solutions for class 12 maths chapter 11 three dimensional geometry article, you will learn how to use vector algebra to study three dimensional geometry. It will be easy for you to solve three dimensional problems with help of vector algebra. The purpose of this approach in the CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry is to make the study simple and effective. In this chapter there are many formulas to solve the problems. You are advised to write formulas when you are solving the problems. So, you can remember formulas very easily. Solutions of NCERT for class 12 maths chapter 11 three dimensional geometry will build your base for many other higherlevel concepts like tensors and manifolds. Check all NCERT solutions , which will help you get a better understanding of concepts in a much easy way. In this chapter, there are three exercises & miscellaneous exercise.
The important topics like direction cosines and direction ratios of a line joining two points, equations of lines in space, the angle between two lines, the angle between two planes, angle between a line and a plane, the shortest distance between two skew lines, equation of a plane in the normal form, etc. are covered in this chapter. Questions from all the topics are covered in t he CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry. A total of 36 questions in 3 exercises are given in this chapter. All these NCERT questions are solved and explained in the CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry article to clear your doubts.
In this chapter we deal with formulas like

Topics and subtopics of NCERT class 12 maths chapter 11 Three Dimensional Geometry
11.1 Introduction
11.2 Direction Cosines and Direction Ratios of a Line
11.2.1 Relation between the direction cosines of a line
11.2.2 Direction cosines of a line passing through two points
11.3 Equation of a Line in Space
11.3.1Equation of a line through a given point and parallel to a given vector b
11.3.2 Equation of a line passing through two given points
11.4 Angle between Two Lines
11.5 Shortest Distance between Two Lines
11.5.1 Distance between two skew lines
11.5.2 Distance between parallel lines
11.6 Plane
11.6.1 Equation of a plane in normal form
11.6.2 Equation of a plane perpendicular to a given vector and passing through a given point
11.6.3 Equation of a plane passing through three noncollinear points
11.6.4 Intercept form of the equation of a plane
11.6.5 Plane passing through the intersection of two given planes
11.7 Coplanarity of Two Lines
11.8 Angle between Two Planes
11.9 Distance of a Point from a Plane
11.10 Angle between a Line and a Plane
NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry Exercise Questions
Solutions of NCERT for class 12 maths chapter 11 three dimensional geometryExercise: 11.1
Question:1 If a line makes angles with the x, y and zaxes respectively, find its direction cosines.
Answer:
Let the direction cosines of the line be l,m, and n.
So, we have
Therefore the direction cosines of the lines are .
Question:2 Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer:
If the line is making equal angle with the coordinate axes. Then,
Let the common angle made is with each coordinate axes.
Therefore, we can write;
And as we know the relation;
or
Thus the direction cosines of the line are
Question:3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?
Answer:
GIven a line has direction ratios of 18, 12, – 4 then its direction cosines are;
Line having direction ratio 18 has direction cosine:
Line having direction ratio 12 has direction cosine:
Line having direction ratio 4 has direction cosine:
Thus, the direction cosines are .
Question:4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.
Answer:
We have the points, A (2, 3, 4),B (– 1, – 2, 1),C (5, 8, 7);
And as we can find the direction ratios of the line joining the points is given by
The direction ratios of AB are i.e.,
The direction ratios of BC are i.e., .
We can see that the direction ratios of AB and BC are proportional to each other and is 2 times.
AB is parallel to BC. and as point B is common to both AB and BC,
Hence the points A, B and C are collinear.
Answer:
Given vertices of the triangle (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).
Finding each side direction ratios;
Direction ratios of side AB are i.e.,
Therefore its direction cosines values are;
SImilarly for side BC;
Direction ratios of side BC are i.e.,
Therefore its direction cosines values are;
Direction ratios of side CA are i.e.,
Therefore its direction cosines values are;
CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometryExercise: 11.2
Question:1 Show that the three lines with direction cosines
Answer:
GIven direction cosines of the three lines;
And we know that two lines with direction cosines and are perpendicular to each other, if
Hence we will check each pair of lines:
Lines ;
the lines are perpendicular.
Lines ;
the lines are perpendicular.
Lines ;
the lines are perpendicular.
Thus, we have all lines are mutually perpendicular to each other.
Answer:
We have given points where the line is passing through it;
Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and line joining the points (0, 3, 2) and (3, 5, 6).is CD.
So, we will find the direction ratios of the lines AB and CD;
Direction ratios of AB are
or
Direction ratios of CD are
or .
Now, lines AB and CD will be perpendicular to each other if
Therefore, AB and CD are perpendicular to each other.
Answer:
We have given points where the line is passing through it;
Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and line joining the points (– 1, – 2, 1) and (1, 2, 5)..is CD.
So, we will find the direction ratios of the lines AB and CD;
Direction ratios of AB are
or
Direction ratios of CD are
or .
Now, lines AB and CD will be parallel to each other if
Therefore we have now;
Hence we can say that AB is parallel to CD.
Answer:
It is given that the line is passing through A (1, 2, 3) and is parallel to the vector
We can easily find the equation of the line which passes through the point A and is parallel to the vector by the known relation;
, where is a constant.
So, we have now,
Thus the required equation of the line.
Answer:
Given that the line is passing through the point with position vector and is in the direction of the line .
And we know the equation of the line which passes through the point with the position vector and parallel to the vector is given by the equation,
So, this is the required equation of the line in the vector form.
Eliminating , from the above equation we obtain the equation in the Cartesian form :
Hence this is the required equation of the line in Cartesian form.
Answer:
Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the ;
The direction ratios of the line, are 3,5 and 6 .
So, the required line is parallel to the above line.
Therefore we can take direction ratios of the required line as 3k , 5k , and 6k , where k is a nonzero constant.
And we know that the equation of line passing through the point and with direction ratios a, b, c is written by: .
Therefore we have the equation of the required line:
or
The required line equation.
Question:7 The cartesian equation of a line is . Write its vector form .
Answer:
Given the Cartesian equation of the line;
Here the given line is passing through the point .
So, we can write the position vector of this point as;
And the direction ratios of the line are 3 , 7 , and 2.
This implies that the given line is in the direction of the vector, .
Now, we can easily find the required equation of line:
As we know that the line passing through the position vector and in the direction of the vector is given by the relation,
So, we get the equation.
This is the required equation of the line in the vector form.
Answer:
GIven that the line is passing through the and
Thus the required line passes through the origin.
its position vector is given by,
So, the direction ratios of the line through and are,
The line is parallel to the vector given by the equation,
Therefore the equation of the line passing through the point with position vector and parallel to is given by;
Now, the equation of the line through the point and the direction ratios a, b, c is given by;
Therefore the equation of the required line in the Cartesian form will be;
OR
Answer:
Let the line passing through the points and is AB;
Then as AB passes through through A so, we can write its position vector as;
Then direction ratios of PQ are given by,
Therefore the equation of the vector in the direction of AB is given by,
We have then the equation of line AB in vector form is given by,
So, the equation of AB in Cartesian form is;
or
Question:10 Find the angle between the following pairs of lines:
Answer:
To find the angle A between the pair of lines we have the formula;
We have two lines :
and
The given lines are parallel to the vectors ;
where and respectively,
Then we have
and
Therefore we have;
or
Question:10 Find the angle between the following pairs of lines:
Answer:
To find the angle A between the pair of lines we have the formula;
We have two lines :
and
The given lines are parallel to the vectors ;
where and respectively,
Then we have
and
Therefore we have;
or
Question:11 Find the angle between the following pair of lines:
Answer:
Given lines are;
and
So, we two vectors which are parallel to the pair of above lines respectively.
and
To find the angle A between the pair of lines we have the formula;
Then we have
and
Therefore we have;
or
Question:11 Find the angle between the following pair of lines:
Answer:
Given lines are;
and
So, we two vectors which are parallel to the pair of above lines respectively.
and
To find the angle A between the pair of lines we have the formula;
Then we have
and
Therefore we have;
or
Question:12 Find the values of p so that the lines and are at right angles.
Answer:
First we have to write the given equation of lines in the standard form;
and
Then we have the direction ratios of the above lines as;
and respectively..
Two lines with direction ratios and are perpendicular to each other if,
Thus, the value of p is .
Question:13 Show that the lines and are perpendicular to each other.
Answer:
First, we have to write the given equation of lines in the standard form;
and
Then we have the direction ratios of the above lines as;
and respectively..
Two lines with direction ratios and are perpendicular to each other if,
Therefore the two lines are perpendicular to each other.
Question:14 Find the shortest distance between the lines
Answer:
So given equation of lines;
and in the vector form.
Now, we can find the shortest distance between the lines and , is given by the formula,
Now comparing the values from the equation, we obtain
Then calculating
So, substituting the values now in the formula above we get;
Therefore, the shortest distance between the two lines is units.
Question:15 Find the shortest distance between the lines
Answer:
We have given two lines:
and
Calculating the shortest distance between the two lines,
and
by the formula
Now, comparing the given equations, we obtain
Then calculating determinant
Now calculating the denominator,
So, we will substitute all the values in the formula above to obtain,
Since distance is always nonnegative, the distance between the given lines is
units.
Question:16 Find the shortest distance between the lines whose vector equations are and
Answer:
Given two equations of line
in the vector form.
So, we will apply the distance formula for knowing the distance between two lines and
After comparing the given equations, we obtain
Then calculating the determinant value numerator.
That implies,
Now, after substituting the value in the above formula we get,
Therefore, is the shortest distance between the two given lines.
Question:17 Find the shortest distance between the lines whose vector equations are
Answer:
Given two equations of the line
in the vector form.
So, we will apply the distance formula for knowing the distance between two lines and
After comparing the given equations, we obtain
Then calculating the determinant value numerator.
That implies,
Now, after substituting the value in the above formula we get,
Therefore, units are the shortest distance between the two given lines.
NCERT solutions for class 12 maths chapter 11 three dimensional geometryExercise: 11.3
Question:1(a) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
Answer:
Equation of plane Z=2, i.e.
The direction ratio of normal is 0,0,1
Divide equation by 1 from both side
We get,
Hence, direction cosins are 0,0,1.
The distance of the plane from the origin is 2.
Question:1(b) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
Answer:
Given the equation of the plane is or we can write
So, the direction ratios of normal from the above equation are, .
Therefore
Then dividing both sides of the plane equation by , we get
So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.
The direction cosines of the given line are and the distance of the plane from the origin is units.
Question:1(c) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
Answer:
Given the equation of plane is
So, the direction ratios of normal from the above equation are, .
Therefore
Then dividing both sides of the plane equation by , we get
So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.
The direction cosines of the given line are and the distance of the plane from the origin is units.
Question:1(d) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
Answer:
Given the equation of plane is or we can write
So, the direction ratios of normal from the above equation are, .
Therefore
Then dividing both sides of the plane equation by , we get
So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.
The direction cosines of the given line are and the distance of the plane from the origin is units.
Answer:
We have given the distance between the plane and origin equal to 7 units and normal to the vector .
So, it is known that the equation of the plane with position vector is given by, the relation,
, where d is the distance of the plane from the origin.
Calculating ;
is the vector equation of the required plane.
Question:3(a) Find the Cartesian equation of the following planes:
Answer:
Given the equation of the plane
So we have to find the Cartesian equation,
Any point on this plane will satisfy the equation and its position vector given by,
Hence we have,
Or,
Therefore this is the required Cartesian equation of the plane.
Question:3(b) Find the Cartesian equation of the following planes:
Answer:
Given the equation of plane
So we have to find the Cartesian equation,
Any point on this plane will satisfy the equation and its position vector given by,
Hence we have,
Or,
Therefore this is the required Cartesian equation of the plane.
Question:3(c) Find the Cartesian equation of the following planes:
Answer:
Given the equation of plane
So we have to find the Cartesian equation,
Any point on this plane will satisfy the equation and its position vector given by,
Hence we have,
Or,
Therefore this is the required Cartesian equation of the plane.
Question:4(a) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given a plane equation ,
Or,
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then finding the coordinates of the foot of the perpendicular are given by .
The coordinates of the foot of the perpendicular are;
or
Question:4(b) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given a plane equation ,
Or,
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then finding the coordinates of the foot of the perpendicular are given by .
The coordinates of the foot of the perpendicular are;
or
Question:4(c) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given plane equation .
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then finding the coordinates of the foot of the perpendicular are given by .
The coordinates of the foot of the perpendicular are;
or ..
Question: 4(d) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given plane equation .
or written as
The direction ratios of the normal of the plane are .
Therefore
So, now dividing both sides of the equation by we will obtain,
This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.
Then finding the coordinates of the foot of the perpendicular are given by .
The coordinates of the foot of the perpendicular are;
or .
Answer:
Given the point and the normal vector which is perpendicular to the plane is
The position vector of point A is
So, the vector equation of the plane would be given by,
Or
where is the position vector of any arbitrary point in the plane.
Therefore, the equation we get,
or
So, this is the required Cartesian equation of the plane.
Question:5(b) Find the vector and cartesian equations of the planes
that passes through the point (1,4, 6) and the normal vector to the plane is .
Answer:
Given the point and the normal vector which is perpendicular to the plane is
The position vector of point A is
So, the vector equation of the plane would be given by,
Or
where is the position vector of any arbitrary point in the plane.
Therefore, the equation we get,
So, this is the required Cartesian equation of the plane.
Question:6(a) Find the equations of the planes that passes through three points.
(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
Answer:
The equation of the plane which passes through the three points is given by;
Determinant method,
Or,
Here, these three points A, B, C are collinear points.
Hence there will be an infinite number of planes possible which passing through the given points.
Question:6(b) Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
Answer:
The equation of the plane which passes through the three points is given by;
Determinant method,
As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.
Finding the equation of the plane through the points,
After substituting the values in the determinant we get,
So, this is the required Cartesian equation of the plane.
Question:7 Find the intercepts cut off by the plane 2x + y – z = 5.
Answer:
Given plane
We have to find the intercepts that this plane would make so,
Making it look like intercept form first:
By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,
So, as we know that from the equation of a plane in intercept form, where a,b,c are the intercepts cut off by the plane at x,y, and zaxes respectively.
Therefore after comparison, we get the values of a,b, and c.
.
Hence the intercepts are .
Question:8 Find the equation of the plane with intercept 3 on the yaxis and parallel to ZOX plane.
Answer:
Given that the plane is parallel to the ZOX plane.
So, we have the equation of plane ZOX as .
And an intercept of 3 on the yaxis
Intercept form of a plane given by;
So, here the plane would be parallel to the x and zaxes both.
we have any plane parallel to it is of the form, .
Equation of the plane required is .
Answer:
The equation of any plane through the intersection of the planes,
Can be written in the form of; , where
So, the plane passes through the point , will satisfy the above equation.
That implies
Now, substituting the value of in the equation above we get the final equation of the plane;
is the required equation of the plane.
Answer:
Here and
and and
Hence, using the relation , we get
or ..............(1)
where, is some real number.
Taking , we get
or
or .............(2)
Given that the plane passes through the point , it must satisfy (2), i.e.,
or
Putting the values of in (1), we get
or
or
which is the required vector equation of the plane.
Answer:
The equation of the plane through the intersection of the given two planes, and is given in Cartesian form as;
or ..................(1)
So, the direction ratios of (1) plane are which are .
Then, the plane in equation (1) is perpendicular to whose direction ratios are .
As planes are perpendicular then,
we get,
or
or
Then we will substitute the values of in the equation (1), we get
or
This is the required equation of the plane.
Question:12 Find the angle between the planes whose vector equations are and .
Answer:
Given two vector equations of plane
and .
Here, and
The formula for finding the angle between two planes,
.............................(1)
and
Now, we can substitute the values in the angle formula (1) to get,
or
or
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
Answer:
Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Clearly, the given planes are NOT parallel.
Perpendicular check:
.
Clearly, the given planes are NOT perpendicular.
Then find the angle between them,
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
Answer:
Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Perpendicular check:
.
Thus, the given planes are perpendicular to each other.
2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
Answer:
Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Thus, the given planes are parallel as
2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
Answer:
Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Therefore
Thus, the given planes are parallel to each other.
4x + 8y + z – 8 = 0 and y + z – 4 = 0
Answer:
Two planes
whose direction ratios are and whose direction ratios are ,
are said to Parallel:
If,
and Perpendicular:
If,
And the angle between is given by the relation,
So, given two planes
Here,
and
So, applying each condition to check:
Parallel check:
Clearly, the given planes are NOT parallel as .
Perpendicular check:
.
Clearly, the given planes are NOT perpendicular.
Then finding the angle between them,
Question:14 In the following cases, find the distance of each of the given points from the corresponding given plane
Answer:
We know that the distance between a point and a plane is given by,
.......................(1)
So, calculating for each case;
(a) Point and Plane
Therefore,
(b) Point and Plane
Therefore,
(c) Point and Plane
Therefore,
(d) Point and Plane
Therefore,
CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometryMiscellaneous Exercise
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.
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 xaxis and passing through the origin.
Answer:
Equation of a line parallel to the xaxis and passing through the origin is itself xaxis .
So, let A be a point on the xaxis.
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 xaxis and passing through origin is
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 YZplane.
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 ZXplane.
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
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 .
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
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 xaxis.
Therefore, its normal is perpendicular to the xaxis.
The direction ratios of the xaxis are 1,0, and 0.
Substituting in equation (1), we obtain
So, the Cartesian equation is
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
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:
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 zaxis 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 yaxis (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)
NCERT solutions for class 12 maths chapter wise
chapter 1 
Solutions of NCERT for class 12 maths chapter 1 Relations and Functions 
chapter 2 
CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions 
chapter 3 

chapter 4 
Solutions of NCERT for class 12 maths chapter 4 Determinants 
chapter 5 
CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 
chapter 6 
NCERT solutions for class 12 maths chapter 6 Application of Derivatives 
chapter 7 

chapter 8 
CBSE NCERT solutions for class 12 maths chapter 8 Application of Integrals 
chapter 9 
NCERT solutions for class 12 maths chapter 9 Differential Equations 
chapter 10 
Solutions of NCERT for class 12 maths chapter 10 Vector Algebra 
chapter 11 
CBSE NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry 
chapter 12 
NCERT solutions for class 12 maths chapter 12 Linear Programming 
chapter 13 
Solutions of NCERT for class 12 maths chapter 13 Probability 
NCERT solutions for class 12 subject wise
Benefits of NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

Solutions of NCERT for class 12 maths chapter 11 three dimensional geometry are very easy for you to understand the concepts as they are explained in a stepbystep manner.

NCERT solutions for class 12 maths chapter 11 three dimensional geometry will give you some new insight into the concepts.

Scoring good marks in the exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 11 three dimensional geometry as these questions are answered by the experts who know how best to answer the questions in the board exam.

You should solve the miscellaneous exercise also, to develop a grip on the concepts. Here, you will get solutions for miscellaneous exercise too.
Happy learning !!!