NCERT Solutions for Class 12 Maths Exercise 11.2 Chapter 11 Three Dimensional Geometry – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 (Ex 11.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks.
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.2) Exercise 11.2
1.Show that the three lines with direction cosines 
are mutually perpendicular.
For first two lines,
= 
= 
Since, it is 0, therefore, the first two lines are perpendicular to each other.
For second and third lines,
= 
= 
Since, it is 0, therefore, second and third lines are also perpendicular to each other.
For First and third lines,
= 
= 
Since it is 0, therefore, first and third lines are also perpendicular to each other.
Hence, given three lines are mutually perpendicular to each other.
2.Show that the line through the points 
is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
and B
are 
Again, direction ratios of the line joining the points C (0, 3, 2) and D (3, 5, 6) are
(say)
For lines AB and CD,
= 
= 6 + 10 – 16 = 0
Since, it is 0, therefore, line AB is perpendicular to line CD.
3.Show that the line through points (4, 7, 8), (2, 3, 4) is parallel to the line through the points
= 
(say)
Again direction ratios of the line joining the points C
and D (1, 2, 5) are
= 
(say)
For the lines AB and CD,
Since, 
Therefore, line AB is parallel to line CD.
4.Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector
Position vector of a point on the required line is 
The required line is parallel to the vector 
direction ratios of the required line are coefficient of 
in 
are
Vector equation of the required line is
Where 
is a real number.
Cartesian equation of this equation is 
5.Find the equation of the line in vector and in Cartesian form that passes through the point with position vector and is in the direction
and is in the direction 
= 
The required line is in the direction of the vector is 
Direction ratios of required line are coefficients of in = 
Equation of the required line in vector form is
Where is a real number.
Cartesian equation of this equation is x−21=y+12=z−4−1x−21=y+12=z−4−1
6.Find the Cartesian equation of the line which passes through the point and parallel to the line given by
and parallel to the line given by 
Equation of the given line in Cartesian form is 
Direction ratios of the given line are its denominators 3, 5, 6
Equation of the required line is 
= 
7.The Cartesian equation of a line is Write its vector form.
Write its vector form.
= (say) 
General equation for the required line is 
Putting the values of 
in this equation,
= 
[Sincer→=a→+λb→][Sincer→=a→+λb→]
8.Find the vector and Cartesian equations of the line that passes through the origin and
= Position vector of a point here O (say) on the line = (0, 0, 0) = 
= A vector along the line
= 
= Position vector of point A – Position vector of point O
= 
Vector equation of the line is 
NowCartesian equation of the line
Direction ratios of line OA are 
And a point on the line is O (0, 0, 0) =
Cartesian equation of the line =
= 
= 
Remark: In the solution of the above question we can also take:
= Position vector of point A = 
for vector form and point A as = 
for Cartesian form.
Then the equation of the line in vector form is 
And equation of line in Cartesian form is 
9.Find vector and Cartesian equations of the line that passes through the points and
and be the position vectors of the points A
and B
respectively. 
and 
A vector along the line = 
= Position vector of point B – Position vector of point A
= 
= 
= 
Vector equation of the line is
And another vector equation for the same line is 
= 
Cartesian equation
Direction ratios of line AB are 
Equation of the line is
=>=>
10. Find the angle between the following pairs of lines:
(i)
and r→=7iˆ−6kˆ+μ(iˆ+2jˆ+2kˆ)r→=7i^−6k^+μ(i^+2j^+2k^)
(ii)
and r→=2iˆ−jˆ−5kˆ+μ(3iˆ−5jˆ−4kˆ)r→=2i^−j^−5k^+μ(3i^−5j^−4k^)
Comparing with ,
and 
(vector is the position vector of a point on line and is a vector along the line)
Again, equation of the second line is r→=7iˆ−6kˆ+μ(iˆ+2jˆ+2kˆ)r→=7i^−6k^+μ(i^+2j^+2k^)
Comparing with 
,
and 
(vector is the position vector of a point on line and is a vector along the line)
Let 
be the angle between these two lines, then
= 
= 
(ii)Comparing the first and second equations with and resp.
and 
Let be the angle between these two lines, then
= 
= 
⇒θ=cos−183√15⇒θ=cos−18315
11.Find the angle between the following pair of lines:
(i)
and 
(ii)
and 
The direction ratios of this line i.e., a vector along the line is
= 
= 
Now, equation of second line is
The direction ratios of this line i.e., a vector along the line is
= 
= 
Let be the angle between these two lines, then
= 
= 
(ii)Given: Equation of first line is
The direction ratios of this line i.e., a vector along the line is
= 
= 
Nowequation of second line is
The direction ratios of this line i.e., a vector along the line is
= 
= 
Let be the angle between these two lines, then
= 
= 
12.Find the values of so that the lines and are at right angles.
so that the lines
Direction ratios of this line are 
(say)
Again, equation of another line 
Direction ratios of this line are 
(say)
Since, these two lines are perpendicular.
Therefore,
13.Show that the lines and are perpendicular to each other.
Direction ratios of this line are 
=
Again equation of another line 
Direction ratios of this line are 1, 2, 3 =
b→=iˆ+2jˆ+3kˆb→=i^+2j^+3k^
Now
= 
= 
Hence, the given two lines are perpendicular to each other.
14.Find the shortest distance between the lines and
and 
and 
, we get 
and 
Since, the shortest distance between the two skew lines is given by
……….(i)
Here,
Putting these values in eq. (i),
Shortest distance 
15.Find the shortest distance between the lines and .
Comparing this equation with 
, we have
Again equation of another line is 
Comparing this equation with 
, we have
= 
Expanding by first row = 
= 
And
= 
= 
= 
Length of shortest distance = 
= 
(numerically)
= 
16.Find the shortest distance between the lines whose vector equations are
and r→=4iˆ+5jˆ+6kˆ+μ(2iˆ+3jˆ+kˆ)r→=4i^+5j^+6k^+μ(2i^+3j^+k^)
Comparing this equation with ,
and 
Again equation of second line 
Comparing this equation with 
,
and b→=2iˆ+3jˆ+kˆb→=2i^+3j^+k^
Now shortest distance 
= 
……….(i)
Here
Putting these values in eq. (i),
Shortest distance 
17.Find the shortest distance between the lines whose vector equations are
and 
= 
= 
Comparing this equation with 
,
Equation of second line is 
= 
= 
Comparing this equation with 
,
Now Shortest distance = 
……….(i)
Here a2→−a1→=(iˆ−jˆ−kˆ)−(iˆ−2jˆ+3kˆ)=jˆ−4kˆa2→−a1→=(i^−j^−k^)−(i^−2j^+3k^)=j^−4k^
b1→×b2→=∣∣∣∣∣iˆjˆkˆ−11−212−2∣∣∣∣∣(−2+4)iˆ−(2+2)jˆ+(−2−1)kˆb1→×b2→=|i^j^k^−11−212−2|(−2+4)i^−(2+2)j^+(−2−1)k^
Putting these values in eq. (i),
Shortest distance 
