Cbse Class 12 Maths Ncert Solutions Chapter 11
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.
Ans.Given: Direction cosines of three lines are
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).
Ans.We know that direction ratios of the line joining the points A
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 
Ans.We know that direction ratios of the line joining the points A (4, 7, 8) and B (2, 3, 4) are
=
(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 
Ans.A point on the required line is A (1, 2, 3) =
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 
Ans.Position vector of a point on the required line is
=
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 − 2 1 = y + 1 2 = z − 4 − 1 x−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 
Ans.Given: A point on the line is
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.
Ans.Given: The Cartesian equation of the line is
= (say)
General equation for the required line is
Putting the values of
in this equation,
=
[ S i n c e r → = a → + λ b → ] [Sincer→=a→+λb→]
8.Find the vector and Cartesian equations of the line that passes through the origin and 
Ans. 
= 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 
10. Find the angle between the following pairs of lines:
(i) 
and r → = 7 i ˆ − 6 k ˆ + μ ( i ˆ + 2 j ˆ + 2 k ˆ ) r→=7i^−6k^+μ(i^+2j^+2k^)
(ii) 
and r → = 2 i ˆ − j ˆ − 5 k ˆ + μ ( 3 i ˆ − 5 j ˆ − 4 k ˆ ) r→=2i^−j^−5k^+μ(3i^−5j^−4k^)
Ans.(i) Equation of the first line is
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 → = 7 i ˆ − 6 k ˆ + μ ( i ˆ + 2 j ˆ + 2 k ˆ ) 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 − 1 8 3 √ 15 ⇒θ=cos−18315
11.Find the angle between the following pair of lines:
(i) 
and 
(ii) 
and 
Ans.(i) Given: Equation of first line is
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.
13.Show that the lines 
and 
are perpendicular to each other.
Ans.Equation of one line
Direction ratios of this line are
=
Again equation of another line
Direction ratios of this line are 1, 2, 3 =
b → = i ˆ + 2 j ˆ + 3 k ˆ b→=i^+2j^+3k^
Now
=
=
Hence, the given two lines are perpendicular to each other.
14.Find the shortest distance between the lines 
and 
Ans.Comparing the given equations with
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 
.
Ans.Equation of one line is
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 → = 4 i ˆ + 5 j ˆ + 6 k ˆ + μ ( 2 i ˆ + 3 j ˆ + k ˆ ) r→=4i^+5j^+6k^+μ(2i^+3j^+k^)
Ans.Equation of the first line is
Comparing this equation with,
and
Again equation of second line
Comparing this equation with
,
and b → = 2 i ˆ + 3 j ˆ + k ˆ b→=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
Ans.Equation of first line is
=
=
Comparing this equation with
,
Equation of second line is
=
=
Comparing this equation with
,
Now Shortest distance =
……….(i)
Here a 2 → − a 1 → = ( i ˆ − j ˆ − k ˆ ) − ( i ˆ − 2 j ˆ + 3 k ˆ ) = j ˆ − 4 k ˆ a2→−a1→=(i^−j^−k^)−(i^−2j^+3k^)=j^−4k^
b 1 → × b 2 → = ∣∣∣∣∣ i ˆ j ˆ k ˆ − 1 1 − 2 1 2 − 2 ∣∣∣∣∣ ( − 2 + 4 ) i ˆ − ( 2 + 2 ) j ˆ + ( − 2 − 1 ) k ˆ b1→×b2→=|i^j^k^−11−212−2|(−2+4)i^−(2+2)j^+(−2−1)k^
Putting these values in eq. (i),
Shortest distance
Cbse Class 12 Maths Ncert Solutions Chapter 11
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