NCERT Solutions for Class 12 Maths 3D Geometry

Hi Students, Welcome to Amans Maths Blogs (AMB). In this post, you will get the NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2.

As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the CBSE NCERT Syllabus. Thus, NCERT Solutions helps the students to solve the exercise questions as given in NCERT Books.

NCERT Solutions for class 12 is highly recommended by the experienced teacher for students who are going to appear in CBSE Class 12 and JEE Mains and Advanced and NEET level exams. Here You will get NCERT Solutions for Class 12 Maths Three Dimensional Geometry Exercise 11.2 of all questions given in NCERT textbooks of class 12 in details with step by step process. 

The PDF books of NCERT Solutions for Class 12 are the first step towards the learning and understanding the each sections of Maths, Physics, Chemistry, Biology as it all help in engineering medical entrance exams. To solve it, you just need to click on download links of NCERT solutions for class 12.

CBSE Class 12^th is an important school class in your life as you take some serious decision about your career. And out of all subjects, Maths is an important and core subjects. So CBSE NCERT Solutions for Class 12^th Maths is major role in your exam preparation as it has detailed chapter wise solutions for all exercise. This NCERT Solutions can be downloaded in PDF file. The downloading link is given at last.

NCERT Solutions for Class 12 Maths are not only the solutions of Maths exercise but it builds your foundation of other important subjects. Getting knowledge of depth concept of CBSE Class 12^th Maths like Algebra, Calculus, Trigonometry, Coordinate Geometry help you to understand the concept of Physics and Physical Chemistry.

NCERT Solutions for Class 12 Maths 3D Geometry 

Note: In this solution, the vector is represented by BOLD font. For example: a, b, OP, AB, i, j, k represent the vectors

. 

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 1.

Show that the three lines with direction cosines (12/13, −3/13, −4/13); (4/13, 12/13, 3/13); (3/13, −4/13, 12/13); are mutually perpendicular.

NCERT Solutions:

Let the line L₁ has direction cosines as l₁ = 12/13, m₁ = −3/13, n₁ = −4/13.

Let the line L₂ has direction cosines as l₂ = 4/13, m₂ = 12/13, n₂ = 3/13.

Let the line L₃ has direction cosines as l₃ = 3/13, m₃ = −4/13, n₃ = 12/13.

Let the angle between the lines L₁ and L₂ is θ₁. Then,

cos θ₁ = |l₁l₂ + m₁m₂ + n₁n₂|

= |(12/13 x 4/13) + (−3/13 x 12/13) + (−4/13 x 3/13)|

= |48/169 − 36/169 −12/169|

= |(48 − 36 − 12)/169| = 0 ⇒ θ₁ = 90^o 

Thus, the angle between the lines L₁ and L₂ is 90^o.

Let the angle between the lines L₂ and L₃ is θ₂. Then,

cos θ₂ = |l₂l₃ + m₂m₃ + n₂n₃| 

= |(4/13 x 3/13) + (12/13 x −4/13) + (3/13 x 12/13)|

= |12/169 − 48/169 + 36/169| = |(12 − 48 + 36)/169| = 0

⇒ θ₂ = 90^o 

Thus, the angle between the lines L₂ and L₃ is 90^o.

Let the angle between the lines L₁ and L₃ is θ₃. Then,

cos θ₃ = |l₁l₃ + m₁m₃ + n₁n₃| 

= |(12/13 x 3/13) + (−3/13 x −4/13) + (−4/13 x 12/13)|

= |36/169 + 12/169 − 48/169| = |(36 + 12 − 48)/169| = 0

⇒ θ₃ = 90^o 

Thus, the angle between the lines L₁ and L₃ is 90^o.

Therefore the given lines are mutually perpendicular.

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 2.

Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

NCERT Solutions:

Direction ratios of the line joining the points A(1, – 1, 2) and B(3, 4, – 2) are

a₁ = (3 – 1) = 2, b₁ = (4 + 1) = 5, c₁ = (–2 – 2) = –4

Direction ratios of the line joining the points C(0, 3, 2) and D(3, 5, 6) are

a₂ = (3 – 0) = 3, b₂ = (5 – 3) = 2, c₂ = (6 – 2) = 4.

Now, two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular if

a₁a₂ + b₁b₂ + c₁c₂ = 0

Since a₁a₂ + b₁b₂ + c₁c₂ = (2 x 3) + (5 x 2) + (–4 x 4) = 6 + 10 – 16 = 0, then the given lines AB and CD are perpendicular.

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 3.

Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5).

NCERT Solutions:

Direction ratios of the line joining the points A(4, 7, 8) and B(2, 3, 4) are

a₁ = (2 – 4) = –2, b₁ = (3 – 7) = –4, c₁ = (4 – 8) = –4

Direction ratios of the line joining the points C(– 1, – 2, 1) and D(1, 2, 5) are

a₂ = (1 + 1) = 2, b₂ = (2 + 2) = 4, c₂ = (5 – 1) = 4.

Now, two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are parallel if

Since , then the given lines AB and CD are parallel.

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 4.

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3i + 2j −2k.

NCERT Solutions:

Given that the line which passes through the point A(1, 2, 3). Then, a = i + 2j + 3k. And, the this line is parallel to the vector b = 3i + 2j −2k.

We know that the vector equation of the line passing through a and parallel to b is r = a + λb. where λ is any real number.

Thus, the required vector equation of the given line is r = (i + 2j + 3k) + λ(3i + 2j −2k).

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 5.

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i− j + 4k and is in the direction i + 2j − k.

NCERT Solutions:

Given that the line which passes through the point with position vector a = 2i− j + 4k and is in the direction b = i + 2j − k. 

We know that the vector equation of the line passing through a and parallel to b is r = a + λb. where λ is any real number.

Thus, the vector equation of the given line is r = (2i− j + 4k) + λ(i + 2j − k).

Since r is the position vector of any point P(x, y, z), then r = xi + yj + zk.

Thus,

xi + yj + zk = (2i− j + 4k) + λ(i + 2j − k)

⇒ xi + yj + zk = (2 + λ)i + (−1 + 2λ)j + (4 − λ)k

⇒ x = (2 + λ), y = (−1 + 2λ), z = (4 − λ)

Eliminating λ, we get the Cartesian equation of line as

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 6.

Find the Cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by .

NCERT Solutions:

Given that the line passes through the point P(– 2, 4, – 5) and parallel to the line given by .

Thus, the line passing through P(x₁, y₁, z₁) where x₁ = – 2, y₁ = 4, z₁ = – 5 and the direction ratios

a₁ = 3, b₁ = 5, c₁ = 6 is

⇒

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 7.

The cartesian equation of a line is . Write its vector form.

NCERT Solutions:

Comparing the given equation with the standard form 

We observe that x₁ = 5, y₁ = –4, z₁ = 6; a = 3, b = 7, c = 2.

Thus, the required line passes through the point P(5, –4, 6) ⇒ a = 5i – 4j + 6k and is parallel to the vector b = 3i + 7j +2k.

Let r be the position vector of any point on the line, then the vector equation of the line is given by

r = a + λb ⇒ r = (5i – 4j + 6k) + λ(3i + 7j +2k).

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 8.

Find the vector and the Cartesian equations of the lines that passes through the origin and (5, – 2, 3).

NCERT Solutions:

Given that the line passes through the origin P(0, 0, 0) ⇒ a = 0i + 0j + 0k and Q(5, – 2, 3) ⇒ b = 5i – 2j + 3k.

Let r be the position vector of any point on the line, then the vector equation of the line is given by

r = a + λ(b – a)

⇒ r = (0i + 0j + 0k) + λ(5i – 2j + 3k – 0i – 0j – 0k)

⇒ r = 0 + λ(5i – 2j + 3k).

Now, put r = xi + yj + zk.

xi + yj + zk = λ(5i – 2j + 3k)

⇒ xi + yj + zk = (5λ)i + (−2λ)j + (3λ)k

⇒ x = (5λ), y = (−2λ), z = (3λ)

Eliminating λ, we get the Cartesian equation of line as

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 9.

Find the vector and the Cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

NCERT Solutions:

Given that the line passes through the origin A(3, – 2, – 5) ⇒ a = 3i – 2j – 5k and B(3, – 2, 6) ⇒ b = 3i – 2j + 6k.

Let r be the position vector of any point on the line, then the vector equation of the line is given by

r = a + λ(b – a)

⇒ r = (3i – 2j – 5k) + λ(3i – 2j + 6k – 3i + 2j + 5k)

⇒ r = (3i – 2j – 5k) + λ(0i + 0j + 11k)

⇒ r = (3i – 2j – 5k) + λ(11k)

Now, put r = xi + yj + zk.

xi + yj + zk = (3i – 2j – 5k) + λ(11k)

⇒ xi + yj + zk = (3 + 0λ)i + (−2 + 0λ)j + (– 5 + 11λ)k

⇒ x = (3 + 0λ), y = (−2 + 0λ), z = (– 5 + 11λ)

Eliminating λ, we get the Cartesian equation of line as

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 10 (i).

Find the angle between the following pairs of lines (i) r = 2i − 5j + k+ λ(3i + 2j + 6k) and r = 7i − 6k + μ(i + 2j + 2k)

NCERT Solutions:

Given that b₁ = 3i + 2j + 6k and b₂ = i + 2j + 2k.

Let θ be the angle between the given lines. Then,

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 10 (ii)

Find the angle between the following pairs of lines (i) r = 3i + j − 2k+ λ(i − j − 2k) and r = 2i − j − 56k + μ(3i − 5j − 4k).

NCERT Solutions:

Given that b₁ = i − j − 2k and b₂ = 3i − 5j − 4k.

Let θ be the angle between the given lines. Then,

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 11(i)

Find the angle between the following pair of lines

(i)

NCERT Solutions:

Direction ratio of the given lines are a₁ = 2, b₁ = 5, c₁ = -3 and a₂ = -1, b₂ = 8, c₂ = 4.

Let θ be the angle between the given lines. Then,

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 12

Find the values of p so that the lines and

 at right angles.

NCERT Solutions:

Given equation can be written as

Then, the direction ratios of the given lines are a₁ = -3, b₁ = 2p/7, c₁ = 2 and a₂ = -3p/7, b₂ = 1, c₂ = -5.

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0.

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 13

Show that the lines and  are perpendicular to each other.

NCERT Solutions:

Given equation of the lines are and  .

Then, the direction ratios of the given lines are a₁ = 7, b₁ = -5, c₁ = 1 and a₂ = 1, b₂ = 2, c₂ = 3.

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0.

Now, a₁a₂ + b₁b₂ + c₁c₂ = (7)(1) + (-5)(2) + (1)(3) = 0. Thus, the given lines are perpendicular.

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 14

Find the shortest distance between the lines r = (i + 2j + k) + λ(i − j + k) and r = 2i − j − k + μ(2i + j + 2k).

NCERT Solutions:

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 15

Find the shortest distance between the lines

NCERT Solutions:

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 16

Find the shortest distance between the lines whose vector equations are r = (i + 2j + 3k) + λ(i − 3j + 2k) and r = 4i + 5j + 6k + μ(2i + 3j + k).

NCERT Solutions:

NCERT Solutions for Class 12 Maths 3D Geometry Exercise 11.2 Ques No 17

Find the shortest distance between the lines whose vector equations are r = (1− t)i + (t − 2)j + (3 − 2t)k and r = (s +1)i + (2s − 1)j − (2s +1)k.

NCERT Solutions:

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