# 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*.

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**CBSE Class 12**^{th}Maths## 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 L_{1} has direction cosines as *l*_{1 }= 12/13, *m*_{1 }= −3/13, *n*_{1 }= −4/13.

Let the line L_{2} has direction cosines as *l*_{2 }= 4/13, *m*_{2 }= 12/13, *n*_{2 }= 3/13.

Let the line L_{3} has direction cosines as *l*_{3 }= 3/13, *m*_{3 }= −4/13, *n*_{3 }= 12/13.

Let the angle between the lines L_{1} and L_{2} is θ_{1}. Then,

cos θ_{1} = |*l*_{1}*l*_{2 }+ *m*_{1}*m*_{2 }+ *n*_{1}*n*_{2}|

= |(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 ⇒ θ_{1} = 90^{o }

Thus, the angle between the lines L_{1} and L_{2} is 90^{o}.

Let the angle between the lines L_{2} and L_{3} is θ_{2}. Then,

cos θ_{2} = |*l*_{2}*l*_{3 }+ *m*_{2}*m*_{3 }+ *n*_{2}*n*_{3}|

= |(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

⇒ θ_{2} = 90^{o }

Thus, the angle between the lines L_{2} and L_{3} is 90^{o}.

Let the angle between the lines L_{1} and L_{3} is θ_{3}. Then,

cos θ_{3} = |*l*_{1}*l*_{3 }+ *m*_{1}*m*_{3 }+ *n*_{1}*n*_{3}|

= |(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

⇒ θ_{3} = 90^{o }

Thus, the angle between the lines L_{1} and L_{3} 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

a_{1} = (3 – 1) = 2, b_{1} = (4 + 1) = 5, c_{1} = (–2 – 2) = –4

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

a_{2} = (3 – 0) = 3, b_{2} = (5 – 3) = 2, c_{2} = (6 – 2) = 4.

Now, two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if

a_{1}a_{2 }+ b_{1}b_{2 }+ c_{1}c_{2 }= 0

Since a_{1}a_{2 }+ b_{1}b_{2 }+ c_{1}c_{2 }= (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

a_{1} = (2 – 4) = –2, b_{1} = (3 – 7) = –4, c_{1} = (4 – 8) = –4

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

a_{2} = (1 + 1) = 2, b_{2} = (2 + 2) = 4, c_{2} = (5 – 1) = 4.

Now, two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} 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 3**i** + 2**j** −2**k**.

**NCERT Solutions:**

Given that the line which passes through the point A(1, 2, 3). Then, **a** = **i** + 2**j** + 3**k**. And, the this line is parallel to the vector **b** = 3**i** + 2**j** −2**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 required vector equation of the given line is **r** = (**i** + 2**j** + 3**k) + 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 2**i**− **j** + 4**k** and is in the direction **i** + 2**j** − **k**.

**NCERT Solutions:**

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

Since r is the position vector of any point P(x, y, z), then **r** = x**i** + y**j** + z**k**.

Thus,

x**i** + y**j** + z**k** = (2**i**− **j** + 4**k****) + **λ**(i + 2j − k) **

** ⇒ **x

**i**+ y

**j**+ z

**k**= (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(x_{1}, y_{1}, z_{1}) where x_{1} = – 2, y_{1} = 4, z_{1} = – 5 and the direction ratios

a_{1} = 3, b_{1} = 5, c_{1} = 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 x_{1} = 5, y_{1} = –4, z_{1} = 6; a = 3, b = 7, c = 2.

Thus, the required line passes through the point P(5, –4, 6) ** ⇒ a** = 5

**i**– 4

**j**+ 6

**k**and is parallel to the vector

**b**= 3

**i**+ 7

**j**+2

**k**.

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** ** ⇒ r** = (5

**i**– 4

**j**+ 6

**k**) + λ(3

**i**+ 7

**j**+2

**k**).

**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** = 0

**i**+ 0

**j**+ 0

**k**and Q(5, – 2, 3)

**= 5**

**⇒**b**i**– 2

**j**+ 3

**k**.

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 **= (

**) + λ**

**0i + 0j + 0k**

**(5i – 2j + 3k – 0i – 0j –**

**0k)**** ⇒ r **=

**0**+ λ(

**).**

**5i – 2j + 3k**Now, put **r** = x**i** + y**j** + z**k**.

x**i** + y**j** + z**k** = λ(** 5i – 2j + 3k**)

** ⇒ **x

**i**+ y

**j**+ z

**k**= (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** = 3

**i**– 2

**j**– 5

**k**and B(3, – 2, 6)

**= 3**

**⇒**b**i**– 2

**j**+ 6

**k**.

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 **= (3

**i**– 2

**j**– 5

**k**) + λ

**(3i – 2j + 6k – 3i + 2j + 5k**

**)**** ⇒ r **= (3

**i**– 2

**j**– 5

**k**) + λ

**(0i + 0j + 11k**

**)**** ⇒ r **= (3

**i**– 2

**j**– 5

**k**) + λ

**(11k**

**)**Now, put **r** = x**i** + y**j** + z**k**.

x**i** + y**j** + z**k** = (3**i** – 2**j** – 5**k**) + λ**(11k****)**

** ⇒ **x

**i**+ y

**j**+ z

**k**= (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** = 2**i** − 5**j** + **k**+ λ(3**i** + 2**j** + 6**k**) and **r** = 7**i** − 6**k** + μ(**i** + 2**j** + 2**k**)

**NCERT Solutions:**

Given that **b**_{1} = 3**i** + 2**j** + 6**k **and **b**_{2} = **i** + 2**j** + 2**k**.

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** = 3**i** + **j** − 2**k**+ λ(**i** − **j** − 2**k**) and **r** = 2**i** − **j **− 56**k** + μ(3**i** − 5**j** − 4**k**).

**NCERT Solutions:**

Given that **b**_{1} = **i** − **j** − 2**k**** **and **b**_{2} = 3**i** − 5**j** − 4**k**.

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 a_{1} = 2, b_{1} = 5, c_{1} = -3 and a_{2} = -1, b_{2} = 8, c_{2} = 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 a_{1} = -3, b_{1} = 2p/7, c_{1} = 2 and a_{2} = -3p/7, b_{2} = 1, c_{2} = -5.

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular, if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 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 a_{1} = 7, b_{1} = -5, c_{1} = 1 and a_{2} = 1, b_{2} = 2, c_{2} = 3.

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular, if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0.

Now, a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (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** + 2**j** + **k**) + λ(**i** − **j** + **k**) and **r** = 2**i** − **j** − **k** + μ(2**i** + **j** + 2**k**).

**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** + 2**j** + 3**k**) + λ(**i** − 3**j** + 2**k**) and **r** = 4**i** + 5**j** + 6**k** + μ(2**i** + 3**j** + **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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