NCERT Solution for class 11 maths Chapter 12 Introduction to Three Dimensional Geometry is one of the most important topic to practice. The NCERT solutions for class 11 maths gives students a basic idea of the types of questions that are usually asked in the examination. NCERT questions are important for the students as practicing these questions are helpful for them to excel in the examination as well in in their competitive examination. Students can also download the NCERT Solution Class 11 Maths Introduction to Three Dimensional Geometry pdf by following the link given. Get more NCERT solutions by visiting our site.

### NCERT Solutions Class 11 Maths Chapter 12 Exercises

- NCERT Solutions Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.1
- NCERT Solutions Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2
- NCERT Solutions Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

**Exercise 12.1**

**For any given point, the sign of its coordinates determines the octant in which it will lie. **

**Now, from ****the following table it can be easily determined in which coordinates the point lies.**

** **

**Q.1: A point is lying on y – axis. What are its ‘x‘ coordinates and ‘z’ coordinates?**

**Sol.**

**‘x’ coordinates and ‘z’ coordinates are zero if any point lies on y –axis.**

**Q.2: A point is lying on YZ – plane. What are its ‘x‘ coordinates?**

**Sol.**

**If any point lies in the YZ plane then its x-coordinates are zero.**

**Q.3: In which of the octant the following points lie:**

**(2, 3, 4), (8, -1, -1), (-4, 9, -8), (-1, -2, -3), (4, -5, 6), (7, -1, -4), (-3, -5, 1), (0, 0, -3)**

**Sol.**

**(2, 3, 4)** are **all** **positive**. Therefore, this point is lying in **octant (I).**

**point** **(8, -1, -1)** are **positive**, **negative** and **negative**. Therefore, this point is lying in **octant (VII).**

**point** **(-4, 9, -8)** are **negative**, **positive** and **negative**. Therefore, this point is lying in **octant (VI).**

**point** **(-1, -2, -3)** are **all negative**. Therefore, this point is lying in **octant (VII).**

**point (4, -5, 6) **are **positive**, **negative** and **positive**. Therefore, this point is lying in **octant (IV).**

**point (7, -1, -4)** are **positive**, **negative** and **negative**. Therefore, this point is lying in **octant (VII).**

**point (-3, -5, 1)** are negative, negative and positive. Therefore, this point is lying in **octant (III).**

**point (-1, 2, 2)** are **negative**, **positive** and **positive**. Therefore, this point is lying in **octant (II).**

** Q.4: Answer the following questions:**

**(i). What is the name of a plane determined by the Z-axis and the Y-axis when taken together?**

**(ii). What is the general form of coordinates of points in the XZ-plane?**

**(iii). Coordinate plane divides the space into how many octants?**

**Sol.**

**(i). The plane determined by the Z-axis and the Y-axis when taken together is known as YZ–plane.**

**(ii). The general form of coordinates of points in the XZ-plane is ( x, 0 , z ).**

**(iii). The coordinate plane divides the space into 8 octants.**

**Exercise: 12.2**

** ****Distance Formula: **

**The distance between point A(x _{1}, y_{1}, z_{1}) and point B(x_{2}, y_{2}, z_{2}) is given by:**

**AB = (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√**

** **

**Q.1: Find the distance between two points whose coordinates are given below:**

**(i). (2, 8, 9) and (4, 5, 8)**

**(ii). (-3, 4, 5) and (2, 6, -1)**

**(iii). (-6, -4, 1) and (5, -2, 6)**

**(iv). (-1, 9, 8) and (6, 5, -3)**

**Sol.**

**The distance between point A (x _{1}, y_{1}, z_{1}) and point B (x_{2}, y_{2}, z_{2}) is given by:**

**AB = (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√**

**(i). The distance between point A (2, 8, 9) and point B (4, 5, 8) is:**

**Therefore, The distance between point A (2, 8, 9) and point B (4, 5, 8) is 14−−√ units**

**(ii).** **The distance between point A (****-3, 4, 5****) and point B (****2, 6, -1****) is:**

**Therefore, The distance between point A ****(****-3, 4, 5) and point B (2, 6, -1) is 65−−√ units**

** **

**(iii). The distance between point A ****(-6, -4, 1****) and point B (****5, -2, 6****) is:**

**Therefore, The distance between point A ****(****-6, -4, 1) and point B (5, -2, 6) is 56–√ units**

** **

**(iv). The distance between point A ****(-1, 9, 8****) and point B (****6, 5, -3****) is:**

**Therefore, The distance between point A ****(****-1, 9, 8) and point B (6, 5, -3) is 186−−−√ units**

** **

**Q.2: Show that the points (7, 0, -1), (-2, 3, 5) and (1, 2, 3) are collinear.**

**Sol.**

Let, the coordinates of **points A, B and C** are **(7, 0, -1), (-2, 3, 5) and (1, 2, 3)** respectively.

**Points A, B and C are collinear if they lie on the same line.**

Now,

**Therefore, AB = 314−−√ units**

Now,

**Therefore, BC = 14−−√ units**

And,

**Therefore, AC = 214−−√ units**

Since, **AC + BC = 214−−√+14−−√ = AB**

**Therefore, the points (7, 0, -1), (-2, 3, 5) and (1, 2, 3) are collinear.**

** **

**Q.3: Prove the following statements:**

**(i). (-4, 9, 6), (0, 7, 10) and (-1, 6, 6) are the vertices of a right angled triangle.**

**(ii). (4, 9, -6), (0, 7, -10) and (1, 6, -6) are the vertices of an isosceles triangle.**

**(****iii). (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram.**

** **

**Sol.**

**The distance between points A (x _{1}, y_{1}, z_{1}) and B (x_{2}, y_{2}, z_{2}) is given by:**

**AB = (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√**

** **

**(i).** Let, the **coordinates** of points **A, B and C** are **(-4, 9, 6), (0, 7, 10) and (-1, 6, 6)** respectively.

Now,

**Therefore, AB = 6 units.**

Now,

**Therefore, BC = 32–√ units.**

And,

**Therefore, AC = 32–√units.**

Now, AB^{2} = 36, BC^{2 }= 18 and AC^{2} = 18

Since, **AB ^{2} = AC^{2} + BC^{2}**. Therefore, by

**Pythagoras**

**theorem, ABC is a right angled triangle.**

**Hence, the given points (-4, 9, 6), (0, 7, 10) and (-1, 6, 6) are the vertices of a right angled triangle.**

**(ii).** Let, the **coordinates** of points **A, B and C** are **(4, 9, -6), (0, 7, -10) and (1, 6, -6) **respectively.

Now,

**Therefore, AB = 6 units.**

Now,

**Therefore, BC = 32–√ units.**

And,

**Therefore, AC = 32–√ units.**

**i.e. AC = BC ****≠**** AB**

Now, a triangle is said to be an **isosceles** triangle if **two sides** of a triangle are **equal **to each other.

Since, **AC = BC**. **Therefore triangle ABC is an isosceles triangle.**

**Hence, the given points (4, 9, -6), (0, 7, -10) and (1, 6, -6) are the vertices of an isosceles triangle.**

**(iii).** Let, the **coordinates** of **points A, B, C and D** are **(1, -2, 5), (4, -7, 8), (2, -3, 4) and (-1, 2, 1) respectively**.

Now,

**Therefore, AB = 43−−√ units.**

Now,

**Therefore, BC = 6 units.**

Now,

**Therefore, CD = 43−−√ units.**

And,

**Therefore, AD = 6 units**.

Since**, BC = AD = 6 units** and **AB = CD = 43−−√ units**

Since, it is proved that in **quadrilateral** **ABCD** **opposite** **sides** are **equal**. Therefore,** ABCD is a parallelogram.**

**Hence, the given points (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram.**

** **

**Q.4: Find the equation of the set of points P which are equidistant from point A (3, 2, 1) and point B (-1, 2, 3).**

**Sol.**

Let, **(x, y, z)** be the **coordinates** of **point P**, which is **equidistant** from **point A (3, 2, 1)** and **point B (-1, 2, 3).**

Now, according to the given condition:

**AP ^{2} = BP^{2}**

**Therefore, the required equation is: 2x – z = 0**

** **

**Q.5: The sum of distance of point P from point A (3, 0, 0) and point B (-3, 0, 0) is equal to 12 units. Find the equation of the set of points.**

**Sol.**

**Let, (x, y, z)** be the **coordinates** of **point P**.

Now, according to the given condition:

**AP + BP = 12**

Now, **on squaring both the side**, we will get:

Again, **on squaring both the side**, we will get:

**Therefore, the required equation is: 3x2+4y2+4z2−108=0**

** **

**Section Formula:**

The **coordinates** of **point P**, which divides the line segment joining two points **A (a _{1}, b_{1}, c_{1})** and

**B (a**

_{2}, b_{2}, c_{2})**internally**in the ratio

**m and n**are:

The **coordinates** of **point Q**, which divides the line segment joining two points A (a_{1}, b_{1}, c_{1}) and B (a_{2}, b_{2}, c_{2}) **Externally** in the ratio **m and n** are:

** **

**Exercise 12.3**

** ****Q.1: Find the coordinates of the point which divides the line segment joining the points (-2, 1, 0) and (1, 3, 6) internally in the ratio of 1:4 and externally in the ratio of 1:4.**

**Sol:**

**Since, the coordinates of point P, which divides the line segment joining two points A (a _{1}, b_{1}, c_{1}) and B (a_{2}, b_{2}, c_{2}) internally in the ratio m and n are:**

Therefore, the **coordinates** of **point P**, which divides the line segment joining two points **A(-2, 1, 0) **and **B (1, 3, 6) internally** in the ratio **1 and 4** are:

**(x, y, z) =**

**(x, y, z) = **

**Therefore, the coordinates of point P are: **

**Now, The coordinates of point Q, which divides the line segment joining two points A (a _{1}, b_{1}, c_{1}) and B (a_{2}, b_{2}, c_{2}) Externally in the ratio m and n are:**

Therefore, the **coordinates** of **point Q**, which divides the line segment joining two points **A(-2, 1, 0) **and **B (1, 3, 6) externally **in the ratio **1 and 4** are:

**(x, y, z) =**

**(x, y, z) =**

**Therefore, the coordinates of point Q are: [−3,13,−2]**

** **** **

**Q.2: Given that the points A (3, 2, -4) B (5, 4, -6) and C (9, 8, -10), are collinear. Find the ratio in which line AC is divided by B.**

**Sol.**

Since, **point A, point B and point C** are **collinear. **Therefore, let us assume that **point B** **divides** the line segment joining **point A** and **point C **in the **ratio k : 1.**

**Hence, by Section Formula, **coordinates of **point B** are:

And, the **coordinates** of **point B** are: **(5, 4, -6) [Given]**

Therefore, on comparing it with **equation (1)** we will get:

Therefore, k =

**Hence, point B divides AB in the ratio of 1 : 2**

** **

**Q.3: Find the ratio in which XZ – plane divides the line segment AB formed by joining the points (3, -5, 8) and (-2, 4, 7).**

**Sol.**

**Let the XZ plane divides the line segment PQ in the ratio k : 1**

**Hence, by Section Formula the coordinates of point of intersection are:**

Now, in XZ – plane, the y – coordinate of any point is zero.

**Therefore, k=54**

**Hence, the XZ plane divides the line segment formed by the joining of given points in the ratio of 2 : 3.**

**Q.4: By using the section formula; Show that the points P (-1, 2, 1), Q (0, 13, 2) and R (2, -3, 4) are collinear.**

**Sol. **

**Let, point A divides line PQ in the ratio k : 1**

**Hence, by Section Formula the coordinates of point A are:**

Therefore, the **coordinates** **of point A** are:

Now, the value of **‘k’** for which **point A** coincides with **point R**:

Therefore, **k =**

Now, for **k =** **point A** are:

**Therefore**, **the** **coordinates** of **point A** are: **(2, -3, 4)**

Hence, the coordinates of **point A** coincides with the coordinates of point R.

**R (2, -3, 4)** is a point that divides **PQ** **externally** and is same as **point A.**

**Hence, the points P, Q and R are colinear.**

**Q.5: The line segment joining points A (5, 3, -6) and B (9, 15, 7) is trisected by the points P and Q, Find the coordinates of points P and Q.**

**Sol.**** **

**Point P divides** the **line segment AB** in the **ratio 1 : 2** and **point Q** **divides** the **line segment AB** in the **ratio 2 : 1.**

Therefore, by **section formula** the coordinates of **point P** and **point Q** are:

**For Point P:**

**Therefore, the coordinates of point P are: [193,7,−53]**

**For Point Q:**

**Therefore, the coordinates of point Q are: [233,11,83]**