# NCERT Solutions Class 11 Chapter12- Introduction to Three Dimensional Geometry Class 11

**Exercise 12.1 : Solutions of Questions on Page Number : 271**

** Q1 :A point is on the x-axis. What are its y-coordinates and z-coordinates?**

**Answer :**

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.

**Q2 :A point is in the XZ-plane. What can you say about its y-coordinate?**

**Answer :
**If a point is in the XZ plane, then its y-coordinate is zero.

**Q3 :Name the octants in which the following points lie:**

**(1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5),**

**(-3, -1, 6), (2, -4, -7)**

**Answer :
**The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, -5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (-3, -1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

**Q4 :Fill in the blanks:**

**Answer :**

**(i)** The x-axis and y-axis taken together determine a plane known as **
(ii)** The coordinates of points in the XY-plane are of the form

**(iii)**Coordinate planes divide the space into octants

**Exercise 12.2 : Solutions of Questions on Page Number : 273**

** Q1 :Find the distance between the following pairs of points:**

**(i) (2, 3, 5) and (4, 3, 1) (ii) (-3, 7, 2) and (2, 4, -1)**

**(iii) (-1, 3, -4) and (1, -3, 4) (iv) (2, -1, 3) and (-2, 1, 3)**

**Answer :**

The distance between points P(x_{1}, y_{1}, z_{1}) and P(x_{2}, y_{2}, z_{2}) is given by

**(i)** Distance between points (2, 3, 5) and (4, 3, 1)

**(ii)** Distance between points (-3, 7, 2) and (2, 4, -1)

**(iii)** Distance between points (-1, 3, -4) and (1, -3, 4)

**(iv)** Distance between points (2, -1, 3) and (-2, 1, 3)

**Q2 :Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.**

**Answer :**

Let points (-, 3, 5), (1, 2, 3), and (7, 0,-1) be denoted by P, Q, and R respectively.

Points P, Q, and R are collinear if they lie on a line.

Here, PQ + QR = PR

Hence, points P(-2, 3, 5), Q(1, 2, 3), and R(7, 0, -1) are collinear.

**Q3 :Verify the following:**

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

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

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

**Answer :**

**(i)** Let points (0, 7, -10), (1, 6, -6), and (4, 9, -6) be denoted by A, B, and C respectively.

Here, AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

**(ii)** Let (0, 7, 10), (-1, 6, 6), and (-4, 9, 6) be denoted by A, B, and C respectively.

Therefore, by Pythagoras theorem, ABC is a right triangle.

Hence, the given points are the vertices of a right-angled triangle.

**(iii)** Let (-1, 2, 1), (1, -2, 5), (4, -7, 8), and (2, -3, 4) be denoted by A, B, C, and D respectively.

Here, AB = CD = 6, BC = AD =

Hence, the opposite sides of quadrilateral ABCD , whose vertices are taken in order, are equal.

Therefore, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

**Q4 :Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).**

**Answer :**

Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2,

1).

Accordingly, PA = PB

⇒ x^{2} – 2x + 1 + y^{2} – 4y + 4 + z^{2} – 6z + 9 = x^{2} – 6x + 9 + y^{2} – 4y + 4 + z^{2} + 2^{z} + 1

⇒ -2x -4y – 6z + 14 = -6x – 4y + 2z + 14

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

Thus, the required equation is x – 2z = 0.

**Q5 :Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10.**

**Answer :**

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

The coordinates of points A and B are (4, 0, 0) and (-4, 0, 0) respectively.

It is given that PA + PB = 10.

On squaring both sides, we obtain

On squaring both sides again, we obtain

25 (x^{2} + 8x + 16 + y2 + z2) = 625 + 16x^{2} + 200x

⇒ 25x^{2} + 200x + 400 + 25y^{2} + 25z^{2} = 625 + 16x^{2} + 200x

⇒ 9x^{2} + 25y^{2} + 25z^{2} – 225 = 0

Thus, the required equation is 9x^{2} + 25y^{2} + 25z^{2} – 225 = 0.

**Exercise 12.3 : Solutions of Questions on Page Number : 277**

** Q1 :Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally.**

**Answer :**

**(i)** The coordinates of point R that divides the line segment joining points P (x_{1}, y_{1}, z1) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n are

Let R (x, y, z) be the point that divides the line segment joining points(-2, 3, 5) and (1, -4, 6) internally in the ratio 2:3

Thus, the coordinates of the required point are

**(ii)** The coordinates of point R that divides the line segment joining points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) externally in the ratio m: n are

Let R (x, y, z) be the point that divides the line segment joining points(-2, 3, 5) and (1, -4, 6) externally in the ratio 2:3

Thus, the coordinates of the required point are (-8, 17, 3).

**Q2 :Given that P (3, 2, -4), Q (5, 4, -6) and R (9, 8, -10) are collinear. Find the ratio in which Q divides PR.**

**Answer :**

Let point Q (5, 4, -6) divide the line segment joining points P (3, 2, -4) and R (9, 8, -10) in the ratio k:1.

Therefore, by section formula,

Thus, point Q divides PR in the ratio 1:2.

**Q3 :Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).**

**Answer :
**Let the YZ planedivide the line segment joining points (-2, 4, 7) and (3, -5, 8) in the ratio k:1.

Hence, by section formula, the coordinates of point of intersection are given by

On the YZ plane, the x-coordinate of any point is zero.

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2:3.

**Q4 :Using section formula, show that the points A (2, -3, 4), B (-1, 2, 1) and are collinear.**

**Answer :
**The given points are A (2, -3, 4), B (-1, 2, 1), and

Let P be a point that divides AB in the ratio k:1.

Hence, by section formula, the coordinates of P are given by

Now, we find the value of k at which point P coincides with point C.

By taking , we obtain k = 2.

For k = 2, the coordinates of point P are

i.e. is a point that divides AB externally in the ratio 2:1 and is the same as point P.

Hence, points A, B, and C are collinear.

**Q5 :Find the coordinates of the points which trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).**

**Answer :
**Let A and B be the points that trisect the line segment joining points P (4, 2, -6) and Q (10, -16, 6)

Point A divides PQ in the ratio 1:2. Therefore, by section formula, the coordinates of point A are given by

Point B divides PQ in the ratio 2:1. Therefore, by section formula, the coordinates of point B are given by

Thus, (6, -4, -2) and (8, -10, 2) are the points that trisect the line segment joining points P (4, 2, -6) and Q (10, -16, 6).

**Exercise Miscellaneous : Solutions of Questions on Page Number : 278**

** Q1 :Three vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4) and C (-1, 1, 2). Find the coordinates of the fourth vertex.**

**Answer** :

The three vertices of a parallelogram ABCD are given as A (3, -1, 2), B (1, 2, -4), and C (-1, 1, 2). Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram ABCD, AC and BD bisect each other.

∴ Mid-point of AC = Mid-point of BD

⇒ x = 1, y = -2, and z = 8

Thus, the coordinates of the fourth vertex are (1, -2, 8).

**Q2 :Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).**

**Answer :
**Let AD, BE, and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid-point of BC.

∴ Coordinates of point D = = (3, 2, 0)

Thus, the lengths of the medians of ΔABC are

**Q3 :If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c), then find the values of a, b and c.**

**Answer :
**

It is known that the coordinates of the centroid of the triangle, whose vertices are (x

_{1}, y

_{1}, z

_{1}), (x

_{2}, y

_{2}, z

_{2}) and (x

_{3}, y

_{3}, z

_{3}), are

Therefore, coordinates of the centroid of ΔPQR

It is given that origin is the centroid of ΔPQR.

Thus, the respective values of a, b, and c are

**Q4 :Find the coordinates of a point on y-axis which are at a distance of from the point P (3, -2, 5).**

**Answer :
**If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.

Let A (0, b, 0) be the point on the y-axis at a distance of from point P (3, -2, 5).

Accordingly,

Thus, the coordinates of the required points are (0, 2, 0) and (0, -6, 0).

**Q5 :A point R with x-coordinate 4 lies on the line segment joining the points P (2, -3, 4) and Q (8, 0, 10). Find the coordinates of the point R.**

**[Hint suppose R divides PQ in the ratio k: 1. The coordinates of the point R are given by **

**Answer :**

The coordinates of points P and Q are given as P (2, -3, 4) and Q (8, 0, 10).

Let R divide line segment PQ in the ratio k:1.

Hence, by section formula, the coordinates of point R are given by

It is given that the x-coordinate of point R is 4.

Therefore, the coordinates of point R are

**Q6 :If A and B be the points (3, 4, 5) and (-1, 3, -7), respectively, find the equation of the set of points P such that PA ^{2} + PB^{2} = k^{2}, where k is a constant.**

**Answer :**

The coordinates of points A and B are given as (3, 4, 5) and (-1, 3, -7) respectively.

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

On using distance formula, we obtain

Now, if PA

^{2}+ PB

^{2}= k

^{2}, then

Thus, the required equation is