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

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

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

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

**3. 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)**

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.

**4. Fill in the blanks:**

(i) The x-axis and y - axis taken together determine a plane known as XY - plane.

(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).

(iii) Coordinate planes divide the space into eight octants.

**5. 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) **

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

**6. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.**

**7. 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.**

(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.

**8. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).**

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} + 2z + 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.

**9. 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.**

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 + y^{2} + z^{2}) = 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.

**10. 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.**

(i) 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}) 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).

**11. 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.**

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.

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

Let the YZ plane divide 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.

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

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.

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

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).**15. Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex.**

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).

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

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

**17. 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.**

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

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

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).

**19. A point R with x-coordinate 4 lies on the line segment joining the pointsP (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**

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

**20. 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.**

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.

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