- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 11

- Subject
- Mathmatics

**Coordinates of a points**

Having chosen a fixed 3D coordinate system in the space, we can associate a given point in the space using three coordinates (x,y,z) and conversely, given a triplet of three numbers (x, y, z), how, we locate a point in the space.

Locate the point “x” on the X-axis. From the point x, moving parallel to the Y-axis, locate the point “y”. Similarly, from the determined point, moving parallel to the Z-axis, locate the point “z”. This is the final coordinate point in the three-dimensional plane, which we are looking for.

when the co-ordinates of the point are given, then we have to draw three planes parallel to XY, YZ and ZX plane meeting the three axes in points A, B and C as shown in the figure. Let OA = x , OB = y and OC = z. Then the coordinates of the point are given as (x,y,z).

** **The above the co-ordinates of P are given by (x, y, z). The coordinates of the origin O is (0, 0, 0) Also the coordinates of the point A is given by (x, 0, 0)as A lies completely on the x-axis. Similarly, the coordinates of any point on y-axis is given as (0, y, 0) and on the z-axis, the coordinates are given as (0, 0, z). Also the coordinates of any point in three planes XY, YZ and ZX will be (x,y,0), (0,y,z) and (x,0,z) respectively.

**Example: The coordinates of a point are (3, -2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.**

**Solution:**

Given:

Point (3, -2, 5)

The Absolute value of any point(x, y, z) is given by,

√(x^{2} + y^{2} + z^{2})

We need to make sure that absolute value to be the same for all points.

So let the point A(3, -2, 5)

*Remaining 7 points are:*

*Point B(3, 2, 5) (By changing the sign of y coordinate)*

*Point C(-3, -2, 5) (By changing the sign of x coordinate)*

*Point D(3, -2, -5) (By changing the sign of z coordinate)*

*Point E(-3, 2, 5) (By changing the sign of x and y coordinate)*

*Point F(3, 2, -5) (By changing the sign of y and z coordinate)*

*Point G(-3, -2, -5) (By changing the sign of x and z coordinate)*

*Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)*

**Example: Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.**

**Solution:**

Given:

The points (2, 4, 5) and (3, 5, 4)

By using the section formula,

We know X coordinate is always 0 on yz-plane

So, let Point C(0, y, z), and let C divide AB in ratio k: 1

Then, m = k and n = 1

A(2, 4, 5) and B(3, 5, 4)

The coordinates of C are:

On comparing we get,

[3k + 2] / [k + 1] = 0

3k + 2 = 0(k + 1)

3k + 2 = 0

3k = – 2

k = -2/3

∴We can say that, C divides AB externally in ratio 2: 3

**Question 1:**

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

**Solution:**

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

Thus, we can say that, PA = PB

Take square on both the sides, we get

PA^{2 }= PB^{2}

It means that,

(x-1)^{2} + (y-2)^{2}+(z-3)^{2} = (x-3)^{2}+(y-2)^{2}+(z+1)^{2}

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

Now, simplify the above equation, we get:

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

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

Hence, the required equation for the set of points is x – 2z = 0.

**Question 2:**

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.

**Solution:**

Assume that the 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 using the section formula, we can write it as:

(5, 4, -6) = [ (k(9)+3)/(k+1), (k(8)+2)/(k+1), (k(-10)-4)/(k+1)]

⇒(9k+3)/(k+1) = 5

Now, bring the L.H.S denominator to the R.H.S and multiply it

⇒9k+3 = 5k+5

Now, simplify the equation to find the value of k.

⇒4k= 2

⇒k = 2/4

⇒k=½

Therefore, the value of k is ½.

Hence, the point Q divides PR in the ratio of 1:2

**Question 3:**

Prove that the points: (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right-angled triangle

**Solution:**

Let the given points be A = (0, 7, 10), B = (–1, 6, 6), and C = (–4, 9, 6)

Now, find the distance between the points

**Finding for AB:**

AB = √ [(-1-0)^{2} + (6-7)^{2 }+(6-10)^{2}]

AB = √ [(-1)^{2} + (-1)^{2 }+(-4)^{2}]

AB = √(1+1+16)

AB = √18

AB = 3√2 …. (1)

**Finding for BC:**

BC= √ [(-4+1)^{2} + (9-6)^{2 }+(6-6)^{2}]

BC = √ [(-3)^{2} + (3)^{2 }+(-0)^{2}]

BC = √(9+9)

BC = √18

BC = 3√2 …..(2)

**Finding for CA:**

CA= √ [(0+4)^{2} + (7-9)^{2 }+(10-6)^{2}]

CA = √ [(4)^{2} + (-2)^{2 }+(4)^{2}]

CA = √(16+4+16)

CA = √36

CA = 6 …..(3)

Now, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2 }…..(4)

Now, substitute (1),(2), and (3) in (4), we get:

6^{2 }= ( 3√2)^{2} + ( 3√2)^{2}

36 = 18+18

36 = 36

The given points obey the condition of Pythagoras Theorem.

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

**Question 4:**

Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.

(a)8 (b)7 (c)6 (d) None of the above

**Solution:**

A correct answer is an **option (A)**

**Explanation:**

Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is z-coordinate of P, i.e., 8 units

Hence, the correct answer is an option (a)

**Question 5:**

If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then find the length of edges of a parallelopiped and length of the diagonal

**Solution:**

Let A = (2, 3, 5), B = (5, 9, 7)

To find the length of the edges of a parallelopiped = 5 – 2, 9 – 3, 7 – 5

It means that 3, 6, 2.

Now, to find the length of a diagonal = √(3^{2} + 6^{2} + 2^{2})

= √(9+36+4)

= √49

= 7

Therefore, the length of a diagonal of a parallelopiped is 7 units.