- 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,
√(x2 + y2 + z2)
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
PA2 = PB2
It means that,
(x-1)2 + (y-2)2+(z-3)2 = (x-3)2+(y-2)2+(z+1)2
⇒ x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 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,
AC2 = AB2 + BC2 …..(4)
Now, substitute (1),(2), and (3) in (4), we get:
62 = ( 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 = √(32 + 62 + 22)
= √(9+36+4)
= √49
= 7
Therefore, the length of a diagonal of a parallelopiped is 7 units.