Question 1: 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.
Solution:
ABCD is a parallelogram, with vertices A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2) and D (x, y, z).
Using the property:
The diagonals of a parallelogram bisect each other,
Midpoint of AC = Midpoint of BD = Point O
Now, by using Midpoint section formula
Coordinates of O for the line segment joining (x1,y1,z1) and (x2,y2,z2) =
(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}) So, Coordinates of O for the line segment joining AC =
(\frac{3+(-1)}{2}, \frac{-1+1}{2}, \frac{2+2}{2}) =
(\frac{2}{2}, 0, \frac{4}{2}) = (1, 0, 2) ............................(1)
and, Coordinates of O for the line segment joining BD =
(\frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2}) .............(2)Using the Eq(1) and Eq(2), we get
\frac{1+x}{2} = 1x = 1
\frac{2+y}{2} = 0y = -2
\frac{z-4}{2} = 2z = 8
Hence, the coordinates of the fourth vertex is D (1, -2, 8).
Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0), and (6, 0, 0).
Solution:
The vertices of the triangle are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
So, let the medians be AD, BE and CF corresponding to the vertices A, B and C respectively.
D, E and F are the midpoints of the sides BC, AC and AB respectively.
Coordinates of mid-point for the line segment joining (x1,y1,z1) and (x2,y2,z2) =
(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}) So, Coordinates of D for the line segment joining BC =
(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}) Coordinates of D = (3, 2, 0)
and, Coordinates of E for the line segment joining AC =
(\frac{6+0}{2}, \frac{0+0}{2}, \frac{0+6}{2}) Coordinates of E = (3, 0, 3)
and, Coordinates of F for the line segment joining AB =
(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}) Coordinates of F = (0, 2, 3)
By using the distance formula for two points, P(x1,y1,z1) and Q(x2,y2,z2)
PQ =
\mathbf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}} So, AD =
\sqrt{(0-3)^2+(0-2)^2+(6-0)^2} AD =
\sqrt{9+4+36} = 7 and, BE =
\sqrt{(0-3)^2+(4-0)^2+(0-3)^2} BE =
\sqrt{9+16+9} = \sqrt{34} and, CF =
\sqrt{(6-0)^2+(0-2)^2+(0-3)^2} CF =
\sqrt{36+4+9} = 7Hence, the lengths of the medians are 7, √34 and 7.
Question 3: 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.
Solution:
The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).
Coordinates of centroid(0, 0, 0) of the triangle having vertices (x1,y1,z1), (x2,y2,z2) and (x3,y3,z3) =
(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}) (0, 0, 0) =
(\frac{2a-4+8}{3}, \frac{2+3b+14}{3}, \frac{6-10+2c}{3}) (0, 0, 0) =
(\frac{2a+4}{3}, \frac{3b+16}{3}, \frac{2c-4}{3}) So,
\frac{2a+4}{3} = 0a = -2
and,
\frac{3b+16}{3} = 0b =
\mathbf{\frac{-16}{3}} and,
\frac{2c-4}{3} = 0c = 2
Hence, the values of a, b and c are a = -2, b =
\mathbf{\frac{-16}{3}} and c = 2
Question 4: Find the coordinates of a point on y-axis which are at a distance of 5√2 from the point P (3, –2, 5).
Solution:
Point on y-axis = A (0, y, 0).
Distance between the points A (0, y, 0) and P (3, -2, 5) = 5√2.
Now, by using distance formula,
Distance of PQ =
\mathbf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}} So, the distance between the points A (0, y, 0) and P (3, -2, 5) will be
Distance of AP = √[(3-0)2 + (-2-y)2 + (5-0)2]
= √[32 + (-2-y)2 + 52]
= √[(-2-y)2 + 9 + 25]
5√2 = √[(-2-y)2 + 34]
Squaring on both the sides, we get
(-2 -y)2 + 34 = 25 × 2
(-2 -y)2 = 50 – 34
4 + y2 + (2 × -2 × -y) = 16
y2 + 4y + 4 -16 = 0
y2 + 4y – 12 = 0
y2 + 6y – 2y – 12 = 0
y (y + 6) – 2 (y + 6) = 0
(y + 6) (y – 2) = 0
y = -6, y = 2
Hence, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.
Question 5: 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
Solution:
Let the coordinates of the required point be (4, y, z).
So now, let the point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) in the ratio k: 1.
Coordinates of the point which divides PQ in the ratio k : 1 =
(\frac{kx_2+x_1}{k+1}, \frac{ky_2+y_1}{k+1}, \frac{kz_2+z_1}{k+1}) So, we have
(\frac{8k+2}{k+1}, \frac{-3}{k+1}, \frac{10k+4}{k+1}) = (4, y, z)
\frac{8k+2}{k+1} = 48k + 2 = 4 (k + 1)
8k + 2 = 4k + 4
8k – 4k = 4 – 2
4k = 2
k =
\frac{2}{4} k =
\mathbf{\frac{1}{2}} Now, substituting the value we get,
y =
\frac{-3}{\frac{1}{2}+1} = \frac{-3}{\frac{3}{2}} = -2z =
\frac{10(\frac{1}{2})+4}{\frac{1}{2}+1} = \frac{5+4}{\frac{3}{2}} = 6Hence, the coordinates of the required point are (4, -2, 6).
Question 6: 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 PA2 + PB2 = k2, where k is a constant.
Solution:
The points A (3, 4, 5) and B (-1, 3, -7)
Let the point be P (x, y, z).
Now by using distance formula,
Distance of point (x1, y1, z1) and (x2, y2, z2) =
\mathbf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}} So, the distance between the points A (3, 4, 5) and P (x,y,z)) will be
Distance of PA = √[(3-x)2 + (4-y)2 + (5-z)2]
Distance of PB = √[(-1-x)2 + (3-y)2 + (-7-z)2]
As, PA2 + PB2 = k2
[(3 – x)2 + (4 – y)2 + (5 – z)2] + [(-1 – x)2 + (3 – y)2 + (-7 – z)2] = k2
[(9 + x2 – 6x) + (16 + y2 – 8y) + (25 + z2 – 10z)] + [(1 + x2 + 2x) + (9 + y2 – 6y) + (49 + z2 + 14z)] = k2
9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z + 1 + x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z = k2
2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = k2
2x2 + 2y2 + 2z2 – 4x – 14y + 4z = k2 – 109
2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109
(x2 + y2 + z2 – 2x – 7y + 2z) =
\frac{(k^2 – 109)}{2} Hence, the required equation is (x2 + y2 + z2 – 2x – 7y + 2z) =
\mathbf{\frac{(k^2 – 109)}{2}}