Introduction to Three Dimensional Geometry Miscellaneous Exercise
Question:
Find the lengths of the medians of the triangle with vertices A(0,0,6),B(0,4,0) and (6,0,0).
Answer:
Step 1: Calculate the midpoints of the sides of the triangle. The midpoint of AB can be calculated by taking the average of the xcoordinates and the average of the ycoordinates of the two points: Midpoint of AB: (0+6)/2, (0+4)/2, (6+0)/2 = (3,2,3) The midpoint of BC can be calculated by taking the average of the xcoordinates and the average of the ycoordinates of the two points: Midpoint of BC: (6+0)/2, (4+0)/2, (0+0)/2 = (3,2,0) The midpoint of AC can be calculated by taking the average of the xcoordinates and the average of the ycoordinates of the two points: Midpoint of AC: (0+6)/2, (4+0)/2, (6+0)/2 = (3,2,3)
Step 2: Calculate the distance between the midpoints and the vertices. The distance between the midpoint of AB and vertex A can be calculated using the distance formula: Distance between midpoint of AB and vertex A = √((30)^2 + (20)^2 + (36)^2) = 5 The distance between the midpoint of BC and vertex B can be calculated using the distance formula: Distance between midpoint of BC and vertex B = √((30)^2 + (24)^2 + (00)^2) = 2 The distance between the midpoint of AC and vertex C can be calculated using the distance formula: Distance between midpoint of AC and vertex C = √((36)^2 + (20)^2 + (30)^2) = 5
Step 3: Calculate the length of the medians. The length of the median from vertex A can be calculated by adding the distances between the midpoint of AB and vertex A, and the midpoint of AC and vertex A: Length of median from vertex A = 5 + 5 = 10 The length of the median from vertex B can be calculated by adding the distances between the midpoint of AB and vertex B, and the midpoint of BC and vertex B: Length of median from vertex B = 2 + 5 = 7 The length of the median from vertex C can be calculated by adding the distances between the midpoint of AC and vertex C, and the midpoint of BC and vertex C: Length of median from vertex C = 5 + 2 = 7
Therefore, the lengths of the medians of the triangle with vertices A(0,0,6), B(0,4,0) and C(6,0,0) are 10, 7 and 7 respectively.
Question:
A point R with xcoordinate 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.
Answer:

To find the coordinates of the point R we need to first find the coordinates of the midpoint of the line segment.

The midpoint of the line segment joining the points P and Q can be found by taking the average of the x, y, and z coordinates of P and Q.

The xcoordinate of the midpoint is (2+8)/2 = 5.

The ycoordinate of the midpoint is (3+0)/2 = 1.5.

The zcoordinate of the midpoint is (4+10)/2 = 7.

Since we know that the point R lies on the line segment, and its xcoordinate is 4, then the coordinates of the point R are (4, 1.5, 7).
Question:
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:

The centroid of a triangle is the point at the intersection of the three medians of the triangle.

The median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

The midpoint of the side PQ is (−1, 2.5b, −8).

The midpoint of the side QR is (4, 8.5b, −4).

The midpoint of the side PR is (5a, 8, 4c).

Equate the xcoordinates of the three midpoints to find the value of a:
−1 = 5a a = −1/5
 Equate the ycoordinates of the three midpoints to find the value of b:
2.5b = 8.5b b = 8/5
 Equate the zcoordinates of the three midpoints to find the value of c:
−8 = 4c c = −2
Question:
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:

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

We need to find the equation of the set of points P such that PA^2 +PB^2 =K^2.

We know that the coordinates of point A and B are (3, 4, 5) and (−1, 3, −7) respectively.

Therefore, the equation of the set of points P can be written as
(x  3)2 + (y  4)2 + (z  5)2 + (1  x)2 + (3  y)2 + (7  z)2 = K2
 Simplifying the above equation, we get
2x2 + 2y2 + 2z2  12x  8y + 12z  28 = K2
 Therefore, the equation of the set of points P such that PA^2 +PB^2 =K^2 is
2x2 + 2y2 + 2z2  12x  8y + 12z  28 = K2
Question:
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 fourth vertex of the parallelogram is D.
 Since the parallelogram is a closed figure, the sum of the xcoordinates of the four vertices must be equal. Therefore, 3 + 1 + (1) + Dx = 0
 Similarly, the sum of the ycoordinates must also be equal. Therefore, 1 + 2 + 1 + Dy = 0
 Lastly, the sum of the zcoordinates must be equal. Therefore, 2 + (4) + 2 + Dz = 0
 Solving the above three equations, we get Dx = 3, Dy = 3 and Dz = 6
 Therefore, the coordinates of the fourth vertex D are (3,3,6).
Question:
Find the coordinates of a point on yaxis which are at a distance of 5√2 from the point P(3,−2,5)
Answer:
Step 1: The coordinates of the point P are given as (3, 2, 5).
Step 2: Since the point is on the yaxis, the xcoordinate of the point will be 0.
Step 3: The distance between the points P and the point on the yaxis is 5√2.
Step 4: We can use the Pythagorean Theorem to calculate the ycoordinate of the point.
Step 5: The equation for the Pythagorean Theorem is a2 + b2 = c2.
Step 6: Substituting the values, we get (2)2 + y2 = (5√2)2
Step 7: Simplifying the equation, we get y2 = 82
Step 8: Taking the square root of both sides, we get y = ±√82
Step 9: The coordinates of the point on the yaxis which are at a distance of 5√2 from the point P(3,−2,5) are (0, ±√82).