Find the angle, in degrees, between two adjacent sides of the pyramid.
P(2, 0, 0), Q(2, 2, 0), R(0, 2, 0), (base points)
S(1, 1, 7) (tip point)
To find the angle between two adjacent sides of the pyramid, we need to consider the triangle formed by the base points and the tip point.
First, let's find the lengths of the three sides of the base triangle. We can use the distance formula to find the distances between the points:
Distance PQ:
d(PQ) = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
= sqrt((2 - 2)^2 + (2 - 0)^2 + (0 - 0)^2)
= sqrt(0 + 4 + 0)
= sqrt(4)
= 2
Distance QR:
d(QR) = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
= sqrt((0 - 2)^2 + (2 - 2)^2 + (0 - 0)^2)
= sqrt((-2)^2 + 0 + 0)
= sqrt(4)
= 2
Distance RP:
d(RP) = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
= sqrt((0 - 2)^2 + (2 - 0)^2 + (0 - 0)^2)
= sqrt((-2)^2 + 4 + 0)
= sqrt(4 + 4)
= sqrt(8)
= 2√2
Now, let's find the height of the pyramid, which is the distance between the tip point S and the base plane formed by the base points P, Q, and R. We can use the distance formula to find the distance:
Distance SH:
d(SH) = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
= sqrt((1 - 2)^2 + (1 - 1)^2 + (7 - 0)^2)
= sqrt((-1)^2 + 0 + 49)
= sqrt(1 + 49)
= sqrt(50)
= 5√2
Now, we can use the law of cosines to find the angle between the sides PQ and QR:
cosine(angle) = (d(PQ)^2 + d(QR)^2 - d(RP)^2) / (2 * d(PQ) * d(QR))
angle = arccos((d(PQ)^2 + d(QR)^2 - d(RP)^2) / (2 * d(PQ) * d(QR)))
angle = arccos((2^2 + 2^2 - (2√2)^2) / (2 * 2 * 2))
= arccos((4 + 4 - 8) / 8)
= arccos(0 / 8)
= arccos(0)
= 90 degrees
Therefore, the angle between the two adjacent sides of the pyramid is 90 degrees.
To find the angle between two adjacent sides of the pyramid, we need to find the lengths of the sides first using the given coordinates of the points.
Let's consider the base triangle with points P, Q, and R. To find the lengths of the sides, we can use the distance formula:
Distance between two points (x1, y1, z1) and (x2, y2, z2) is given by:
Distance = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
Using this formula, we can calculate the lengths of the sides of the base triangle:
Side PQ:
Distance(PQ) = √((2 - 2)^2 + (2 - 0)^2 + (0 - 0)^2) = √(0 + 4 + 0) = √4 = 2
Side QR:
Distance(QR) = √((0 - 2)^2 + (2 - 2)^2 + (0 - 0)^2) = √((-2)^2 + 0 + 0) = √4 = 2
Side RP:
Distance(RP) = √((0 - 2)^2 + (2 - 0)^2 + (0 - 0)^2) = √((-2)^2 + 4 + 0) = √8
Now, let's consider one of the triangle's sides to calculate the angle.
For example, let's find the angle between sides PQ and PS. To do this, we need to find the dot product of the vectors formed by these sides, and then use the dot product formula to find the angle.
The dot product of vectors A = (x1, y1, z1) and B = (x2, y2, z2) is given by:
A · B = (x1 * x2) + (y1 * y2) + (z1 * z2)
Using the coordinates of points P, Q, and S, we can find the vectors PQ and PS:
Vector PQ = (2 - 2, 2 - 0, 0 - 0) = (0, 2, 0)
Vector PS = (1 - 2, 1 - 0, 7 - 0) = (-1, 1, 7)
Now, we can find their dot product:
PQ · PS = (0 * -1) + (2 * 1) + (0 * 7) = -2 + 2 + 0 = 0
Since the dot product is 0, we know that the vectors PQ and PS are perpendicular to each other. Therefore, the angle between them is 90 degrees.
Note: We assumed that the points P, Q, R, and S are connected in the given order. If this is not the case, the calculation of the angle may vary.
use cross product to get one side (n1)
n1= 2,0,0 x 1,1,7 = 0,-14, 2
use cross product to get another side (n2)
n2 = 0,2,0 x 1,1,7 = 14,0,-2
cos theta = (n1 x n2) / sqrt(sum of n1square) x sqrt(sum of n2 squares
-4 / sqrt(196+4) x sqrt(196+4)
-4/200
cos theta = - .02
theta = arccos (-.02) = 91.1