1. Two edges PQ, RS of a tetrahedron PQRS are perpendicular; show that the distance between the mid-points of PS and QR is equal to the distance between the mid-points of PR and QS.
You can set up the vertices PQRS as points on the x,y,z axes such that
P = (0,0,c)
Q = (0,b,0)
R = (0,0,0)
S = (a,0,0)
Then the midpoints are:
PS: (a/2,0,c/2)
QR: (0,b/2,0)
PR: (0,0,c/2)
QS: (a/2,b/2,0)
You can easily show that the lengths of those segments are equal.
There's probably some sly similar-triangles way to show it as well, but I'm too lazy just now.
To show that the distance between the midpoints of PS and QR is equal to the distance between the midpoints of PR and QS, we can use the concept of vector equations.
Let's consider the tetrahedron PQRS. PQ and RS are perpendicular, which means they are orthogonal vectors.
Step 1: Define vectors and points:
Let A be the midpoint of PS.
Let B be the midpoint of QR.
Let C be the midpoint of PR.
Let D be the midpoint of QS.
Step 2: Represent vectors in terms of known points:
The vectors PQ, QR, RP, and SR can be expressed in terms of the known points as follows:
PQ = Q - P
QR = R - Q
RP = P - R
SR = R - S
Step 3: Determine midpoint vectors:
The vectors representing the midpoints can be calculated using their respective coordinates:
AC = (C - A)/2
BD = (D - B)/2
Step 4: Show that AC and BD are equal:
To prove that AC and BD are equal, we need to show that they have the same magnitude.
Step 5: Calculate AC:
AC = (C - A)/2
= (1/2)(P + R)/2 - (1/2)(P + S)/2
= (1/4)(2P + 2R - P - S)
= (1/4)(P + R - S)
Step 6: Calculate BD:
BD = (D - B)/2
= (1/2)(Q + S)/2 - (1/2)(Q + R)/2
= (1/4)(2Q + 2S - Q - R)
= (1/4)(S + Q - R)
Step 7: Show that AC and BD are equal:
To prove AC = BD, we need to show that their magnitudes are equal.
Magnitude of AC:
|AC| = sqrt((1/4)(P + R - S) · (P + R - S))
= sqrt((1/4)(P + R - S) · (P + R - S))
= sqrt((1/4)(P + R - S) · (P + R - S))
Magnitude of BD:
|BD| = sqrt((1/4)(S + Q - R) · (S + Q - R))
= sqrt((1/4)(S + Q - R) · (S + Q - R))
= sqrt((1/4)(S + Q - R) · (S + Q - R))
Step 8: Simplify and conclude:
Both |AC| and |BD| have the same expression and hence have the same magnitude.
Therefore, the distance between the midpoints of PS and QR (represented by AC) is equal to the distance between the midpoints of PR and QS (represented by BD), which confirms the result.
To prove that the distance between the midpoints of PS and QR is equal to the distance between the midpoints of PR and QS in a tetrahedron PQRS where PQ and RS are perpendicular, we can use the concept of vectors.
Let A and B be the midpoints of PS and QR, respectively. We need to prove that the length of vector AB is equal to the length of vector CD, where C is the midpoint of PR and D is the midpoint of QS.
We can represent the position vectors of P, Q, R, and S as vectors p, q, r, and s, respectively. Therefore, the position vectors of A, B, C, and D can be expressed as:
A = (p + s) / 2
B = (q + r) / 2
C = (p + r) / 2
D = (q + s) / 2
Now, let's find the vector AB by subtracting vector A from vector B:
AB = B - A = (q + r) / 2 - (p + s) / 2
= (q - p + r - s) / 2
Similarly, we can find vector CD by subtracting vector C from vector D:
CD = D - C = (q + s) / 2 - (p + r) / 2
= (q - p + s - r) / 2
To prove that vector AB and vector CD have the same length, we need to show that their magnitudes are equal:
|AB|^2 = (AB · AB) (Dot product of AB with itself)
= [(q - p + r - s) / 2] · [(q - p + r - s) / 2]
= [(q - p) · (q - p) + (r - s) · (r - s) + (q - p) · (r - s) + (r - s) · (q - p)] / 4 (Expanding the dot product)
Similarly,
|CD|^2 = (CD · CD)
= [(q - p + s - r) · (q - p + s - r)] / 4
We need to show that |AB|^2 = |CD|^2.
To simplify the expression, we can focus on the cross products:
(q - p) · (r - s) + (r - s) · (q - p) = (r - s) · (q - p) + (q - p) · (r - s)
= 2[(r - s) · (q - p)]
Substituting this back into the expression for |AB|^2:
|AB|^2 = [(q - p) · (q - p) + (r - s) · (r - s) + 2[(r - s) · (q - p)]] / 4
Similarly,
|CD|^2 = [(q - p) · (q - p) + (r - s) · (r - s) - 2[(r - s) · (q - p)]] / 4
Now, it is evident that |AB|^2 = |CD|^2.
Since the magnitudes of vector AB and CD are equal, it follows that the distance between the midpoints of PS and QR is equal to the distance between the midpoints of PR and QS. Thus, the claim is proven.
Note: The proof provided uses vector algebra and dot products to show the equality of lengths. This approach is commonly used in geometry to establish relationships between various points, lines, or vectors within a figure.