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
To prove 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 vectors.
Let A be the midpoint of PS, B be the midpoint of QR, C be the midpoint of PR, and D be the midpoint of QS.
We'll start by considering the vectors AP and RB. Since A is the midpoint of PS, we can express AP as PS/2. Similarly, we can express RB as QR/2, as B is the midpoint of QR.
Now, we know that PQRS is a tetrahedron and PQ is perpendicular to RS. Therefore, PQ and RS are orthogonal, which means that the dot product between the vectors PQ and RS is zero.
Mathematically, this can be expressed as:
PQ · RS = 0
Expanding the dot product in terms of vectors, we have:
(P - Q) · (R - S) = 0
Now, let's rewrite P - Q as PA + AQ and R - S as RD + DS:
(PA + AQ) · (RD + DS) = 0
Expanding further, we get:
PA · RD + PA · DS + AQ · RD + AQ · DS = 0
Since AP = -PA (the negative of vector PA), we can rewrite the equation as:
-AP · RD + AP · DS + AQ · RD + AQ · DS = 0
Using the distributive property of the dot product, we have:
(AP + AQ) · RD + (AP + AQ) · DS = 0
Simplifying, we get:
AB · RD + AB · DS = 0
Since AB = RD/2 (as mentioned earlier) and AB = DS/2, we can rewrite the equation as:
(RD/2) · RD + (DS/2) · DS = 0
Expanding, we have:
RD^2/2 + DS^2/2 = 0
RD^2 + DS^2 = 0
This implies that RD = 0 and DS = 0, which means that R and D coincide, and S and D coincide. Therefore, CD = 0, and C is the midpoint of DS and R.
Similarly, by using a similar procedure, you can prove that C is also the midpoint of AQ and S.
So, we have shown that the midpoint of PS (A) coincides with the midpoint of QR (B), and the midpoint of PR (C) coincides with the midpoint of QS (D).
Therefore, the distance between the midpoints of PS and QR (AB) is equal to the distance between the midpoints of PR and QS (CD).