Line 1 = -4i + j +3k + s(i + 5j - 2k)
Line 2 = 3i - 7j + 9k + t(2i - 3j + k)
The lines are skew.
Question: Find the position vector of the point C on line 2 such that ∠ABC = 90°.
Annnd...i have no idea how...I've tried random things but nothing works.
The answer is -7i + 8j + 4k
Any help or prods in the right direction would be appreciated.
Thank-you very much!
Then A and B are not on line 1 or line 2? And, in fact, line 1 has nothing to do with this problem? C, since it is on line 2, can be written with position vector (3+2t)i- (7+ 3t)j+ (9+ t)k.
Then the vector from C to B is (3+2t- 3)i- (7+ 3t- 6)j+ (9+t- 1)k= 2ti- (1+ 3t)j+ (8+ t)k. The vector from A to B is (3- 4)i+ (6- 1)j+ (1- 3)k= -i+ 5j- 2k. The angle ABC will be 90 degrees if and only if those two vectors are orthogonal which itself is true if and only if their dot product is 0:
-1(2t)- 5(1+ 3t)- 2(8+ t)= -25t- 21= 0 or t= -21/25. That does NOT give the C that you say is the answer. Are you sure you have copied everything correctly?