find the distance from point Q to line l (matrix)
Q=(0, 1, 0)
line = < x y z > = < 1 1 1 > + t< -2 0 3 >
Formula:
__
||PQ × u||
D = ------------
||u||
where u is the direction vector for the line and
P is a point on the line
Using the direction numbers -2, 0, and 4, you know
that the direction vector for the line is
u = < -2, 0, 4 >
To find a point on the line, let t = 0 and obtain
P = (1, 1, 1)
Thus, since Q = (0, 1, 0)
__
PQ = < 0-1, 1-1, 0-1 > = < -1,0,-1 >
__
Now get the cross product PQ × u
__ | i j k|
PQ × u = |-1 0 -1| = i(0+0) - j(-4-2) + k(0+0) = -6j
|-2 0 4|
__
||PQ × u|| = ||-6j|| = 6
____________ ______ __ _
||u|| = Ö(-2)²+0²+ 4² = Ö4+0+16 = Ö20 = 2Ö5
__ _
||PQ × u|| 6 3 3Ö5
D = ------------ = ----- = ---- = -----
||u|| 2Ö5 Ö5 5
Edwin
