✔ 最佳答案
(a) RP should be a line passing through (13, 9, -17) and having a direction ratio of [2, 3, -6]
So, in parametric form, the line can be represented as:
x = 2t + 13, y = 3t + 9 and z = -6t - 17
Sub this relation into the equation of the plane:
2(2t + 13) + 3(3t + 9) - 6(-6t - 17) = 8
4t + 26 + 9t + 27 + 36t + 102 = 8
49t = -147
t = -3
x = 7, y = 0 and t = 1
So R is (7, 0, 1) and RP = 6i + 9j - 18k
And the distance between P and the plane = √(62 + 92 + 182) = 21
(b) QR = 6i + 2j + 3k
2010-06-16 13:04:54 補充：
Direction ratio in 3D coord geom. is something like slope in 2D geom, indicating the direction of the line.
If a plane has an eqn Ax + By + Cz + D = 0, then its normal line has a dir. ratio of [A, B, C]
2010-06-18 00:38:17 補充：
Alternatively, we can interpret as follows:
Suppose P is (a, b, c), then vector RP should be = p(2i + 3j - 6k) where p is a non-zero constant since it is perp. to the plane.
In another form:
vec RP = (13 - a)i + (9 - b)j + (-17 - c)k
2010-06-18 00:39:37 補充：
Therefore, we have:
(13 - a)/2 = (9 - b)/3 = (-17 - c)/(-6)
Which can eventually give:
a = 2t + 13, b = 3t + 9 and c = -6t - 17
which is the same as the interpretation above.
