# 3D Vector Algebra

Update:

What's meant by "having a direction ratio of [2, 3, -6]"?? thx~

Update 2:

Actually, I have not learnt about "direction ratio" in the Matrix Chapter, is there any alternatives to solve part (a)?

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(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.

Source(s): Myself