Advanced Math/3d vectors

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Expert: - 2/9/2009


Question
I dont understand this topic. Please help me, i need the whole solutions. The points A,B,C and D have position vectors a,b,c,d respectively,given by a=i-j+2k,b=3i+7j-k,c=4i+15j+10k, & d=2i+ 18j +2k.a) find,in vector form,an equation for the line l which passes through A & B.b) find a vector equation of the plane ¥ which passes through D & is perpendicular to line l. Find also the position vector of the point E in which l & ¥ intersect. Deduce the position vector of the image of the point B under a reflection in the plane ¥.c) find the acute angle that the CD makes with the plane ¥. Thanks

Answer
A is at (1,-1,2)
B is at (3,7,-1)
C is at (4,15,10)
D is at (2,18,2)
L passes through A and B
¥ is the plane that passes thru D and is perpendicular to l.

a) an equation passing through A and B would be (1+2t, -1+8t, 2-3t).
We can see when t=0, A is found.  When t=1, B is found.
So the line L is (1+2t, -1+8t, 2-3t).

b) To pass through D, we need to find a point (x0,y0,z0) with two conditions: 1) That point is on the line 2) The cross product of a vector in that direction with the direction of just line is 0.

The direction of the point D from the line would be
(x0-2, y0-19, z0-2).

The direction of the line is (2,9,-3).

For the cross product to be zero, we need each of the compnets to be zero.  

The x0 componet is 9z0-18 + 3ty0+57.
The y0 componet is 2z0-4 + 3x0+6.
The z0 componet is 2y0-38 + 3x-6.

Set all of these equations to 0 and solve.
Using x0, y0, and z0, put them back into (x0-2, y0-19, z0-2)
to find the direction.

Ask again if you need more help.  That's all the farther I can take
it right now.

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