Calculus/Vectors

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Expert: - 2/9/2009


Question
i dont know how to get started with this as i dont understand this topic. Please guide me. the line l has equation r=(7i+3k)+ t(5i+3j+2k), & the plane ¥ has equation r* (i+j+2k)=1. Find the coordinates of thd point A where l meets ¥. a)the point B has coordinates (7,0,3) & C is the foot of the perpendicular from B to ¥. Find the coordinates of C. b)find a vector equation for the line AC, & calculate the angle BAC,giving your answer to the nearest degree. Thanks a lot

Answer
Line L has the equation
r = (7i+0j+3k)+ t(5i+3j+2k) = (7,0,3) + (5,3,2)t.

Plane ¥ has equation i+j+2k = 1.

Find coordinates where A meets L.
In other words, find a variable t such that 7+5t + 0+3t + 3+2t = 1.
This reduces to 10t + 10 = 1, or t = -9/10.
The point would then be (7-45/10, 0+27/10, 3+19/10) =
(25/10, 27/10, 49/10) = (2.5, 2.7, 4.9).  This is A.

B is at (7,0,3) and C is at the foot when B is perpendicular to ¥.

We know that (1,1,2) is perpendicular to the plane.
The cross product of (7-Cx,0-Cy,3-Cz) and a vector in the plane
must be 0 in order for them to be perpendicular.

Take C - A to get the directional componet between the two.

It is know that AxB = ab sin(Θ)n where n is the perpendicular
to the plane with A and B.

Use this to calculate the angle.  Remember that your calculator
should be in degree mode.  If not, 180° = π radians.  

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