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Please help me solve this question,

The straight line l has equation r = i +6j-3k + s(i-2j+2k). The plane p has equation (r-3i) . (2i-3j+6k)=0. The line l intersects the plane p at point A.

1) Find the position vector of A

2) Find the acute angle between l and p

3) Find the vector equation for the line which lies in p, passes through A and is perpendicular to l.

Thx....

I need to make sure I understand your notation before proceding. I assume that you're using the notation i, j, k to represent the unit vectors in the x, y, z directions. Also, I normally represent the eqns for lines and planes in 3D as

L = (x0,y0,z0) + s<v1,v2,v3> where x0, y0 and z0 represent some point and v1, v2, v3 are components of some 3-D vector. Likewise, as plane is represented by

P: a(x-x0)+b(y-y0)+c(z-z0) = 0.

Interpreting the coefficients in your notation with the x, y and z directions, I get for you line

(1,6,3) + s<1,-2,2>

and for the plane

2(x+2)-3(y-6)+6(z+3) = 0.

Please confirm this interpretation and I'll take it from there.

Randy

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