a position particle is given as r=<(4+3t)i + t^(3)j - 5tk>. at what time does it

Question

a position particle is given as r=<(4+3t)i + t^(3)j - 5tk>. at what time does it pass through the point (1,-1,5)? find its velocity at this time . fid the equations of the line tangent to its path and the plane normal to the path at (1,-1,5).

i found the time to be t=-1 and velocity =<3i+ 3t^(2)j - 5k> ... i just dont know how to solve the last part...thanks

Explanation / Answer

4+3t = 1 => t=-1 so the particle passes through the given point at t=-1. at A(1,-1,5) dr/dt= 3i + 3t^2 j -5k at t=-1 => dr/dt at t=-1 = 3i +3j-5k (tangent) so equation of tangent is given by y= (1,-1,5) + k(3i+3j-5k) where k is a real number. to get the equation of normal.. at t=0, the particle passes through B(4,0,0) the normal is perpendicular to AB, tangent.. so it's the cross product of the two vectors AB= (4,1,-5) tangent =(3,3-5) normal is parallel to (4i+j-5k)*(3i+3j-5k) =10 i +5j +9k so equation of normal is y=i-j+5k + t(10i+5j+9k) where t is a real number

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