Question: 1 Solve the initial value problem for r as a vector function of t. dr Differential equation: dt = -7ti-t j - 5t k Initial

 1 Solve the initial value problem for r as a vectorfunction of t. dr Differential equation: dt = -7ti-t j - 5tk Initial condition: r(0) = 6i + 8j + 9k r(t) =it kSolve the initial value problem for r as a vector function

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of t. dr 9 Differential Equation: E dt 5 (t+1)"i+4elj+ t+ 1Initial condition: r(0) = k r(t) = + kSolve the initial valueproblem for r as a vector function of t. 2 dr Differentialequation: = - 4k 2 dt dr Initial conditions: r(0) = 10kand dt = 10i + 10j It= 0 r(t) = + kSolvethe following initial value problem for r as a function of t.

Solve the initial value problem for r as a vector function of t. dr Differential equation: dt = -7ti-t j - 5t k Initial condition: r(0) = 6i + 8j + 9k r(t) = it kSolve the initial value problem for r as a vector function of t. dr 9 Differential Equation: E dt 5 (t+1)"i+4elj+ t+ 1 Initial condition: r(0) = k r(t) = + kSolve the initial value problem for r as a vector function of t. 2 dr Differential equation: = - 4k 2 dt dr Initial conditions: r(0) = 10k and dt = 10i + 10j It= 0 r(t) = + kSolve the following initial value problem for r as a function of t. Differential equation: 2 =2eli-e -tj + 12e 2 k dt Initial conditions: r(0) = 61 + 4j + 6k dr dt = - 2i + 5j It = 0 r(t) = it + kFind the plane determined by the intersecting lines. L1 x= -1+2t y = 2 + 3t z =1-t L2 x=1-4s y = 1+2s z=2-2s III Using a coefficient of - 1 for x, the equation of the plane is (Type an equation.)At time t= 0, a particle is located at the point (6,6,9). It travels in a straight line to the point (4,9,1), has speed 3 at (6,6,9) and constant acceleration - 2i + 3j- 8k. Find an equation for the position vector r(t) of the particle at time t. The equation for the position vector r(t) of the particle at time t is r(t) = + K. (Type exact answers, using radicals as needed.)

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