(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

f(x,y) = x^{2}+ y^{2}

P(9,6)

v=<2,4,0>

1)Find a vertical plane that passes through the point P(9,6) and has the vector v=<2,4,0>

2)What is the vector function of the curve of intersection of the vertical plane and z=f(x,y)

2. Relevant equations

A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0})

normal vector

n= <0,0,1>

given point and vector:

P(9,6)

v = < 2, 4 0>

3. The attempt at a solution

I tried plugging the give point and and the normal vector into A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0})

and I got 3x+2y but that's not a vertical plane.

I also tried plugging in the point and the given vector and got x+2y-21 and that doesn't look right either.

I found the unit vector to be <1/√3 , 2/√3 , 0>. I'm not sure if I need to use that find the vertical plane.

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# Homework Help: The vertical plane through a point and containing a vector.

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