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rac_se
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f(x, y, z) = x2 + sin(y) - 2z2 = 0 defines a surface in 3 dimensions. First I need to find the equation of the normal line to the surface at point P0(2pi, 0, 3/2). Then, I need to find the point which is at a distance of 4 from the tangent plane at the point P0
Equation for the normal line at a point P0 is given by:
x = x0 + fx(Po)t
y = y0 + fy(Po)t
z = z0 + fz(Po)t
Equation for the tangent plane is given by:
fx(P0)(x-xo) + fy(P0)(y-yo) + fz(P0)(z-zo)
I used the first equation to find the normal line at P0 and the answer I got is:
x = 2pi-4pi*t
y= t
z = 3/2 - 6t
I know that the distance between a point P0 on a plane with normal n and a point P is given by the dot product of the vector projection of P0P and n/|n|. I get the equation for normal n from first part. But, |n| =sqrt(x2 + y 2 + z2) gives very complicated result. Then, I am not sure how to use the distance formula to find the point. Or, there is some other way to solve this problem.
I hope someone could guide me on how to approach this problem. This is the first time I am posting to this forum. So, please correct me if I made any mistake
Thanks!
Equation for the normal line at a point P0 is given by:
x = x0 + fx(Po)t
y = y0 + fy(Po)t
z = z0 + fz(Po)t
Equation for the tangent plane is given by:
fx(P0)(x-xo) + fy(P0)(y-yo) + fz(P0)(z-zo)
I used the first equation to find the normal line at P0 and the answer I got is:
x = 2pi-4pi*t
y= t
z = 3/2 - 6t
I know that the distance between a point P0 on a plane with normal n and a point P is given by the dot product of the vector projection of P0P and n/|n|. I get the equation for normal n from first part. But, |n| =sqrt(x2 + y 2 + z2) gives very complicated result. Then, I am not sure how to use the distance formula to find the point. Or, there is some other way to solve this problem.
I hope someone could guide me on how to approach this problem. This is the first time I am posting to this forum. So, please correct me if I made any mistake
Thanks!
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