Find the points on the given surface at which the tangent plane is parallel to the indicated plane. x2+y2+z2=7 2x+4y+6z=1

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Find the points on the given surface at which the tangent plane is parallel to the indicated plane.   x2+y2+z2=7  2x+4y+6z=1

David Tyson · Accepted Answer

The given function is F(x,y,z)=x2+y2+z2 Hence, the gradient of this elipsoid is △F(x,y,z)=2ξ+2yj+2zk So normal vector to the surface at (x0,y0,z0) is 2x0i+2y0j+2z0k Also, obtaining the given equation of the plane 2x+4y+6z a normal vector to this plane would be 2i+4j+6k Since we need to find points on the given surface at which the tangent plane, let us take P(x0,y0,z0) for the desired point on the given surface. Then, normal vector a multiple of a normal vector to the indicate plane, let us say c times. Then we have this system of equations: 2x0=2c⇒x0=c 2y0=4c⇒y0=2c 2z0=6c⇒z0=3c Pulling back these values instead of (x0,y0,z0) at the given surface, we get the equation c2+4c2+9c2=7 ⇒14c2=7 ⇒c2=12 ⇒c=±12 Since we got the value for themultiple constant c, we can take it back to the obtained points and get their values: P1(x0,y0,z0)=(22,2⋅22,3⋅22)=(22,2,322) and P2(x0,y0,z0)=(−22,2⋅−22,3⋅(−22))=(−22,−2,−322)

Find the points on the given surface at which the

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