Question

g Find the equation for the plane containing the points (0, 0, 1), (1, 2, 3), (2, 4, 5), and (4, 8, 11). (b) Consider the tangent plane to the surface z = x 2+xy−3y at the point (2, 3, 1). Compute the angle of intersection between this plane and the plane from part (a

Answer 1

Answer:

Step-by-step explanation:

Any 3 non collinear points would determine a unique plane. The four points given are:

P(0, 0, 1),Q (1, 2, 3), R(2, 4, 5), and S (4, 8, 11).

Direction ratios are listed below

PQ = 1,2,2

QR = 1,2,2

RS = 2,4,6

Hence we find that P,Q,R are collinear so let us select P,Q and S to determine the plane equation.

Hence we have plane equation

$\left[\begin{array}{ccc}x-0&y-0&z-1\\1&2&2\\4&8&10\end{array}\right] \\=x(20-16)-y(10-8)+(z-1)(8-8)=0\\4x-2y=0\\2x-y=0$

b) Given $z=x^2+xy-3y\\w = x^2+xy-3y-z$

Del w = (2x+y)i+(x-3)j-1k

At the point (2,3,1) normal would have direction ratios as

(7,-1,-1)

Hence tangent plane is

$7(x-2)-1(y-3)-1(z-1)=0\\7x-y-z-10=0$

c) Angle of intersection would be the angle between normals

i.e. cos t =$\frac{ (2,-1,0).(7,-1,-1)}{\sqrt{5}\sqrt{51}  } =\frac{15}{\sqrt{255} }$ where t is the angle.