Recitation 3

Section 10.5 Problem 10: Find the parametric equation and symmetric equation for the line of intersection of the planes x + 2y + 3z = 1 and x + y + z = 1.

Comment: To find the parametric equation and symmetric equation of the line, it is enough to find a point (x_0,y_0,z_0) on the line and the direction \langle A,B,C\rangle of the line. Then the parametric equation is x(t)=x_0+At, y(t)=y_0+Bt, z(t)=z_0+Ct and the symmetric equation is \frac{x-x_0}{A}=\frac{y-y_0}{B}=\frac{z-z_0}{C} unless ABC=0. For this problem, it is easy to observe that (1,0,0) is on the line and the cross product of the normal vectors of the planes is a direction of the line.

Section 10.5 Problem 27: Find the equation of the plane that passes through the point (6, 0, 2) and contains the line x = 4 + 2t, y = 3 + 5t, z = 7 + 4t.

Comment: The point (6,0,2) is on the plane for free. It is easy to see the point (4,3,7) is also on the plane since it is on the line. So the vector between those two points \langle -2,3,5 \rangle is parallel to the plane. Also the direction of the line \langle 2, 5, 4 \rangle is also parallel to the plane. Thus their cross product is a normal vector of the plane.

Section 10.5 Problem 45: Which of the following four planes are parallel? Are any of them identical? P1: 3x+6y-3z=6, P2: 4x-12y+8z=5, P3: 9y=1+3x+6z, P4: z= x+2y-2.

Comment: Compare the normal vectors of the planes to see whether they are parallel.

Section 10.5 Problem 49: Find the distance from the point to the given plane. (1,-2,4), 3x+2y+6z=5.

Comment: Use the formula for the distance from a point to a plane.

Section 10.5 Problem 51: Find the distance between the given parallel planes. 2x-3y+z=4, 4x-6y+2z=3.

Comment: It is enough to find the distance from a point on the first plane to the second plane.

Section 10.6 Problem 7: Describe and sketch the surface. xy=1

Comment: Draw the curve in the x-y plane and use it to form a cylinder.

Section 10.6 Problem 18: Use traces to sketch and identify the surface. 4x^2 -16y^2 +z^2 =16.

Comment: The x-sections and the z-sections are hyperbolas and the y-sections are ellipses. Thus the surface is hyperboloid with one sheet.

Section 10.6 Problem 25: Reduce the equation to one of the standard forms, classify the surface, and sketch it. 4x^2 +y^2 +4z^2 -4y-24z+36=0.

Comment: Complete the squares and see it is an ellipsoid.

Section 10.6 Problem 31: Find an equation for the surface consisting of all points that are equidistant from the point (-1, 0, 0) and the plane x = 1. Identify the surface.

Solution: Let P=(x,y,z) be a point on the surface. Then (x+1)^2+y^2+z^2=(x-1)^2. Notice that the left hand side is the square of the distance from P to the point (-1, 0, 0) and the right hand side is the square of the distance form P to the plane x = 1. This equation gives us 2x+y^2+z^2=0 which is an elliptic paraboloid.

Leave a Reply

Your email address will not be published. Required fields are marked *