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(MATH100)midterm-rev-sol.pdf
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MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Midterm Review Solutions (March 28, 2007)
1.1 (Rectangular Coordinates in 3-space) Find an equation of sphere with center in the xz-plane and passing through the points P (0, 8, 0), Q(4, 6, 2) and R(0, 12, 4).
. . .
. Let C(a, 0,c) be the center. Then .PC. = .QC. = .RC..

2
So, a2 + 64 + c2 = .(a . 4)2 + 36 + (c . 2)2 = .a2 + 144 + (c . 4)2. Hence, a2 + 64 + c=
(a.4)2+36+(c.2)2 = a2+144+(c.4)2 . Equate the .rst and third terms : 64+c2 = 144+(c.4)2 , 2
c= 80+ c2 . 8c + 16, 8c = 96, c = 12. Substitute c = 12 in the .rst equation: a2 +64+144 = (a . 4)2 + 36 + 100, a2 +72 = a2 . 8a + 16, 8a = .56, a = .7. So, the
.
center is (.7, 0, 12). The radius .PC. = 49 + 64 + 144 = 257. The equation of the circle is therefore (x + 7)2 + y2 +(z . 12)2 = 257.
.
1.2 (Vectors) Find the length of the vector PQ from P = (1, .2, 4) to Q = (3, 4, 3).
. . .
. PQ = OQ . OP = . 3, 4, 3 ... 1, .2, 4 . = . 2,6, .1 . and the length of the vector is given
.
by .PQ. = .22 +62 +(.1)2 = 4+36+1= 41.

1.3 (Dot Product and Projections) Find the vector projection of A = . 1, 2, 4 . on B = . 4, .2, 4 ..
BB . 4,.2,4 .. 4,.2,4 .
. The projection of A on B is given by .A . = .. 1, 2, 4 . .
.B..B. 16+4+16 16+4+16 2 2 282 . = . 16 16
=(. 1, 2, 4 .. 2 , .1 , .). 2 , .1 , . =( 2 . 2 + 8 ). 2 , .1 , . = . 2 , .1 ,, .8 , ..
333 333 333333 3333 9 99

1.4 (Cross Product) Find the area of P QR, when P = (1, .1, 4), Q = (2, 0, 1) and R = (0, 2, 3).
. . . . . .
. PQ = . 1, 1, .3 ., PR = ..1, 3, .1 .. Recall that .PQ PR. = .PQ..PR. sin is
. . . .
the area of the parallelogram formed by PQ and PR. So, the area is .PQ PR. =

.. 1(.1) . (.3)3, (.3)(.1) . 1(.1), 1(3) . 1(.1) .. = .. 8, 4, 4 .. = 64+16+16= 96=4 6.

The area of P QR is half the area of the parallelogram, that is, 2 6.

1.5 (Parametric Equations of Lines) Write down the parametric equations of the line through
Q = (0, 3, 4) which is perpendicular to both lines:
x . 1 y . 2 z . 2 x . 3 z . 1

L1 : == and L2 : =, y = 3.
1 .1 .2 .11
. A vector parallel to L1 is . 1, .1, .2 . and a vector parallel to L2 is ..1, 0, 1 .. Thus we can .nd a vector n that is perpendicular to both lines, n = . 1, .1, .2 . ..1, 0, 1 . = x . 0 y . 3 z . 4
..1, 1, .1 .. Thus the equation of line through Q and parallel to n is ==.
.11 .1
1.6 (Planes in 3-space) Determine the line which is the intersection of the two planes x + y + z =2 and 2x . y + z = 6.
. Since the normal vectors . 1, 1, 1 . and . 2, .1, 1 . are not parallel, the planes are not parallel and must intersect. Their line of intersection L is parallel to the cross product . 1, 1, 1 . . 2, .1, 1 , . = . 2, 1, .3 .. To .nd a point on L, set x = 0 in the equations of planes: y+z =2 and .y + z = 6. After solving for y and z, we have a point (0, .2, 4) on L, and L ha