32A-3 Practice Midterm 1 Solution

Name: , UID: , Section:

I read, understood, and will follow the instructions given below.

Signature:

April 13, 2015

Instructions

This is a 50-minute closed-book exam. The textbook, lecture notes, and all personal scratch papersare not allowed during the test. Discussing or cheating is strictly prohibited. The use of electronic devicesis also not allowed. To receive the full credit, you must explain each step thoroughly. You arenot allowed to use your own scratch paper. There are blank pages at the end of the exam. You can usethe blank pages or back side of any page, but please make a note when you do so.

Problem Points

1

2

3

4

5

Total

Problem 1 Find the following where i = 1, 0, 0, j = 0, 1, 0, and k = 0, 0, 1:(a)

(i+ j) (j+ k). (6 pts)Solution.This is the dot product of two vectors 1, 1, 0 and 0, 1, 1. Their dot product is 1 0 + 1 1 + 0 1 = 1.(b)

(i+ j) (j+ k) (6 pts)Solution.The cross product of two vectors 1, 1, 0 and 0, 1, 1 is

i j k

1 1 0

0 1 1

=1 01 1

i1 00 1

j+1 10 1

k = i j+ k = 1,1, 1.Problem 2 Let v = 3, 5 and w = 1, 2.

(a) Show that v and w are not colinear. (6 pts)Solution .We find the cross product of 3, 5, 0 and 1, 2, 0. This is

i j k

3 5 0

1 2 0

=3 51 2

k = k = 0, 0, 1.Since the result is a nonzero vector, the two vectors are not colinear.

(b) Find a and b in the expression 1, 1 = av + bw. (6 pts)

Solution.Checking componentwise, we have 1 = 3a + b, and 1 = 5a + 2b. This gives 2a + b = 0. Then b = 2a.

Then 1 = 3a 2a, a = 1, and b = 2a = 2.

Problem 3 Find an equation for the intersection of two planes x y + z = 1 and 2x 3y + 4z = 2.

Solution.The intersection is a line which is contained in both planes. Therefore the direction vector v of the

line must be perpendicular to the normal vectors of two planes. A normal vector n1 for the first plane, is1,1, 1. The second one is n2 = 2,3, 4. Then we can take v = n1 n2, which is

i j k

1 1 12 3 4

=1 13 4

i1 12 4

j+1 12 3

k = (1)i 2j+ (1)k = 1,2,1.We need a point that the line is through. Set z = 0, and find x, y. We have x y = 1 and 2x 3y = 2.Then 2x 2y = 2 gives y = 0 and x = 1. Thus, we use (1, 0, 0). We can also use v = 1, 2, 1 as adirection vector. Then

r(t) = 1, 0, 0+ t1, 2, 1.

Problem 4 Suppose that you have two lines r1(t) = 0, 1, 0+ t1, 1, 0 and r2(s) = 0, 1, 1+ s1,1, 1(a) Find the intersection of the two lines. (6 pts)Solution.Set up the equation r1(t) = r2(s) componentwise, we have t = s, 1 + t = 1 s, and 0 = 1 + s. This

does have a solution t = 1, s = 1. Therefore, the intersection is

r1(1) = 0, 1, 0+ 1, 1, 0 = 1, 2, 0.

(b) Find an equation of a plane which contains both lines. (6 pts)Solution.From (a), we have a point 1, 2, 0 on the plane. We need a normal vector of the plane. Since both lines

are contained in the plane, the direction vectors v1 = 1, 1, 0 and v2 = 1,1, 1 are both perpendicularto the normal vector n of the plane. We can take n = v1 v2.

i j k

1 1 0

1 1 1

= 1 01 1

i 1 01 1

j+ 1 11 1

k = 1i 1j+ 0k = 1,1, 0.An equation of the plane is thus,

1(x 1) + (1)(y 2) + 0(z 0) = 0

orx y = 1.

Problem 5 Let r(t) = t, t2, t4 1, t, t2. Find a parametrization of the tangent line at t = 1.

Solution .We need r(1) as a direction vector of the tangent line, and r(1) as a point the line is through.By the product rule,

r(t) = t, t2, t4 0, 1, 2t+ 1, 2t, 4t3 1, t, t2.

Thus, r(1) = 1, 1, 1 0, 1, 2+ 1, 2, 4 1, 1, 1 = 1, 1, 1 (0, 1, 2 1, 2, 4).This is 1, 1, 1 1,1,2

i j k

1 1 1

1 1 2

=1 11 2

i1 11 2

j+1 11 1

k = 1i 1j+ 0k = 1,1, 0,r(1) = 1, 1, 1 1, 1, 1 gives 0, 0, 0.

Then an equation of the tangent line is

l(s) = 0, 0, 0+ s1,1, 0

orl(s) = s,s, 0.