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Mathematics

MATH 125

Scott

Fall

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MATH 125 (R1) Winter 2017
Sample Midterm Exam { Solutions
Note: Row reduction of matrices is omitted from the solutions below. In your exam, however, you
should show all you work, indicating which row operations were applied at each step in the process.
1. Solve the following system of linear equations by ▯rst forming its augmented matrix and then
bringing it to its reduced row echelon form. Give your answer in vector form.
x1 + x2 + 5x 4 = ▯4
2x2 + 4x 3 + 2x 4 = ▯2
▯x + 2x ▯ 4x = 3
1 3 4
Solution: Applying the row reduction algorithm, we ▯nd that the reduced row echelon form
of the system is 2 3
1 0 ▯2 4 ▯3
6 7
4 0 1 2 1 ▯1 5
0 0 0 0 0
The leading variables are the1efore2x and x , and the free 3ariab4es are x and x . Assigning
the parametric values s and t to the free 3ariab4es x and x , respectively, and re-writing the
above matrix as a system of equations, we see that the general solution of the system is
x1 = ▯3 + 2s ▯ 4t
x2 = ▯1 ▯ 2s ▯ t
x3 = s (s;t 2 R);
x4 = t
or, in vector form,
2 3 2 3 2 3
▯3 2 ▯4
6 7 6 7 6 7
x =6 ▯1 7 + s6 ▯2 7 + t6 ▯1 7 (s;t 2 R):
4 05 4 15 4 0 5
0 0 1
2 ▯! ▯!
2. a) Let A = (2;3) and B = (▯1;4) be points in R . Compute the length jjABjj of AB.
n
b) Letu,v be orthogonal unit vectors in R . Compute theu +vjj ou +v.
▯!
Solution a) AB = [▯1 ▯ 2;4 ▯ 3] = [▯3;1], so
▯! p 2 2 p
jjABjj = (▯3) + 1 = 10: b) To say thau and v are orthogonal unit vectors means tujj = vjj = 1 andu v = 0.
Now
2 2 2 2 2
ju +~vjj = ujj + 2(u v) + jvjj = 1 + 2(0) + 1 = 2;
and so p
ju +~vjj = 2:
3. Consider the plane P in R with general equation 2x + y ▯ 3z = 1.
3
a) Find a point in R that does not lie on P. Justify your answer.
~
b) Show that the vector d is parallel to P, where
2 3
▯5
~ 6 7
d = 4 4 5 :
▯2
c) Find a vector equation of a line ‘ in R which is parallel to P, but not contained in P.
Solution: a) We need to ▯nd a point (a;b;c) with the property that 2a + b ▯ 3c 6= 1. The
point (0;0;0) will do, since 2(0)+0▯3(0) = 0 6= 1. Of course, this is just one example { there
are in▯nitely many points which don’t lie on P.
2 3
6 2 7
b) Note ▯rst that the vecn = ~4 1 5 is a normal vector to P (we just read this o▯ the
▯3
given equation for P). Now, to say that d is parallel to P is equivalent to saying that d is
orthogonal to the normal vecn. But
2 3 2 3
6 ▯5 7 6 27
d ▯n = 4 4 5 ▯4 15 = ▯5(2) + 4(1) ▯ 2(▯3) = ▯10 + 4 + 6 = 0;
▯2 ▯3
~ ~
so d andn are indeed orthogonal, and hence d is parallel to P.
c) There are arbitrary many lines with these properties. Di▯erent choices of a point on the line
or a direction vector to the line will give di▯erent answers. We give here the natural solution
based on parts a) and b) above.2Let ‘3be the line which passes through the origin (0;0;0) and
▯5
which is parallel to the vector d =4 7. Then ‘ is parallel to P by part b), but is not
4 5
▯2
contained in P by part a).
A vector equation of ‘ is given by: 2 3 2 3
6 0 7 6 ▯5 7

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