UBC Mathematics 152 Section 206
Midterm Exam I (Feb 10, 2012)
Instructor: Yue-Xian Li
Answers
Duration: 50 min
I.
(30 marks)
Consider the plane P defined by x − 2y + z = 1. (No intermediate step required in
(a), (b), (d)).
(a) Find its normal direction n.
(b) Find the point of intersection between P and the x-axis.
(c) Show that the point p= [1 0 − 1] is not on the plane P .
(d) Write the equation of the line connecting P and p that is perpendicular to
P in parametric form.
(e) Find the point of intersection between the line in (d) and P .
(f) Find the shortest distance between P and p.
Solutions:
(a) n= [1 − 2 1].
(b) [1 0 0].
(c) Substitute p= [1 0 − 1] into the equation for P , we obtain lhs = 1 − 1 = 0 6=
1 = rhs. Thus, it is not on the plane.
(d) Using the point-direction formula, we obtain
x = p + tn = [1 0 − 1] + t[1 − 2 1].
(e) Substitute x = t + 1, y = −2t, z = t − 1 into the equation for P :
t + 1 + 4t + t − 1 = 1
⇒
t = 1/6.
Thus, the point of intersection is [7/6 − 2/6 − 5/6].
(f)
p The shortest distancepis k[7/6 − 1, −2/6, −5/6 + 1]k = k[1/6, −2/6, 1/6]k =
1/36 + 4/36 + 1/36 = 1/6.
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II.
(a)
(b)
(c)
(d)
(30 marks)
Find the equation form of the plane P with normal vector n = [0 1 − 1] and
passing through the point p = [0 1 0].
Find the equation of the line L that is perpendicular to the plane Q with
equation x − y + z = 1 and passing through the origin.
Find the system of linear equations that determines the point of intersection
between the plane P and the line L.
Use Gaussian elimination to solve the system in (c) to find the point of
intersection.
Solution:
(a) The equation form of the plane P is
n·x=n·p
⇒
y − z = 1.
(b) The normal direction of the the plane Q is q = [1 − 1 1]. Using point-direction
formula
x = 0 + tq = t[1 − 1 1].
We can use these to find the equation form of the line L. In this case we obtain
x+y =0
x − z = 0.
(c) The intersection of the plane P and the line L is the solution of the system of
equation
x+y =0
x−z =0
y − z = 1.
(c) The augmented matrix associated to this system is


1 1 0 | 0
1 0 −1 | 0
0 1 −1 | 1
Applying Gaussian elimination we obtain

1 1 0
0 1 1
0 0 −2
the echelon form

| 0
| 0
| 1
We can use this system using backwards substitution. In this case we get z =
−1/2, y = 1/2, x = −1/2. Thus the intersection between P and L is the point
− 21 [1 − 1 1].
III. These questions do not require lengthy calculations. Put your answers in
the boxes.
(20 marks, 4 for each)
(a) Find the direction of the line of intersection between the following two planes
x − y + z = 1 and x + y − 2z = 5 (do not find the equation of this line).
n1 = [1 − 1 1], n2 = [1 1 − 2], l = n1 × n2 = [1 − 1 1] × [1 1 − 2] = [1 3 2].
(b) Determine if the vectors a = [1 0 1], b = [1 2 0] and c = [1 1 1] are linearly
independent or linear dependent.


1 0 1
a · (b × c) = det 1 2 0 = 1 6= 0. They are linearly independent.
1 1 1
(c) Calculate the volume of
and c = [1 1 1].

1 0
a · (b × c) = det 0 2
1 1
the parallelepiped with edges a = [1 0 1], b = [0 2 1]

1
1 = −1. Thus, the volume is |a · (b × c)| = | − 1| = 1.
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(d) Find the projection of a = [1 1 1] on b = [0 3 4] , i.e. Projb a.
Projb a=
a·b
b
kbk2
=
7
25 [0
3 4].
(e) Given the augmented matrix of a system of linear
1
0
1 |
echelon form and find all solutions.  1 −2 3 |
−2 2 −4 |
equations,
find the reduced

−2
2 .
0

 



 
   
1 0 1 | −2
1 0 1 | −2
x1
−1
2
1 0 1 | −2
1 −2 3 | 2 → 0 −2 2 | 4  → 0 1 −1 | −2 ⇒ x2  = α  1  − 2 .
0 0 0 | 0
0 0 0 | 0
0 0 0 | 0
x3
1
0