Math 152 Test #1 version B
B2: The points the A = (1, 5, 2), B = (0, 2, 3), and C = (1, 1, 1) are the
vertices of a triangle.
(a) [2 marks] Find a normal vector to the plane that contains the
triangle.
(b) [1] What is the area of the triangle?
(c) [1] Write an equation form for the plane P that contains the triangle.
(d) [1] Is the point ( 2, 5, 7) on the plane P from part (c) above?
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Test #1, Version B, Math
B2 (a) To find a normal vector to the plane containing the triangle, we can find two linearly independent
vectors in that plane, and take their cross product.
Any two vertices of the triangle will give us a vector in the plane. For example:
AC: (1, 5, 2) (1, 1, 1) = (0, 4, 3) and
BC: (0, 2, 3) (1, 1, 1) = ( 1, 1, 2) 2

3
4
3
7
2
3 6 det 1 2
2 3
6
7

i
j k
11
6
7
0
3
7= 435
35 = 6
AC ⇥ BC = det 4 0 4
det
6
7
6
 1 2 7
1 1 2
4
4
0 4 5
det
1 1
Other answers are possible: they will be a scalar multiple of this vector.
(b) Given two vectors a, b in R3 , the quantity ka ⇥ bk gives the area of the parallelogram with a and b as
two of 2its sides.
So, the area of the triangle with those two sides will be 12 ka ⇥ bk. That is,
3
11
p
1p
1 4 5
3
= 12 112 + 32 + 42 =
146 .
2
2
4
(c) A plane with normal vector [a, b, c] will have equation ax+by+cz = d for some constant d. So, our plane
will have equation 11x + 3y + 4z = d for some constant d. Since we know points on the plane, each of them
will have to fulfill the equation. So d = 11(1)+3(1)+4(1) = 18. (Equivalently: d = 11(1)+3(5)+4( 2) = 18,
or d = 11(0) + 3(2) + 4(3) = 18.)
So the equation of the plane is 11x+3y+4z=18 (or a scalar multiple of this)
(d) To test whether the point ( 2, 5, 7) is on the plane, we plug in x = 2, y = 5, and z = 7 and see
whether we get 18.
11( 2) + 3(5) + 4(7) = 21 6= 18
so no , it is not on the plane.
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TTh Version 2. Consider the line L in R2 that passes through the points (4, 3) and
(8, 6).
(a) [1 mark] Write a parametric form for L.
(b) [1 mark] Find a vector n perpendicular to L.
(c) [1 mark] Write an equation form for L.
(d) [1 mark] Find the projection of (1, 1) in the direction n from part (b).
(e) [1 mark] Find the smallest distance from the point (1, 1) to points on the line L.
Solution: (a) This line passes through a = (4, 3) and has direction vector b = (8, 6)
(4, 3) = (4, 3). The parametric for of this line is
(x, y) = a + tb = (4, 3) + t(4, 3) .
If we replace t by s
1, we can simplify this to (x, y) = s(4, 3) or equivalently,
x = 4s
y = 3s .
(b) n = (3, 4). Check: n · b = (3, 4) · (4, 3) = 0.
(c) n · ((x, y)
a) = 0. Equivalently, (3, 4) · (x
3(x
4, y + 3) = 0 or
4) + 4(y + 3) = 0 .
Multiplying out, we obtain 3x 12 + 4y + 12 = 0 or 3x + 4y = 0.
Answer: 3x + 4y = 0.
To check that this works, note that both points given in the problem, (4, 3) and (8, 6),
satisfy this equation.
(d) Projn (1, 1) =
(1, 1) · n
3+4
21 28
n= 2
(3, 4) = ( , ).
n·n
3 + 42
25 25
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(e) The smallest distance is the length of the vector computed in part (d), i.e., ||(3, 4)|| =
25
7p 2
7
2
3 +4 = .
25
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