Solution
Robert answered on
Dec 25 2021
Name: Solutions to Practice Midterm
1.
(a) Are the vectors 〈1, 5, 3〉 and 〈−3, 0, 2〉 perpendicular?
(b) Are the points P = (1,−1, 1), Q = (2, 1,−1), R = (0, 3,−1), S = (−1, 0, 1) coplanar?
(c) Find the area of the triangle with vertices A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 2).
Solution.
(a) Their dot product is
−3 + 0 + 6 = 3,
which is not zero, therefore they are not perpendicular.
(b) We just need to check if the triple product |~a · (~b×~c)| of the vectors
−−→
PQ,
−→
PR and
−→
PS is zero.
We have
−−→
PQ = 〈1, 2,−2〉,
−→
PR = 〈−1, 4,−2〉,
−→
PS = 〈−2, 1, 0〉, and so
−→
PR×
−→
PS = 〈2, 4, 7〉 and
−−→
PQ · (
−→
PR×
−→
PS) = −4.
Since this is not zero, the 4 points are not coplanar.
(c) The area of the triangle is half the area of the parallelogram spanned by the vectors
−−→
AB =
〈−1, 1, 0〉 and
−→
AC = 〈−1, 0, 2〉. This area is the length of the vecto
−−→
AB ×
−→
AC = 〈2, 2, 1〉,
which is
√
4 + 4 + 1 = 3, so the area of the triangle is 3/2.
2.
(a) Find an equation for the plane through the points P = (1, 0, 1), Q = (1, 2, 2), R = (3, 1, 2).
(b) Find the intersection of the line
~r(t) = 〈1, 1, 0〉+ t〈3, 1, 1〉
with the plane x+ y + z = 1.
Solution.
(a) We need to find 2 things: a vector ~n normal to the plane, and a point on the plane. To do
this, we can just take ~n =
−−→
PQ×
−→
PR. We have
−−→
PQ = 〈0, 2, 1〉,
−→
PR = 〈2, 1, 1〉, so
~n =
−−→
PQ×
−→
PR = 〈1, 2,−4〉.
Since the plane passes through P = (1, 0, 1), we take this to be our point, and therefore the
plane has equation
(x− 1) + 2y − 4(z − 1) = 0.
1
(b) Write the line as
x = 1 + 3t, y = 1 + t, z = t.
To find the intersection points we substitute these into the equation for the plane x+y+z = 1
and solve for t:
1 + 3t+ 1 + t+ t = 1,
so t = −1/5, and the intersection point is
(1− 3/5, 1− 1/5,−1/5) = (2/5, 4/5,−1/5).
3. Consider the following two lines:
L1 : x =
y − 1
2
=
z − 1
3
L2 : 6(x− 1) = 3y + 1 = z.
(a) Show that L1 and L2 are skew.
Solution.
(a) We...