Fall 2022 32A-4 First Midterm
Full Nam e:
UIN:
Circle the nam e of your TA and the day of your discussion:
Gurkiran Dhaliwal Benjamin Major Advika Rajapakse
Tuesday Thursday
Instrucitons
• Read each Problem carefully.
• To get credit for a Problem, you must show all of your reasoning and calculations.
• Box your final answer.
• No calculators are allowed.
• You are allowed to use your formula sheet. The sheet should be hand-written, one-
sided, and of letter size.
• You may use the blank pages 3, 5, 7, 9, 11 for scratch work. Work found on those
pages will not be graded unless clearly indicated in the exam.
• Th er e w ill be 6 Pr oblem s in the exam . Problem 6 is True or False.
Problem Points Scores
1 15
2 10
3 10
4 10
5 10
6 10
Total 65
1
, Pr oblem 1 (15 points). Consider the 4 points
P = (−2, 0, 3), Q = (−1, −1, 4), R = (−1, 1, 2), S = (−3, 1, 4).
1. (5 points) Find a vector parametrization for the line L passing through P and Q.
2. (5 points) Find a vector parametrization for the line M passing through R and S.
3. (5 points) Determine whether the line L (of Part 1) and the line M (of Part 2) are
parallel, intersecting at a single point, or skew.
−→
Solutions. 1. The line L passes through P = (− 2, 0, 3) and has direction of P Q⟨ =−1, 1,
1. ⟩
A vector parametrization is given by
r⃗(t) = ⟨−2, 0, 3⟩ + t⟨1, −1, 1⟩ (−∞ < t + ∞).
2. The line passes through
M − ⟨−
⟩ of R−→S =
R = ( 1, 1, 2) and has direction 2, 0, 2 . A
vector parametrization is given by
s⃗(t) = ⟨−1, 1, 2⟩ + t⟨−2, 0, 2⟩ (−∞ < t < +∞).
3. To determine whether L, M are parallel, we ask if the direction vectors ⟨1, −1, 1⟩, ⟨−2, 0, 2⟩
of L, M are parallel, or equivalently, if the following equation can be solved for λ:
⟨1, −1, 1⟩ = λ⟨−2, 0, 2⟩.
Comparing the x, y, z components of the above equation yields three equations:
1 = −2λ (1)
−1 = 0 (2)
1 = 2λ. (3)
Since (2) is impossible, L, M are NOT parallel.
To determine whether L, M intersect, we ask if r⃗(t1) = s⃗(t2) can be solved for (t1, t2).
Comparing the x, y, z components of the equation yields three equations:
−2 + t1 = −1 − 2t2 (4)
−t1 = 1 (5)
3 + t1 = 2 + 2t2. (6)
Solving (5) for t1 yields:
t1 = −1. (7)
Substituting (7) to (4) and solving it for t2 yield t2 = 1, while substituting (7) to (6) and
solving it for t2 yield t2 = 0. This is a contradiction, hence L, M do NOT intersect.
We conclude that L, M are skew.
2
Full Nam e:
UIN:
Circle the nam e of your TA and the day of your discussion:
Gurkiran Dhaliwal Benjamin Major Advika Rajapakse
Tuesday Thursday
Instrucitons
• Read each Problem carefully.
• To get credit for a Problem, you must show all of your reasoning and calculations.
• Box your final answer.
• No calculators are allowed.
• You are allowed to use your formula sheet. The sheet should be hand-written, one-
sided, and of letter size.
• You may use the blank pages 3, 5, 7, 9, 11 for scratch work. Work found on those
pages will not be graded unless clearly indicated in the exam.
• Th er e w ill be 6 Pr oblem s in the exam . Problem 6 is True or False.
Problem Points Scores
1 15
2 10
3 10
4 10
5 10
6 10
Total 65
1
, Pr oblem 1 (15 points). Consider the 4 points
P = (−2, 0, 3), Q = (−1, −1, 4), R = (−1, 1, 2), S = (−3, 1, 4).
1. (5 points) Find a vector parametrization for the line L passing through P and Q.
2. (5 points) Find a vector parametrization for the line M passing through R and S.
3. (5 points) Determine whether the line L (of Part 1) and the line M (of Part 2) are
parallel, intersecting at a single point, or skew.
−→
Solutions. 1. The line L passes through P = (− 2, 0, 3) and has direction of P Q⟨ =−1, 1,
1. ⟩
A vector parametrization is given by
r⃗(t) = ⟨−2, 0, 3⟩ + t⟨1, −1, 1⟩ (−∞ < t + ∞).
2. The line passes through
M − ⟨−
⟩ of R−→S =
R = ( 1, 1, 2) and has direction 2, 0, 2 . A
vector parametrization is given by
s⃗(t) = ⟨−1, 1, 2⟩ + t⟨−2, 0, 2⟩ (−∞ < t < +∞).
3. To determine whether L, M are parallel, we ask if the direction vectors ⟨1, −1, 1⟩, ⟨−2, 0, 2⟩
of L, M are parallel, or equivalently, if the following equation can be solved for λ:
⟨1, −1, 1⟩ = λ⟨−2, 0, 2⟩.
Comparing the x, y, z components of the above equation yields three equations:
1 = −2λ (1)
−1 = 0 (2)
1 = 2λ. (3)
Since (2) is impossible, L, M are NOT parallel.
To determine whether L, M intersect, we ask if r⃗(t1) = s⃗(t2) can be solved for (t1, t2).
Comparing the x, y, z components of the equation yields three equations:
−2 + t1 = −1 − 2t2 (4)
−t1 = 1 (5)
3 + t1 = 2 + 2t2. (6)
Solving (5) for t1 yields:
t1 = −1. (7)
Substituting (7) to (4) and solving it for t2 yield t2 = 1, while substituting (7) to (6) and
solving it for t2 yield t2 = 0. This is a contradiction, hence L, M do NOT intersect.
We conclude that L, M are skew.
2
