MATH1014 Calculus II, 2016-17 Spring
Week 12 Worksheet: Vectors.
Name:
(L11)
ID No.:
Tutorial Section:
1. Find out the length of the following vectors.
(a) a = (1, 4)
√
√
Solution: |a| = 12 + 42 = 17.
(b) a = (3, 5, 8) √
√
√
Solution: |a = 32 + 52 + 82 = 98 = 7 2.
(c) a = (2, −4, 0, 5) √
√
Solution: |a| = 22 + (−4)2 + 02 + 52 = 3 5.
−
→ −
→
−
→
2. Given forces F1 , F2 , find F3 such that when the three forces act on the same object, this object remains
stationary.
−
→ −
→ −
→ →
−
The object remains stationary when the forces cancels each other: F1 + F2 + F3 = 0
−
→
−
→
(a) F1 = 3i + 9j, F2 = −4i + 2j.
Solution:
→ →
−
→ →
→
−
→
−
−
→
−
→ −
−
→
−
−
3 i + 9 j − 4 i + 2 j + F3 = 0 =⇒ F3 = i − 11 j
−
→
−
→
(b) F1 = 6i − 3j, F2 = −4i + 8j.
Solution:
→ −
−
→
−
→
−
→
−
→
−
→ −
−
→
−
→
→
6 i − 3 j − 4 i + 8 j + F3 = 0 =⇒ F3 = −2 i − 5 j
3. Find the distance between the given points.
(a) P = (1, 0, 5), Q = (3, −4, 9)
√
−−→
−−→
Solution: P Q =< 3 − 1, −4 − 0, 9 − 5 >=< 2, −4, 4 >, then |P Q| = 22 + (−4)2 + 42 = 6.
(b) P = (26, 5, −7), Q = (22, −5, 0)
√
−−→
−−→
Solution:
P Q =< 22−26, −5−5, 0−(−7) >=< −4, −10, 7 >, then |P Q| = (−4)2 + (−10)2 + 72 =
√
165.
4. Find the distance of a point P to the line through the points A and B.
(a) P = (0, 0, 1), A = (1, 0, 0), B = (0, 1, 0)
−−→
Solution: The distance is |P P ′ | where P ′ is the projection of the point P onto the line AB. So
−−→′ −−→′ −→
P P = AP − AP
−→ −→
−
→ AP − AP
= proj−
AB
−−→ −→
AB · AP −−→ −→
= −−→ −−→ AB − AP
AB · AB
< −1, 1, 0 > · < −1, 0, 1 >
=
< −1, 1, 0 > − < −1, 0, 1 >
< −1, 1, 0 > · < −1, 1, 0 >
1 1
=< − , , 0 > − < −1, 0, 1 >
2 2
1 1
=< , , −1 >
2 2
Then,
−−→
|P P ′ | =
√
1
1 1
+ +1=
4 4
√
3
2
(b) P = (1, 0, 1), A = (1, 1, 0), B = (0, 1, 1)
−−→
Solution: The distance is |P P ′ | where P ′ is the projection of the point P onto the line AB. So
−−→′ −−→′ −→
P P = AP − AP
−→ −→
−
→ AP − AP
= proj−
AB
−−→ −→
AB · AP −−→ −→
= −−→ −−→ AB − AP
AB · AB
< −1, 0, 1 > · < 0, −1, 1 >
=
< −1, 0, 1 > − < 0, −1, 1 >
< −1, 0, 1 > · < −1, 0, 1 >
1
1
=< − , 0, > − < 0, −1, 1 >
2
2
1
1
=< − , 1, − >
2
2
Then,
−−→
|P P ′ | =
√
1
1
+1+ =
4
4
√
3
2
5. Find the projection of the vector u onto v – that is projv u. And then find out the angles between the
two vectors.
(a) u = (1, 0, 5), v = (0, 10, 3)
Solution:
−
→
−
15
75
v ·→
u−
< 0, 10, 3 > · < 1, 0, 5 >
→
−
→
−
proj→
< 1, 0, 5 >=<
, 0,
>
v =
v u = −
→
→
−
< 1, 0, 5 > · < 1, 0, 5 >
26
26
v · v
(b) u = (−1, 1, 5), v = (−1, 2, 3)
Solution:
→
−
→
9 18 27
v ·−
u−
< −1, 1, 5 > · < −1, 2, 3 >
−
→
→
−
proj→
< −1, 2, 3 >=< − , ,
>
v =
v u = →
−
−
< −1, 2, 3 > · < −1, 2, 3 >
7 7 7
v ·→
v
6. Calculate the cross product of the vectors given in the previous question.
(a) u = (1, 0, 5), v = (0, 10, 3)
Solution:
−
→
→
u ×−
v =
→
−
→ −
→ −
i
j
k
1 0
5
0 10 3
−
→
−
→
−
→
= −50 i − 3 j + 10 k
(b) u = (−1, 1, 5), v = (−1, 2, 3)
Solution:
−
→
→
u ×−
v =
−
−
→ −
→ →
i
j
k
−1 1
5
−1 2
3
2
→
−
→
−
→ −
= −7 i − 2 j − k