Worksheet 3
8/4/2014
Please work in groups and use the given time to think and try to do the problems. If you cannot
solve a question please ask me for a hint or just pass that question.
If v = ha1 , b1 , c1 i and w = ha2 , b2 , c2 i then
i j k a1 b1 a1 c1 b1 c1 k
j+
i−
v × w = a1 b1 c1 = a2 b2 a2 c2 b2 c2 a2 b2 c2 Here the bars mean take the determinant, and
a b
c d
the determinant of a 2 × 2 matrix is
= ad − bc
1. Calculate the determinant:
5 10
9 −8
=
Solution: (5)(-8)+(10)(9)=-40+90=50
2. Find the cross product of v = h1, 2, 1i and w = h3, 1, 1i. Verify that v and v × w are
orthogonal (Show v · (v × w) = 0).
Solution:
i j k 2 1 1 1 1 2 1 2 1 =
1 1 i − 3 1 j + 3 1 k
3 1 1 = (2 − 1)i − (1 − 3)j + (1 − 6)k
= i + 2j − 5k = h1, 2, −5i
3. If kf k = 5, kgk = 10 and the angle between the vectors is π/6, find kf × gk
Solution:
π kf × gk = kfkkgk sin
6
1
= 5 · 10 ·
2
= 25
4. Calculate (−3u + 6w) × w assuming that u × w = h3, −8, 7i.
Solution:
(−3u + 6w) × w = −3u × w + 6w × w
= −3(u × w) + 6(w × w)
= −3 h3, −8, 7i + 6 h0, 0, 0i
= h−9, 24, −21i
1
5. Find the area of the parallelogram spanned by u = h1, 0, 3i and v = h2, 1, 1i. (For parallelograms, the area is given by ku × vk.)
Solution:
i j k 0 3 1 3 1 0 1 0 3 =
1 1 i − 2 1 j + 2 1 k
2 1 1 = (0 − 3)i − (1 − 6)j + (1 − 0)k
= −3i + 5j + k = h−3, 5, 1i
k h−3, 5, 1i k =
p
√
√
(−3)2 + 52 + 12 = 9 + 25 + 1 = 35
6. If a = i + 14j and b = i + 8j + 4k, find a unit vector with positive first coordinate orthogonal
to both a and b.
Solution: We know a × b is orthogonal to both a and b. So first we find a × b and then
find the unit vector ea×b , which has a positive first coordinate.
i j k 1 14 0 = 14 0 i − 1 0 j + 1 14 k
8 4 1 4 1 8 1 8 4 = (56 − 0)i − (4 − 0)j + (8 − 14)k
= 56i − 4j − 6k = h56, −4, −6i
k h56, −4, −6i k =
ea×b =
D
p
√
(56)2 + (−4)2 + (−6)2 = 3188
√ 56 , √−4 , √−6
3188
3188
3188
E
7. Calculate the force (in newtons) required to push a 40-kg wagon up a 10◦ incline. Solution:
We want to decompose the downward gravity vector into its part that is in the direction of
the incline and the direction perpendicular to that. So we project the gravity force vector f
onto the incline. Since all we want is the magnitude of this projection we have:
kfk k = (40)(9.8) cos(80◦ )
kfk k = 392 cos(80◦ )
kfk = 68.07N
8. Compute the component and find the projection of u = 5i + 7j + k along v = k.
Solution: u = h5, 7, 1i and v = h0, 0, 1i. Notice that v is a unit vector.
uk = (u · ev )ev
= (u · v)v
= (0 + 0 + 1) h0, 0, 1i
= h0, 0, 1i
2
9. Let a = h1, −3, 1i and b = h4, 1, 1i, find the decomposition a = ak + a⊥ with respect to b.
Solution:
a·b
b
b·b
4 + −3 + 1
=
h4, 1, 1i
16 + 1 + 1
1
= h4, 1, 1i
9
4 1 1
, ,
=
9 9 9
ak =
a⊥ = a − ak
= h1, −3, 1i −
5 −28 8
=
,
,
9 9 9
4 1 1
, ,
9 9 9
−28
8
0
a⊥ are
Notice b · a⊥ = 20
9 + 9 + 9 = 9 = 0, so
b
and
orthogonal, as they should be.
5 −28 8
4 1 1
We get the decomposition: a = 9 , 9 , 9 + 9 , 9 , 9 .
3