Section 14.2



1. If u ×v = 2u, what can one conclude about u ? About v ? Explain.
Solution Note that the vector, u ×v, is perpendicular to both u and v. Therefore, 2 u·u = 0 and 2 u ·v = 0 which implies that u = 0 (if u ·u = 0, u must be a zero vector).
3. Use the dot product to derive the law of cosine,
c2 = a2 + b2 − 2 a b cos θ,
where c, a, b, θ are shown in the figure. HINT : Regard the sides of the triangle as vectors. oriented such that C = a −b, and note that C ·C = (a − b) ·(a − b).
Solution Note that (a − b) ·(a − b) = a2 + b2−2 a ·b = a2 + b2− 2 a b cos θ which is already the law of cosine.
5. Derive the Lagrange identity
|| u ×v||2 = ||u||2||v||2 − (u ·v)2.
Solution Remember that S ≡ || u ×v || represents the area of a parallelogram spanned by u and v. Therefore, S2 = u2 v2 sin 2 θ = u2 v2 ( 1− cos 2 θ) = u2 v2 − (u v cos θ)2 = ||u||2 ||v||2 −(u·v)2.
8. Is u ×(v ×w) necessarily equal to (u ×v) ×w ? Prove or disprove ?
Solution No. Expand the both sides using the vector triple product identity and you will see they are not equal each other.


Section 14.3



3. Use a suitable cross product to determine whether or not the following points lie on a straight line

(a) (2,3,1), (1,-1,4), (2,2,1).
Solution Try computing (b−a) ×(c−a) to see if it is zero (if two vectors are parallel, the angle between the two vectors is 0). In this case a=(2,3,1), b=(1, −1, 4), c=(2, 2, 1) and b−a = (−1, −4, 3), c− a = (0, −1, 0). Therefore, (b−a) ×(c−a) = (3,0,1). Thus they don't lie on a straight line.
4. Use a suitable property of the cross product to fine the area of the triangle with the following vertices.

(a) (1,4,3), (2,0,-1), (0,0,5)
Solution (a) a = (1,4,3), b = (2,0,−1), c=(0,0,5). Compute || (b−a) ×(c−a)||. S = (1,−4, −4) ×(−1, −4, 2) = (−24, 2, −8). Thus, the absolute value is √{(−24)2 + 22 + (−8)2} = 2√{161}. The area is
√{161}
.
5. (a) Determine a unit normal vector to the plane x + 2y − z = 5 by finding three distinct (noncollinear) point A, b, c in the plane, crossing ab with aC, and normalizing.
Solution (a) A bit simpler method than the given hint is to rewrite the equation as a·x = 5 where a=(1,2,−1). Normalizing this vector yields n = (1,2,−1)/√6.
7. Solution (a) Divide both sides by √{14} so that the equation of the plane is expressed as n·x = 4/√{14} where n = (2/√{14}, 1/√{14}, −3/√{14}). Thus, according to the remark, the shortest distance is
4/√{14}
.
8. The vector u = (1, 1, 2) and v = (3, 2, −1) determine a plane (by their span). Find a nonzero vector in that plane that is perpendicular to w = (2, 4, 3). As usual, explain your reasoning.
Solution Note that an arbitrary vector on the plane spanned by the two vecors, u and v, is expressed as
x = αu + βv,
where α and β are some numbers (scaling). The numbers, α and β, can be chosen so that x is perpendicular to w, i.e.
x ·w = 0,
or
αu ·w + βv ·w = 0.
Using
u ·w = 12,    v ·w = 11,
the above equation becomes
12 α+ 11 β = 0,
so for example (by setting α = 1)
x = u − 12
11
 v = ( −25
11
 , −13
11
 , 34
11
 ),
which is perpendicular to w (verify!).


Section 14.4



1.
If
u =
^
i
 
 −
^
j
 
 ,     v = 2
^
i
 
 +
^
j
 
 + 3
^
k
 
 ,    w =
^
j
 
 −
^
k
 
 ,
evaluate
(a) u ·v ×w and u ×v ·w
Solution (a) v ×w = (−4,2,2) and u ×v = (−3,−3,3). u ·(v ×w) = −6 and (u ×v)·w = −6.
4. Find the volume of the parallelepiped having the following vectors as adjacent edges.
(d)

3
^
j
 
 , 2
^
i
 
 −
^
j
 
 ,
^
i
 
 + 5
^
k
 
 .
Solution (d) u = 3 j, v=2i−j, w=i+5k. The volume is V = (u ×v)·w = −30. Take the absolute value, i.e. 30.
5. Show that (u ×v) ×w = (w ·u)v− (w ·v)u You may use(9).
Solution (u ×v) ×w = − w ×(u ×v) = − ( (w·v) u − (w ·u) v).
9. Prove the following identities involving quadruple products. HINT: Use (7) and (9).

(a) (a ×b) ·(C ×D) = (a ·C)(b ·D) − (a ·D)(b ·C) This is Lagrange's identity, of which equation (5,1) of Exercise 5 in Section 14.2 is a special case.

(b) (a ×b) ×(C ×D) = (a ·b ×D)C − (a ·b ×C)D
Solution (a)
(A ×B)·(C ×D) (1)
=

 
 C·( D ×(A ×B)) (2)
=

 
 C ·( (B ·D) A − (D ·A) B) = (B ·D) (C ·A) − (D ·A) (C ·B)
. Notes:(1) Scalar triple product. (2) Vector triple product.
(b) (A ×B) ×(C ×D) = ( (A ×B) ·D) C −((A ×B) ·C) D = (A ·(B ×D)) C −(A ·(B ×C)) D.


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On 01 Oct 2022, 10:49.