,Web Instructor’s Manual — Andrilli/Hecker, 5th Ed. Lines and Planes
Lines and Planes and the Cross Product in R3
(1) (a) = 3 = −1 + = −4 ( ∈ R)
(b) = −4 − 2 = 2 − 3 = 8 ( ∈ R)
(c) = 6 − 2 = 2 − 5 = 1 + 6 ( ∈ R).
(Another valid form: = 4 + 2 = −3 + 5 = 7 − 6 ( ∈ R))
(d) = 4 − = −2 = 9 + 6 ( ∈ R).
(Another valid form: = 3 + = −2 = 15 − 6 ( ∈ R))
(e) = 1 − 2 = −5 + = −7 ( ∈ R)
(f) = −2 − 3 = 4 = −5 − ( ∈ R)
(2) (a) The lines intersect at a single point: (0 1 7)
(b) The lines intersect at a single point: (8 −5 10)
(c) The lines do not intersect.
(d) The lines do not intersect.
(e) The lines are identical, so the intersection of the lines is the set of all points on (either) line; that
is, the intersection consists of all points on the line = 4 − 7 = 3 + 2 = − 5 ( ∈ R).
(f) The lines are identical, so the intersection of the lines is the set of all points on (either) line; that
is, the intersection consists of all points on the line = 3 − 2 = + 1 = 2 − 7 ( ∈ R).
(3) (a) = 90◦ (b) = 45◦ (c) = 60◦ (d) = 150◦
(4) (a) There are two parts to the proof. First, we show that if a point lies on the given line , then
it satisfies the symmetric equations for the line . The given line has the parametric equations
= 0 + = 0 + and = 0 + Suppose (1 1 1 ) is a point on . Then 1 = 0 +
1 = 0 + and 1 = 0 + , for some value of . Solving for this value of in each expression,
we find that:
1 − 0 1 − 0 1 − 0
= = =
and thus (1 1 1 ) satisfies the symmetric equations for .
To finish the proof, we show that if a point satisfies the symmetric equations for , then that
point lies on . Suppose that (1 1 1 ) is a point satisfying the symmetric equations
1 − 0 1 − 0 1 − 0
= =
Let this common value be , and then solve for to find that
1 = 0 + 1 = 0 + and 1 = 0 +
Thus, (1 1 1 ) lies on .
−6 −2 −1 −4 +3 −7
(b) −2 = −5 = 6 (Another valid form: 2 = 5 = −6 )
+2 +5
(c) −3 = 4 = −1
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Copyright ° 1
,Web Instructor’s Manual — Andrilli/Hecker, 5th Ed. Lines and Planes
(5) (a) 6 + + 6 = 1 (c) 9 − 2 = −35 (e) = 0
(b) 7 + 3 − 3 = −6 (d) = 0
£2 1 2
¤ £1 4 8
¤
(6) (a) 3 −3 3 (b) 9 9 −9
h √ √ √ i
4 21 −2 21 − 21
(c) 21 21 21 ≈ [0872 −0436 −0218]
(7) (a) = 60◦ (b) = 45◦ (c) = 90◦
(8) (a) [7 −3 1] (d) [−1 1 −4] (g) −8
(b) [2 4 −5] (e) [0 0 0] (h) 0
(c) [−1 2 −3] (f) [0 0 0] (i) [0 0 0]
(9) In what follows, let x = [1 2 3 ], y = [1 2 3 ], and z = [1 2 3 ].
(a) (x) × y
= [(2 )3 − (3 )2 (3 )1 − (1 )3 (1 )2 − (2 )1 ]
= [2 (3 ) − 3 (2 ) 3 (1 ) − 1 (3 ) 1 (2 ) − 2 (1 )]
= x × (y)
(b) (x) × y
= [(2 )3 − (3 )2 (3 )1 − (1 )3 (1 )2 − (2 )1 ]
= [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= (x × y)
(c) x × 0 = [2 (0) − 3 (0) 3 (0) − 1 (0) 1 (0) − 2 (0)] = [0 0 0] = 0
(d) x × x = [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ] = [0 0 0] = 0
(e) x × (y + z)
= [2 (3 + 3 ) − 3 (2 + 2 ) 3 (1 + 1 ) − 1 (3 + 3 )
1 (2 + 2 ) − 2 (1 + 1 )]
= [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
+ [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= (x × y) + (x × z)
(f) y · (x × y)
= [1 2 3 ] · [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= 1 2 3 − 1 3 2 + 2 3 1 − 2 1 3 + 3 1 2 − 3 2 1
= (1 2 3 − 3 2 1 ) + (−1 3 2 + 2 3 1 )
+ (−2 1 3 + 3 1 2 )
= 0+0+0 = 0
(g) (x × y) · z
= [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ] · [1 2 3 ]
= 2 3 1 − 3 2 1 + 3 1 2 − 1 3 2 + 1 2 3 − 2 1 3
= 1 2 3 − 1 3 2 + 2 3 1 − 2 1 3 + 3 1 2 − 3 2 1
= [1 2 3 ] · [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= x · (y × z)
c 2016 Elsevier, Ltd. All Rights Reserved.
Copyright ° 2
, Web Instructor’s Manual — Andrilli/Hecker, 5th Ed. Lines and Planes
(10) x · (y × z) is a scalar (since it is the dot product of two vectors), while (x · y) × z is undefined, since
it is not possible to take the cross product of a scalar (x · y) and a vector (z).
(11) The different answers to Exercises 8(c) and 8(d) furnish a counterexample that shows that the cross
product operation is not associative. For a simpler example, consider i × (j × j) = i × 0 = 0, while
(i × j) × j = k × j = −i.
(12) In what follows, let x = [1 2 3 ], y = [1 2 3 ], z = [1 2 3 ] and w = [1 2 3 ].
(a) x × (y × z)
= [1 2 3 ] × [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= [2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
1 (3 1 − 1 3 ) − 2 (2 3 − 3 2 )]
y × (z × x)
= [1 2 3 ] × [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= [2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
1 (3 1 − 1 3 ) − 2 (2 3 − 3 2 )]
z × (x × y)
= [1 2 3 ] × [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= [2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
1 (3 1 − 1 3 ) − 2 (2 3 − 3 2 )]
Now, consider (x × (y × z)) + (y × (z × x)) + (z × (x × y))
The first entry of this sum is
2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
+ 2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
+ 2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
= 1 (−2 2 − 3 3 + 2 2 + 3 3 )
+ 2 (1 2 − 2 1 + 2 1 − 1 2 )
+ 3 (−3 1 + 1 3 + 3 1 − 1 3 )
= 1 (0) + 2 (0) + 3 (0) = 0
Similarly, the second entry of this sum is
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
+ 3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
+ 3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
= 1 (−1 2 + 2 1 + 1 2 − 2 1 )
+ 2 (−3 3 − 1 1 + 3 3 + 1 1 )
+ 3 (2 3 − 3 2 + 3 2 − 2 3 )
= 1 (0) + 2 (0) + 3 (0) = 0
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Copyright ° 3
Lines and Planes and the Cross Product in R3
(1) (a) = 3 = −1 + = −4 ( ∈ R)
(b) = −4 − 2 = 2 − 3 = 8 ( ∈ R)
(c) = 6 − 2 = 2 − 5 = 1 + 6 ( ∈ R).
(Another valid form: = 4 + 2 = −3 + 5 = 7 − 6 ( ∈ R))
(d) = 4 − = −2 = 9 + 6 ( ∈ R).
(Another valid form: = 3 + = −2 = 15 − 6 ( ∈ R))
(e) = 1 − 2 = −5 + = −7 ( ∈ R)
(f) = −2 − 3 = 4 = −5 − ( ∈ R)
(2) (a) The lines intersect at a single point: (0 1 7)
(b) The lines intersect at a single point: (8 −5 10)
(c) The lines do not intersect.
(d) The lines do not intersect.
(e) The lines are identical, so the intersection of the lines is the set of all points on (either) line; that
is, the intersection consists of all points on the line = 4 − 7 = 3 + 2 = − 5 ( ∈ R).
(f) The lines are identical, so the intersection of the lines is the set of all points on (either) line; that
is, the intersection consists of all points on the line = 3 − 2 = + 1 = 2 − 7 ( ∈ R).
(3) (a) = 90◦ (b) = 45◦ (c) = 60◦ (d) = 150◦
(4) (a) There are two parts to the proof. First, we show that if a point lies on the given line , then
it satisfies the symmetric equations for the line . The given line has the parametric equations
= 0 + = 0 + and = 0 + Suppose (1 1 1 ) is a point on . Then 1 = 0 +
1 = 0 + and 1 = 0 + , for some value of . Solving for this value of in each expression,
we find that:
1 − 0 1 − 0 1 − 0
= = =
and thus (1 1 1 ) satisfies the symmetric equations for .
To finish the proof, we show that if a point satisfies the symmetric equations for , then that
point lies on . Suppose that (1 1 1 ) is a point satisfying the symmetric equations
1 − 0 1 − 0 1 − 0
= =
Let this common value be , and then solve for to find that
1 = 0 + 1 = 0 + and 1 = 0 +
Thus, (1 1 1 ) lies on .
−6 −2 −1 −4 +3 −7
(b) −2 = −5 = 6 (Another valid form: 2 = 5 = −6 )
+2 +5
(c) −3 = 4 = −1
c 2016 Elsevier, Ltd. All Rights Reserved.
Copyright ° 1
,Web Instructor’s Manual — Andrilli/Hecker, 5th Ed. Lines and Planes
(5) (a) 6 + + 6 = 1 (c) 9 − 2 = −35 (e) = 0
(b) 7 + 3 − 3 = −6 (d) = 0
£2 1 2
¤ £1 4 8
¤
(6) (a) 3 −3 3 (b) 9 9 −9
h √ √ √ i
4 21 −2 21 − 21
(c) 21 21 21 ≈ [0872 −0436 −0218]
(7) (a) = 60◦ (b) = 45◦ (c) = 90◦
(8) (a) [7 −3 1] (d) [−1 1 −4] (g) −8
(b) [2 4 −5] (e) [0 0 0] (h) 0
(c) [−1 2 −3] (f) [0 0 0] (i) [0 0 0]
(9) In what follows, let x = [1 2 3 ], y = [1 2 3 ], and z = [1 2 3 ].
(a) (x) × y
= [(2 )3 − (3 )2 (3 )1 − (1 )3 (1 )2 − (2 )1 ]
= [2 (3 ) − 3 (2 ) 3 (1 ) − 1 (3 ) 1 (2 ) − 2 (1 )]
= x × (y)
(b) (x) × y
= [(2 )3 − (3 )2 (3 )1 − (1 )3 (1 )2 − (2 )1 ]
= [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= (x × y)
(c) x × 0 = [2 (0) − 3 (0) 3 (0) − 1 (0) 1 (0) − 2 (0)] = [0 0 0] = 0
(d) x × x = [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ] = [0 0 0] = 0
(e) x × (y + z)
= [2 (3 + 3 ) − 3 (2 + 2 ) 3 (1 + 1 ) − 1 (3 + 3 )
1 (2 + 2 ) − 2 (1 + 1 )]
= [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
+ [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= (x × y) + (x × z)
(f) y · (x × y)
= [1 2 3 ] · [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= 1 2 3 − 1 3 2 + 2 3 1 − 2 1 3 + 3 1 2 − 3 2 1
= (1 2 3 − 3 2 1 ) + (−1 3 2 + 2 3 1 )
+ (−2 1 3 + 3 1 2 )
= 0+0+0 = 0
(g) (x × y) · z
= [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ] · [1 2 3 ]
= 2 3 1 − 3 2 1 + 3 1 2 − 1 3 2 + 1 2 3 − 2 1 3
= 1 2 3 − 1 3 2 + 2 3 1 − 2 1 3 + 3 1 2 − 3 2 1
= [1 2 3 ] · [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= x · (y × z)
c 2016 Elsevier, Ltd. All Rights Reserved.
Copyright ° 2
, Web Instructor’s Manual — Andrilli/Hecker, 5th Ed. Lines and Planes
(10) x · (y × z) is a scalar (since it is the dot product of two vectors), while (x · y) × z is undefined, since
it is not possible to take the cross product of a scalar (x · y) and a vector (z).
(11) The different answers to Exercises 8(c) and 8(d) furnish a counterexample that shows that the cross
product operation is not associative. For a simpler example, consider i × (j × j) = i × 0 = 0, while
(i × j) × j = k × j = −i.
(12) In what follows, let x = [1 2 3 ], y = [1 2 3 ], z = [1 2 3 ] and w = [1 2 3 ].
(a) x × (y × z)
= [1 2 3 ] × [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= [2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
1 (3 1 − 1 3 ) − 2 (2 3 − 3 2 )]
y × (z × x)
= [1 2 3 ] × [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= [2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
1 (3 1 − 1 3 ) − 2 (2 3 − 3 2 )]
z × (x × y)
= [1 2 3 ] × [2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 ]
= [2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
1 (3 1 − 1 3 ) − 2 (2 3 − 3 2 )]
Now, consider (x × (y × z)) + (y × (z × x)) + (z × (x × y))
The first entry of this sum is
2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
+ 2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
+ 2 (1 2 − 2 1 ) − 3 (3 1 − 1 3 )
= 1 (−2 2 − 3 3 + 2 2 + 3 3 )
+ 2 (1 2 − 2 1 + 2 1 − 1 2 )
+ 3 (−3 1 + 1 3 + 3 1 − 1 3 )
= 1 (0) + 2 (0) + 3 (0) = 0
Similarly, the second entry of this sum is
3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
+ 3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
+ 3 (2 3 − 3 2 ) − 1 (1 2 − 2 1 )
= 1 (−1 2 + 2 1 + 1 2 − 2 1 )
+ 2 (−3 3 − 1 1 + 3 3 + 1 1 )
+ 3 (2 3 − 3 2 + 3 2 − 2 3 )
= 1 (0) + 2 (0) + 3 (0) = 0
c 2016 Elsevier, Ltd. All Rights Reserved.
Copyright ° 3
