Solutions
Block 1: Vector A r i t h m e t i c
U n i t 6: Equations of Lines and P l a n e s
1.6.1(L)
+
From E x e r c i s e 1.5.2 w e know t h a t a v e c t o r N normal t o t h e d e s i r e d
W e a l s o know t h a t ( 1 , 2 , 3 ) is i n t h e p l a n e .
p l a n e i s (-14,10,9).
i
s
i
n
t h e p l a n e i f and o n l y i f 8
Aif = 0.
Hence, P ( x , y , z)
Pictorially,
Since
Ab
= (x
-
1, y
-
2, z
-
3) and
8
= (-14,10,9),
N
AP = 0
implies t h a t
A s a "quick" check of
( l ) ,w e know t h a t i t must b e s a t i s f i e d when
x = 1, y = 2, z = 3 [ s i n c e ( 1 , 2 , 3 ) b e l o n g s t o t h e p l a n e ] , when
x = 3 , y = 3, z = 5 [ s i n c e (3,3,5) b e l o n g s t o t h e p l a n e ] , and when
x = 4 , y = 8, z = 1 [ s i n c e ( 4 , 8 , 1 ) b e l o n g s t o t h e p l a n e ] .
W e obtain
and w e see t h a t e q u a t i o n (1) i s s a t i s f i e d i n each c a s e .
.You may have n o t i c e d t h a t w e w e r e s o l v i n g problems o f t h i s t y p e
i n U n i t 4.
One of t h e major d i f f e r e n c e s between t h e n and now i s
t h a t e a r l i e r we t a l k e d a b o u t a v e c t o r b e i n g p e r p e n d i c u l a r t o a
,
Solutions
Block 1: Vector Arithmetic
Unit 6: Equations of Lines and Planes
1.6.1 (L) continued
p l a n e even though we may n o t have been a b l e t o compute such a
v e c t o r . Now, by use of t h e c r o s s product, we may a c t u a l l y d e t e r mine a v e c t o r which' i s perpendicular t o a given plane.
QuiTe i n g e n e r a l , t o f i n d t h e equation of a p l a n e determined by
+
t h r e e non-collinear p o i n t s A, B, and C , we form t h e v e c t o r s AB
and AE, whereupon 6 X AC y i e l d s a normal v e c t o r $ t o t h e plane.
We then p i c k any p o i n t P ( x , y , z ) i n space and form, s a y , t h e v e c t o r
Rb.
The equation of t h e plane i n v e c t o r form i s then simply
The "standard equation"fol1ows from (2) merely by e x p r e s s i n g (2)
i n Cartesian coordinates.
I n summary, i n C a r t e s i a n c o o r d i n a t e s
r e p r e s e n t s t h e equation of t h e plane which h a s t h e v e c t o r
+
+
+
a i + b j + ck a s t h e normal v e c t o r and which p a s s e s through t h e
+
p o i n t (xotyo,zo), f o r i n t h i s c a s e 6
AP = ( a , b , c ) ( x - r o t
y
yo, z
zO) = a ( x - xo) + b ( y
yo) + c ( z
z0)
-
-
-
-
.
1.6.2
We a r e probably conditioned t o t h i n k of y = 2x a s denoting t h e
equation of a l i n e . I n terms of what w e c a l l t h e u n i v e r s e of
d i s c o u r s e , y = 2x denotes a l i n e i f we t h i n k of it a s an abbreviation for
On t h e o t h e r hand, i f we t h i n k of it a s an a b b r e v i a t i o n f o r
then w e should have a p l a n e r a t h e r than a l i n e . That i s , t h e
b a s i c d i f f e r e n c e between (1) and (2) i s t h a t i n (1) our elements
Solutions
Block 1: Vector A r i t h m e t i c
Unit 6: Equations of Lines and P l a n e s
1.6.2
continued
a r e o r d e r e d p a i r s (two-dimensional) w h i l e i n ( 2 ) they a r e o r d e r e d
t r i p l e t s (three-dimensional).
With a l i t t l e g e o m e t r i c i n t u i t i o n w e may r e c o g n i z e t h a t e q u a t i o n
( 2 ) i s t h e p l a n e which p a s s e s through t h e l i n e y = 2x and i s p e r -
p e n d i c u l a r t o t h e xy-plane.
That i s , e q u a t i o n ( 2 ) seems t o s a y
t h a t a p o i n t b e l o n g s t o o u r c o l l e c t i o n a s soon a s y = 2x r e g a r d less of t h e c h o i c e of z.
Our main aim i n t h i s s e c t i o n i s t o show how t h i s r e s u l t a c t u a l l y
f o l l o w s from t h e s t a n d a r d e q u a t i o n of a plane.
rewrite y = 2x a s
Specifically, we
and t o b r i n g i n t h e t h i r d dimension, we r e w r i t e t h i s a s
2x
-
y
+ Oz
= 0.
F i n a l l y , t o p u t t h i s i n t o t h e s t a n d a r d form, we r e w r i t e it a s
and t h i s i s t h e p l a n e which p a s s e s through ( 0 , 0 , 0 ) and has t h e
7
'
v e c t o r 21
-
a s a normal.
A s a check n o t i c e t h a t t h e s l o p e of 2 1
- 3 is
-1/2
and t h i s i s t h e
n e g a t i v e r e c i p r o c a l o f . 2 which i s t h e s l o p e of t h e l i n e y = 2x.
1.6.3(L)
a . The l i n e p a s s e s through (-2,5,-1)
31
+
43
and i s p a r a l l e l t o t h e v e c t o r
+ 2g ( s e e n o t e a t t h e end of t h i s e x e r c i s e f o r a f u r t h e r
discussion).
P i c t o r i a l l y , w e have:
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 6: Equations of L i n e s and P l a n e s
1.6.3 (L) c o n t i n u e d
-----).
t.
Now P ( x , y , z ) i s on !t i f and o n l y i f POP i s p a r a l l e l t o
In
and 3 must be s c a l a r m u l t i p l e s
v e c t o r language t h i s s a y s t h a t
PT
o f one a n o t h e r . That i s , P i s on R i f and o n l y i f t h e r e e x i s t s a
number ( s c a l a r ) t s u c h t h a t
W r i t i n g (1) i n C a r t e s i a n c o o r d i n a t e s w e have
(x
-
(-2), y
o r (x
+
2, y
-
5, z
-
(-1))= t ( 3 , 4 , 2 )
-
5, z
+
1) = ( 3 t , 4 t , 2 t ) .
Hence, by comparing components i n ( 2 ) , w e have
Thus ( x , y , z ) b e l o n g s t o o u r l i n e i f and o n l y i f
Quite i n general x-xo
y-yo
2 - 2 ,
r e p r e s e n t s t h e l i n e which p a s s e s through t h e p o i n t (xo,yo,zo)
i s p a r a l l e l t o t h e v e c t o r a x + bf + cg. b. The p o i n t i n q u e s t i o n i s determined from ( 4 ) w i t h z = 0.
event
and In t h i s
I
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 6 : Equations of Lines and P l a n e s
1.6.3 ( L ) c o n t i n u e d
.
.
r
.
1
Therefore, t h e required p o i n t i s (-2,7,0),
,
THE SLOPE OF A LINE I N THREE-DIMENSIONAL CARTESIAN COORDINATES
The s t a t e m e n t t h a t a l i n e is determined by two d i s t i n c t p o i n t s o r
by one p o i n t and t h e s l o p e i s independent of any p a r t i c u l a r coord i n a t e system. I t i s a l s o independent of whether w e a r e d e a l i n g
i n two-dimensional s p a c e o r three-dimensional space.
In the solut i o n o f t h e e x e r c i s e j u s t concluded, w e t a l k e d a b o u t t h e s l o p e o f
a l i n e i n t h r e e - d i m e n s i o n a l s p a c e i n d i r e c t l y by s p e c i f y i n g t h a t
t h e l i n e be p a r a l l e l t o a g i v e n v e c t o r .
This i d e a i s v e r y much a k i n t o a t r a d i t i o n a l t o p i c known a s
d i r e c t i o n a l cosines.
Given a l i n e i n s p a c e , w e c o n s i d e r e d t h e
l i n e p a r a l l e l t o t h e g i v e n l i n e t h a t passed through t h e o r i g i n
( t h e advantage t o v e c t o r n o t a t i o n i s t h a t i f Ge view t h e l i n e a s
v e c t o r i z e d w e may assume t h a t it p a s s e s through t h e o r i g i n witho u t having t o t h i n k o f i t a s a d i f f e r e n t v e c t o r ) . I n any e v e n t ,
t h e s l o p e of t h e l i n e was w e l l - d e f i n e d a s soon a s w e c o u l d measure
t h e a n g l e it made w i t h each of t h e t h r e e c o o r d i n a t e axes. Tradit i o n a l l y , t h e convention i s t o l e t a d e n o t e t h e a n g l e t h e l i n e
makes w i t h t h e p o s i t i v e x - a x i s , f3 t h e a n g l e i t makes w i t h t h e
p o s i t i v e y - a x i s , and y t h e a n g l e i t makes w i t h t h e p o s i t i v e z-axis.
The p o i n t i s t h a t i f w e u s e v e c t o r n o t a t i o n i t i s p a r t i c u l a r l y e a s y
t o f i n d t h e c o s i n e s of t h e s e t h r e e a n g l e s (even w i t h o u t v e c t o r s
t h e r e a r e simple formulas f o r cosines of t h e s e angles i n t e r m s of
denote
Namely, w e have a l r e a d y s e e n t h a t i f A and
x , y , and 2).
+
-+
u n i t vectors, then A
B i s t h e c o s i n e o f t h e a n g l e between A
and B.
That i s ,
-
and by d e f i n i t i o n o f u n i t v e c t o r s , b o t h
J A J and lq
equal 1 i n t h i s
c a s e , and t h e r e s u l t f o l l o w s .
-f
-
I n o t h e r words, t h e n , i f u d e n o t e s a u n i t v e c t o r p a r a l l e l t o a
+
-t
given l i n e t h e n u
1 r e p r e s e n t s t h e c o s i n e o f t h e a n g l e between
.
Solutions
Block 1: Vector A r i t h m e t i c
Unit 6: Equations o f Lines and P l a n e s
1.6.3 (L) continued
-+
-+
-+
u and 1; and s i n c e u h a s t h e d i r e c t i o n of t h e g i v e n l i n e and s i n c e
? has t h e d i r e c t i o n o f t h e p o s i t i v e x - a x i s , i f f o l l o w s t h a t 2 f
i s a l s o t h e c o s i n e o f t h e a n g l e between t h e l i n e and t h e p o s i t i v e
x-axis.
That is,
+
cosa = u
X.
Similarly,
and
cosy =
2.
G
1.6.4
Since the l i n e
is p a r a l l e l t o t h e v e c t o r 6 1
3j
+
c o s a = +u
-
+
+
35
+
22, and s i n c e
$3
221 =
= 7 , w e have t h a t 6 =
+
+ +%
i s a u n i t v e c t o r along t h e given l i n e . In accord w i t h o u r - n o t e - a t
t h e end o f E x e r c i s e 1.6.3, w e have:
16:
+
cosy = u
t
6
= = 7
2
i; = 7
Therefore, a = cos
$%
31.
Solutions
Block 1: Vector A r i t h m e t i c
Unit 6: Equations of Lines and P l a n e s
1.6.5
The l i n e i n q u e s t i o n i s g i v e n by t h e e q u a t i o n
T h i s e q u a t i o n may be w r i t t e n p a r a m e t r i c a l l y by
I n E x e r c i s e 1.6.1(L)
we saw t h a t t h e p l a n e was given by t h e e q u a t i o n
S i n c e t h e p o i n t we s e e k b e l o n g s t o b o t h t h e l i n e and t h e p l a n e
i t s c o o r d i n a t e s must s a t i s f y b o t h e q u a t i o n s ( 2 ) and ( 3 ) .
t h e v a l u e s of x, y , and z from ( 2 ) i n t o ( 3 ) we o b t a i n :
and t h i s y i e l d s
P u t t i n g t h e v a l u e of t i n ( 4 ) i n t o (2), w e s e e t h a t
Putting
Solutions
Block 1: Vector Arithmetic
Unit 6: Equations of Lines and Planes
---.-----------1.6.5 continued
-33
s o t h a t (T,
plane.
162 -59
T,-?n)
i s t h e p o i n t a t which t h e l i n e meets t h e
1.6.6(L)
a. Since t h e equation of t h e plane i s 2x + 3y + 62 = 8, w e see t h a t
+
+
2 3 6
N = (2,3,6) i s normal t o t h e plane. * I n p a r t i c u l a r uN = f7t1t7)
'
i s a , u n i t v e c t o r normal t o t h e plane. We can now l o c a t e a p o i n t i n t h e plane by a r b i t r a r i l y p i c k i n g values f o r two of t h e v a r i a b l e s i n 2x + 3y + 62 = 8 and then s o l v - For example, i f w e l e t y = 2 e 0, ing t h e equation f o r t h e t h i r d .
then x .= 4.
Hence, A ( 4 , O ,0) is i n t h e plane,
Pictorially, We seek t h e l e n g t h of PB which i n t u r n i s t h e l e n g t h of AE,
V e c t o r i a l l y , t h i s i s t h e magnitude of t h e p r o j e c t i o n of A$ o n t o
i. [We
say "magnituden s i n c e i f t h e measure of +PAC exceeds 90°
( t h a t i s , i f P i s ' b e l o w " t h e plane) t h e p r o j e c t i o n w i l l f a l l i n
t h e o p p o s i t e s e n s e of 8.1
* I n g e n e r a l t o c o n v e r t a x + b y + c z = d i n t o t h e s t a n d a r d form we
may ( i f a # 0) r e w r i t e t h e e q u a t i o n a s
-
(ax d)
+
by
+
cz = 0
Hence, our p l a n e p a s s e s through
a normal v e c t o r .
(d
a
0,O) and h a s a;
+
b;
+
cg as
Solutions
Block 1: V e c t o r A r i t h m e t i c
U n i t 6: Equations o f L i n e s and P l a n e s
1.6.6 ( L ) c o n t i n u e d
-+
Now w e have a l r e a d y s e e n t h a t t o p r o j e c t a v e c t o r A o n t o a v e c t o r
3
t h e magnitude of t h e p r o j e c t i o n i s g i v e n by
.
.
-+
i s a u n i t v e c t o r i n t h e d i r e c t i o n of B.
s e e k is g i v e n by
IA'
IA
-f
+
-+
uB 1 where uB
Hence, t h e d i s t a n c e w e
. -+unl
which i s e q u a l t o
b.
The l i n e t h r o u g h P p e r p e n d i c u l a r t o t h e p l a n e i s , o f c o u r s e , t h e
Hence t h e
l i n e thfough P p a r a l l e l t o 8 (where fi i s a s i n a . ) .
e q u a t i o n of t h i s l i n e i s :
I n p a r a m e t r i c form t h i s becomes:
I f ~ ( x , y , z ) i s t h e p o i n t a t which t h e l i n e through P and p a r a l l e l
t o 8 i n t e r s e c t s t h e p l a n e 2x + 3y + 6 2 = 8, t h e n B must s a t i s f y
b o t h t h e e q u a t i o n of t h e p l a n e and e q u a t i o n ( 2 ) .
have :
That i s , w e must
Hence,
-
Putting (3) i n t o ' (2) y i e l d s
. From ( 4 ) w e f i n d t h a t
25 20 -9
B i s the p o i n t (-7,'7,7).
Again p i c t o r i a l l y ,
By t h e way, s i n c e BP = E,the r e s u l t o f b. should supply us with
an a l t e r n a t i v e s o l u t i o n f o r a . Namely,
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 6: Equations o f Lines and P l a n e s
-
1.6.6 ( L ) c o n t i n u e d
which checks w i t h o u r p r e v i o u s r e s u l t .
c.
Looking a t
and
one might be tempted t o s u b t r a c t ( 6 ) from ( 5 ) and o b t a i n 1 4 a s an
answer.
Rather t h a n a r g u e ( a t l e a s t f o r t h e moment) a g a i n s t t h i s r e a s o n i n g
l e t us a c t u a l l y d e r i v e t h e c o r r e c t answer.
Since the.planes a r e p a r a l l e l (i.e., both a r e perpendicular t o
2;
+
3f
-
+
6g) i t i s s u f f i c i e n t t o f i n d t h e ( p e r d e n d i c u l a r ) d i s t a n c e
from a p o i n t i n one p l a n e t o t h e o t h e r p l a n e .
For example P ( l l t O , O ) b e l o n g s t o 2x
+
3y
+
62 = 22.
Hence i m i t a t -
i n g o u r approach i n p a r t a. we have:
.
Notice o u r o r i e n t a t i o n .
Certainly
(11,0,0) a n d ( 4 , 0 , 0 ) s h o u l d l i e on
t h e x-axis.
That i s AP is a segment o f t h e x - a x i s .
The p o i n t i s
t h a t PB does n o t r e p r e s e n t t h e
d i r e c t i o n of t h e z-axis i n o u r
diagram.
I n o t h e r words t h e d i s t a n c e between t h e p l a n e s i s 2 u n i t s ;
not
14 u n i t s .
To s e e what happened h e r e , l e t ' s look a t t h e g e n e r a l s i t u a t i o n :
Solutions
Block 1: Vector Arithmetic
Unit 6: Equations of Lines and Planes
1 . 6 . 6 (L) continued
Equation of plane i s ax + by + cz = d. Therefore,we may l e t
d
A(Ei.,O,O) denote a p o i n t i n t h e plane ( i f a = 0 we work with
d
(o,glO), e t c . ) .
+
Now a v e c t o r N normal t o our plane is ( a , b , c ) .
normal i s given by
-a x +
bf
+
cE
=
.
Hence a u n i t
Thus, t h e d i s t a n c e from
P t o t h e plane is
I£ we now assume t h a t P l i e s i n t h e plane ax
s i n c e p ( x O , y O , z O )i s i n t h i s plane we have
+ by + c z
=.e, then
P u t t i n g ( 8 ) i n t o ( 7 ) y i e l d s t h a t t h e d i s t a n c e between t h e planes
ax + by + cz = e and ax + by + cz = d i s
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 6: E q u a t i o n s of L i n e s and P l a n e s
1.6.6(L) c o n t i n u e d
From ( 9 ) w e s e e t h a t t h e c o r r e c t answer i s ( e
I n o u r example a = 2, b = 3,
and c = 6.
- d1
i f and o n l y i f
Hence a
= 7.
N o t i c e t h a t had w e w r i t t e n e q u a t i o n s (5) and (5) i n t h e e q u i v a l e n t
form
then a2
22
-
T8I
+
b2
+
c2 = 1, and t h e c o r r e c t answer would t h e n b e
= 2 which checks w i t h t h e p r e v i o u s l y o b t a i n e d r e s u l t .
I n t e r m s of a s i m p l e diagram
t h i s i s t h e d i s t a n c e w e want
t h i s is the distance w e g e t by s u b t r a c t i o n W e have A 6 = ( 3
-
1, 4
-
1, 5
-
4) = (2,3,1) is p a r a l l e l t o our
l i n e and s i n c e ( 1 , 1 , 4 ) is on t h e l i n e [ t h i n g s w i l l work o u t t h e
same i f we choose ( 3 , 4 , 5 ) r a t h e r t h a n (1,1,4)] our l i n e L h a s t h e
equation
o r i n p a r a m e t r i c form
Solutions
Block 1: Vector A r i t h m e t i c
Unit 6: Equations of Lines and P l a n e s
1.6.7 c o n t i n u e d
A v e c t o r normal t o o u r p l a n e M is given by
Hence t h e e q u a t i o n of M i s given by
The p o i n t ( x , y , z ) w e s e e k , s i n c e it belongs t o b o t h t h e l i n e and
t h e p l a n e , must s a t i s f y both ( 2 ) and ( 3 ) . Hencer
8
t h e r e f o r e , t = 3.
P u t t i n g (4) i n t o ( 2 ) y i e l d s
19
20
Thus, t h e p o i n t of i n t e r s e c t i o n i s (T, 9 , --J).
Check :
Solutions
Block 1: Vector A r i t h m e t i c
U n i t 6: Equations of Lines and P l a n e s
1.6.7
continued
and
1.6.8
a.
The a n g l e between t h e two p l a n e s i s e q u a l t o t h e a n g l e between
t h e i r normals.
(From a " s i d e view"
NOW,
-+
A = 3x
-+
+
2J
+
+
25
-2
6k i s normal t o one p l a n e
while
3
= 2x
i s normal t o t h e o t h e r .
Now t h e a n g l e u between
ii
i$
= /A/
Therefore,
or 6
+
4
-
]BI
( 3 , 2 , 6 ) - ( 2 , 2 , - 1 ) = 171 131 cosu
6 = 2 1 cosu.
4
Ti
u = cos
-1
-.241
The l i n e w e s e e k l i e s i n 3x
c u l a r t o 3;
i s determined by
and
cosu
T h e r e f o r e , cosu =
b.
Zf
+
2;
+
6%.
a l s o p e r p e n d i c u l a r t o 2;
+
2y
+
62 = 8.
I t a l s o l i e s i n 2x
+
25
- 2.
Hence i t i s perpendi-
+
2y
-
2
= 5, s o it i s
Solutions
Block 1: Vector Arithmetic
Unit 6: Equations of Lines and Planes
1.6.8 continued
Now one v e c t o r perpendicular t o both
31 +
25
+
6g and 2 1
+
23
-2
is t h e i r c r o s s product
Therefore, - 1 4 f
+
153
+
2% i s p a r a l l e l
t o t h e l i n e w e seek.
To f i n d a p o i n t on t h e l l n e we may t a k e any s o l u t i o n of
L e t t i n g , f o r example, z = 0 w e o b t a i n
3x
2x
+
+
2y = 8
2y = 5
I
therefore, x = 3
Therefore, 3 , , 0
- 1 4 f + 153 + 2k.
1
y = - z .
i s on our l i n e and o u r l i n e i s p a r a l l e l t o
Therefore, t h e equation of our l i n e i s
o r , i n parametric form:
Check :
Solutions Block 1: Vector Arithmetic Unit 6: Equations of Lines and Planes 1.6.8 continued and Solutions
Block 1: Vector A r i t h m e t i c
1.
-
Since 1
1 = 0 and w e have proven t h e theorem t h a t b ( 0 ) = 0 f o r
a l l numbers b, i f f o l l o w s t h a t
-
Next, r e c a l l i n g t h a t 1
butive r u l e
$0
-
1 means 1 + ( - I ) , w e may u s e t h e d i s t r i -
conclude
Using s u b s t i t u t i o n [e.g.,
r e p l a c i n g (-1)(1
-
1) i n ( 2 ) by i t s
v a l u e i n (1)1 w e may conclude from (1) and ( 2 ) t h a t
By t h e p r o p e r t y of 1 b e i n g a m u l t i p l i c a t i v e i d e n t i t y ( i . e . ,
b x 1 = b f o r a l l numbers b ) , it f o l l o w s t h a t
Putting (4) i n t o (3) yields
whereupon adding 1 t o b o t h s i d e s of
(5) yields the desired result.
More f o r m a l l y , by s u b s t i t u t i o n using t h e e q u a l i t y i n ( S ) ,
By v i r t u e o f 0 b e i n g t h e a d d i t i v e i d e n t i t y
w h i l e by t h e a s s o c i a t i v i t y of a d d i t i o n
Solutions
bloc^ 1: Vector Arithmetic
Quiz
'.
.
continued
By t h e a d d i t i v e i n v e r s e r u l e 1
i + [ - i + (-1)(-1)j =
o +
+
(-1) = 0, whence ( 8 ) becomes
(-1)(-1).
(9)
since 0
+
(-1)(-1) = (-1)(-1) + 0 = (-1)(-11,
1 + [-1
+
(-1)(-1)I = (-1)(-1).
we may rewrite ( 9 ) a s
P u t t i n g t h e r e s u l t s of ( 7 ) and ( 1 0 ) i n t o (6) y i e l d s
2.
Since
$=
3x 2
+
1, t h e s l o p e of t h e l i n e t a n g e n t t o our curve a t
(1.2) i s given a t
1
$
= 4.
m e n , s i n c e t h e normal v e c t o r i s a t
x=1
r i g h t a n g l e s t o t h e t a n g e n t v e c t o r (by d e f i n i t i o n of normal), i t s
s l o p e must be t h e r e c i p r o c a l of t h e s l o p e of t h e t a n g e n t v e c t o r .
1 a s i t s s l o p e . I n C a r t e s i a n coorHence, t h e r e q u i r e d normal has -a
d i n a t e s , t h e s l o p e of a v e c t o r i s t h e q u o t i e n t of i t s f-component
d i v i d e d by i t s f-component.
S i n c e lijl = V ( 4 )
given by
2
+ (-112
-
=
Therefore, one normal v e c t o r would be
m, we
s e e t h a t a u n i t normal i s
-t
Finally, since 4 i
J and - 4 f + f have t h e same d i r e c t i o n and
unit
magnitude b u t o p p o s i t e s e n s e , we s e e t h a t t h e r e a r e
normals given by
-+
two
Solutions
Block 1: Vector Arithmetic
Quiz
3.
Now
A8 + AZ
i s t h e v e c t o r which i s t h e diagonal of t h e p a r a l l e l o -
gram determined by A, B , and C.
I n g e n e r a l , such a diagonal
does n o t b i s e c t t h e v e r t e x angle. I n f a c t i t b i s e c t s t h e angle
i f and only i f t h e parallelogram i s a rhombus ( a l l f o u r s i d e s
--
have equal l e n g t h ) .
d and
So consider i n s t e a d t h e parallelogram determined by I AZ I
AZ. This i s a rhombus ( s i n c e each v e c t o r has ~ A Z(I AI = lA8 IAE
a s i t s magnitude)
.
1
Therefore,
b i s e c t s 4BAC. (Notice t h a t I A Z I d i s p a r a l l e l t o P$ and 1 Aifl AE
i s p a r a l l e l t o AZ; hence, i n both paralellograms, QBAC i s a
v e r t e x angle.)
From (1)
Hence one b i s e c t i n g v e c t o r i s
1
,
,
Solutions
Block 1: Vector Arithmetic
Quiz
3.
I
cantinued
Hence, any s c a l a r m u l t i p l e of 4 *1
- 3-
Since 14;
2zl = dl6
QBAC i s given by
- 3 - 2 z is a b i s e c t i n g vector.
+ 1 + 4 = m, a u n i t v e c t o r which b i s e c t s
( I n t h i s type of problem, i t i s o f t e n customary t o r e j e c t one of
t h e s i g n s [though t h i s i s n o t mandatory]. The reason i s t h a t
OBAC, u n l e s s some convention i s made, i s ambiguous s i n c e it does
n o t d i s t i n g u i s h between
However, we s h a l l n o t q u i b b l e about t h i s d i s t i n c t i o n h e r e and w e
s h a l l a c c e p t e i t h e r of t h e two a n g l e s . )
We have
I
I
Solutions
Block 1: Vector A r i t h m e t i c
Quiz
4.
continued
a.
W e have t h a t
Therefore,
(2,1,2)
Therefore,
(2) (3)
( 3 , 2 , 6 ) = ( 3 ) ( 7 ) c o s d BAC.
+
(1)( 2 )
+
( 2 ) ( 6 ) = 2 1 c o s 9 BAC.
n.
6 + 2 + 1 2
20
T h e r e f o r e , c o s 4 BAC = q
l=
T h e r e f o r e , QBAC = c o s
b.
-'n20: % l B O .
A normal t o P i s g i v e n by
S i n c e ( 1 , 2 , 3 ) i s i n P [ ( 3 , 3 , 5 ) and ( 4 , 4 , 9 ) could have been chosen
a s w e l l ] , the equation f o r P is
[I.e.,
r e c a l l t h a t A(x
-
xo)
+
B(y
- yo)
+
C(z
-
zO) = 0 is t h e
e q u a t i o n of t h e p l a n e p a s s i n g through ( x O , y O , z O )and having
A; + B) + C% a s a normal v e c t o r . 1
The e q u a t i o n of P i s
[Check
Solutions Block 1: Vector Arithmetic Quiz 4. continued c. 1
6 x AEI is the area of the parallelogram which has A6 and if as
consecutive sides. Since the area of the triangle is half that of
the parallelogram, we have
[from b. 1
d. The key here is that if A(al,a2,a3), B(bl,b2,b3), and C(cltc2,c3)
are the given points then the medians of AABC intersect at the
+
+
c1 a2 + b2 + c2 a3 + b3
+ C3)
r
3
,-I
example we find that the medians intersect at bl
point e. Since C(4,4,9) is on the line and & = 21
the line, the equation of the line is
or, in parametric form That is, +
3+
.
~n our present
22 is parallel to
Solutions
Block 1: Vector ~ r i t h m e t i c
Quiz
4.
continued
-
x
[The key h e r e i s t h a t
Xo
---
- y -..Yo
7
-
Z
B
-
Z0
C
represents the
e q u a t i o n o f t h e l i n e which p a s s e s through t h e p o i n t ( x O , y O , z O )and
i s p a r a l l e l t o A:
5.
+
83
+ ~8.1
There a r e , of c o u r s e , i n f i n i t e l y many ways t h a t w e may choose a
p o i n t i n t h e p l a n e . Our c h o i c e , q u i t e a r b i t r a r i l y , i s t o l e t
x = y = 0 , whence 4x + 5y + 22 = 6 i m p l i e s t h a t z = 3. I n o t h e r
words Q ( 0 , 0 , 3 ) i s i n t h e p l a n e .
Thus, t h e d i s t a n c e from Po t o t h e p l a n e i s t h e l e n g t h of t h e proj e c t i o n o f P;Q
i n t h e d i r e c t i o n o f t h e normal t o t h e p l a n e and t h i s
i n t u r n i s g i v e n by ~
plane.
-+
~ uNI
Q
, where
+
+
-+
uN i s a u n i t normal t o t h e
S i n c e t h e e q u a t i o n of t h e p l a n e i s 4x + 5y + 22 = 6 , a
P
t
normal t o t h e p l a n e i s 4 1
t
5j
-+
2k.
Hence, a u n i t normal i s
Therefore, t h e d i s t a n c e is
The l a t t e r p a r t of t h i s problem a l s o p r o v i d e s an a l t e r n a t i v e f o r
solving t h e f i r s t part.
S i n c e t h e l i n e i s p e r p e n d i c u l a r t o t h e p l a n e , it i s p a r a l l e l t o
-+
N (= 4 1 + 53 + 283
Then, s i n c e Po (2,3,4) i s on t h e l i n e i t s
.
equation i s
S.l.Q.7
Solutions
Block 1: Vector A r i t h m e t i c
Quiz
5.
continued
o r , i n p a r a m e t r i c form
Thus, t h e r e q u i r e d p o i n t R must s a t i s f y t h e above system o f equat i o n s ( s i n c e R l i e s on t h i s l i n e ) and i t must a l s o s a t i s f y
4x + 5y + 2 2 = 6 ( s i n c e R a l s o l i e s on t h e p l a n e ) .
Hence, w e must have t h a t
4 5 t = -25,
-5
o r t = -p.
P u t t i n g t h i s v a l u e of t i n t o (1) y i e l d s
2 2 26
T h a t i s , R i s g i v e n by (- Y ~ Y , T )
[As a check o u r answer t o t h e f i r s t . p a r t of t h e problem s h o u l d a l s o
b e g i v e n by 1 doI
2 2 26
(-ytgry-)
Now do= ( 2 , 3 , 4 )
.
-
Solutions Block 1: Vector Arithmetic Quiz 5.
continued Therefore,
I doI
= 9 a00
=
+
625
+
100
m= .m
9
.
9 = 5 J S
3 which checks with our previous result.] MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Multivariable Calculus
Prof. Herbert Gross
The following may not correspond to a particular course on MIT OpenCourseWare, but has been
provided by the author as an individual learning resource.
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