Question 1202643: given that a=2i+3j-k, b=i-j+2k, and c=3i+4j+k, find (a) a+2b-c, (b) a vector d such that a+b+c+d=0, and (c) a vector d such that a-b+c+3d=0
Found 2 solutions by Alan3354, Edwin McCravy: Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! given that a=2i+3j-k, b=i-j+2k, and c=3i+4j+k,
find a+2b-c
----
2i + 6j - 1k ---- a
2i - 2j + 4k ---- 2b
-3i - 4j - 1k ---- -c
-------------------------- Add
1i + 0j + 2k
= i + 2k
=======================
(b) a vector d such that a+b+c+d=0
Add a, b and c
d = -1 * the sum
=============================
and (c) a vector d such that a-b+c+3d=0
Add a, -b and c
d = (-1/3) time the sum
Answer by Edwin McCravy(20060) (Show Source):
You can put this solution on YOUR website!
I think you'll find that it's easier to use the < p,q,r > notation in
vector calculus than the pi+qj+rk notation. Otherwise, you are likely
to get letters used for vectors and letters used for scalars confused.
Ordinary letters are used for scalars and letters in bold-face italics are used
for vectors. The < p,q,r > avoids having to use so many bold-face italic
letters.
given that a=2l+3j-k, b=l-j+2k, and c=3l+4j+k, find
(a) a+2b-c
a = < 2,3,-1 >, b = < 1,-1,2 >, c = < 3,4,1 >
a+2b-c = < 2,3,-1 > + 2< 1,-1,2 > - < 3,4,1 > =
< 2,3,-1 > + < 2,-2,4 > + < -3,-4,-1 > =
< 2+2-3,3-2-4,-1+4-1 > = < 1,-3,2 > = l-3j+2k
(b) a vector d such that a+b+c+d=0,
Let d = pl+qj+rk = < p,q,r >
a+b+c+d = < 2,3,-1 > + < 1,-1,2 > + < 3,4,1 > + < p,q,r > = 0 = < 0,0,0 >
< 2+1+3+p,3-1+4+q,-1+2+1+r > = < 0,0,0 >
< 6+p,6+q,2+r > = < 0,0,0 >
6+p=0; 6+q=0; 2+r=0
p=-6; q=-6; r=-2
d = pl+qj+rk = < p,q,r > = < -6,-6,-2 >
and (c) a vector d such that a-b+c+3d=0
Let d = pl+qj+rk = < p,q,r >
a-b+c+3d = < 2,3,-1 > - < 1,-1,2 > + < 3,4,1 > + 3< p,q,r > = 0 = < 0,0,0 >
= < 2,3,-1 > + < -1,1,-2 > + < 3,4,1 > + < 3p,3q,3r > = 0 = < 0,0,0 >
< 2-1+3+3p,3+1+4+3q,-1-2+1+3r > = < 0,0,0 >
< 4+3p,8+3q,-3+3r > = < 0,0,0 >
4+3p=0; 8+3q=0; -2+3r=0
3p=-4; 3q=-8; 3r=2
p=-4/3; q=-8/3; r=2/3
d = pl+qj+rk = < p,q,r > = < -4/3,-8/3,2/3 >
Edwin
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