Q: 2a + 5b
a = 4i + 3j
b = 2i - j
A: From what I read since I'm a bit rusty:
(a+b) = (a1+b1)i + (a2+b2)j + (a3 + b3)k
for:
(2a + 5b) = (2xa1+5xb1)i + (2ax2+5xb2)j + (2xa3 + 5xb3)k
is this method correct?
2006-09-24 06:58:39 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Cheers guys, just solved it.
2006-09-24 07:19:23 · update #1
Yes you add in each axis separately and then put it together
2006-09-24 07:13:22 · answer #1 · answered by class4 5 · 1⤊ 0⤋
at the initiating i'm hoping there is no confusion between C^2=A^2+B^2 and C^2=A^2+B^2-2ABCOS(theta) because the first is a simplified version for perfect angles of the 2d one (cos of ninety = 0 so the 2d 0.5 of the equation is 0) as for its use in fixing vectors, at the same time as vectors are further at the same time, its commonly conventional the lengths of the perimeters (A and B) and from that with using trigonometry, theta will be discovered, and plugged into the above equation to allow you to recognize what the consequent is its all fixing and searching the hypotenuse wish this helped =]
2016-11-23 19:16:05 · answer #2 · answered by chafton 3 · 0⤊ 0⤋
Yeah, that looks right.
2006-09-24 07:01:25 · answer #3 · answered by Pascal 7 · 1⤊ 0⤋