Express the unit vectors iˆ, jˆ, and kˆ as ordered triples and show that:
a) iˆ x� jˆ = �kˆ b) kˆ x� jˆ� = -�iˆ
To begin, I subbed in
iˆ= (1,0,0)
jˆ= (0,1,0)
kˆ = (0,0,1)
for a) I got (0,-1,1) = (0,0,1) after crossing i x j
What do I do next?
(there are no answers for this)^^
Well , according to whatever method of cross-products you use
(1,0,0) x (0,1,0) = ( 0,0,1) which was vector k
then k x j
= (0,0,1) x (0,1,0) = (-1,0,0) = -i
For part a), you have correctly computed the cross product of iˆ and jˆ. The result you obtained, (0, -1, 1), is the correct value for iˆ × jˆ. However, it seems that you made a small mistake in stating the result as (0, 0, 1), which is the unit vector kˆ.
To show that iˆ × jˆ = kˆ, you can compare the components of the computed cross product with the components of the unit vector kˆ:
iˆ × jˆ = (0, -1, 1) ≠ (0, 0, 1)
Therefore, the correct result is iˆ × jˆ = (0, -1, 1).
For part b), let's go through the steps:
kˆ × jˆ = (-1, 0, 0)
To show that kˆ × jˆ = -iˆ, compare the components:
kˆ × jˆ = (-1, 0, 0) = -iˆ
Therefore, kˆ × jˆ = -iˆ.
In summary, the correct answers are:
a) iˆ × jˆ = (0, -1, 1)
b) kˆ × jˆ = -iˆ
To solve part (a), you have correctly substituted the unit vectors as i^ = (1,0,0), j^ = (0,1,0), and k^ = (0,0,1). When you take the cross product of i^ and j^, you get:
i^ × j^ = (0,0,1)
So far, you have the correct result. The cross product of the i^ and j^ unit vectors is equal to the k^ unit vector.
To solve part (b), you can follow a similar process. The given cross product is k^ × j^, which yields:
k^ × j^ = (0,0,-1)
Therefore, the cross product of the k^ and j^ unit vectors is equal to -i^.
In summary:
a) i^ × j^ = k^
b) k^ × j^ = -i^