bb(ul hata), bb(ul hatb) and bb(ul hatc) are unit vectors, and the angle between bb(ul hatb) and bb(ul hatc) is pi/6. Show that bb(ul hata) = +-2 (bb(ul hatb) xx bb(ul hatc))?
2 Answers
For Proof, refer to the Explanation.
Explanation:
Given that,
for unit vectors
Clearly,
Therefore, let us suppose that,
Hence, by
This completes the Proof.
Enjoy Maths.!
|| bb(ul hata) || = || bb(ul hatb) || = || bb(ul hatc) || = 1
We know that:
|| bb(ul U) xx bb(ul V) || = ||bb(ul U)|| * ||bb(ul V)|| * sin theta
And so
|| bb(ul b) xx bb(ul c) || = ||bb(ul b)|| \ ||bb(ul c)|| \ sin (pi/6)
" " = 1 * 1 * 1/2
" " = 1/2
We also have:
bb(ul hat(a)) * bb(ul hat(b)) = 0 => bb(ul hata) perpendicular tobb(ul hatb)
bb(ul hat(a)) * bb(ul hat(c)) = 0 => bb(ul hata) perpendicular tobb(ul hatc)
We know that the cross product of any two vectors is perpendicular to both those vectors, and so it must be that:
bb(ul hata) = lamda (bb(ul hatb) xx bb(ul hatc))
And it follows that also:
|| bb(ul hata)|| = || lamda (bb(ul hatb) xx bb(ul hatc))||
:. 1 = |lamda | \ || hatb xx hatc||
:. 1 = |lamda | \ 1/2
:. |lamda | =2 => lamda =+-2
And therefore:
bb(ul hata) = +-2 (bb(ul hatb) xx bb(ul hatc)) \ \ QED