Question

`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))`?

Answer 1

For Proof, refer to the Explanation.

Explanation:

Given that,

for unit vectors `veca & vec b, veca.vecb=0, &, veca.vecc=0.`

`:. veca bot vecb, and, veca bot vecc," i.e., to say,"`

`veca" is "bot "both "vecb and vecc.`

Clearly, `veca` is along `vecbxxvecc.`

Therefore, let us suppose that,

`veca=k(vecbxxvecc) ; kne0............(ast)`

`rArr ||veca||=||k(vecbxxvecc)||`

`rArr ||veca||=|k|||vecb||||vecc||sin/_(vecb,vecc)`

`rArr 1=|k|*1*1*sin(pi/6)`

`rArr |k|=2, or, k=+-2.`

Hence, by `(ast), veca=+-2(vecbxxvecc).`

This completes the Proof.

Enjoy Maths.!

Answer 2

`bb(ul hata)`, `bb(ul hatb)` and `bb(ul hatc)` are unit vectors and so:

`|| 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 to `bb(ul hatb)`
`bb(ul hat(a)) * bb(ul hat(c)) = 0 => bb(ul hata)` perpendicular to `bb(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