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
Apr 28, 2017

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.!

Apr 28, 2017

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