#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