The answer is?

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1 Answer
Mar 12, 2018

#6 hat i-6hat j+9/2 hat k#

Explanation:

Since #vec{b}_1# and #vec{b}_2# are parallel and perpendicular to #vec a#, respectively, we have #vec{b}_2 cdot vec {a} = 0# and

#|vec{b}_1| = {vec{b}_1cdot vec a}/|vec a|= {vec{b}cdot vec a}/|vec a|#

So that

#vec{b}_1 = {vec{b}cdot vec a}/|vec a| {vec a}/|vec a|= {(vec{b}cdot vec a)vec a}/|vec a|^2#

and

#vec{b}_2 = vec b - vec{b}_1=vec b -{(vec{b}cdot vec a)vec a}/|vec a|^2#

So

#vec{b}_1 times vec{b}_2 = vec{b}_1 times (vec b-vec{b}_1) = vec{b}_1 times vec {b}={(vec{b}cdot vec a)(vec a times vec b)}/|vec a|^2#

Now, given #vec b = 3 hat j +4 hat k# and #vec a = hat i + hat j#, we have

#vec b cdot vec a = 3#
#vec a times vec b = (hat i + hat j)times (3 hat j +4 hat k) = 3 (hat i times hat j)+4 (hat i times hat k)+4 hat j times hat k = 3hat k-4 hat j+4 hat i#
# |vec a|^2 = 2#

So that

#vec{b}_1 times vec{b}_2 = {3(4 hat i-4hat j+3 hat k)}/2 = 6 hat i-6hat j+9/2 hat k#