Question #118540
Given α=i+2j+3k, β=2i+j−k find α×β
a.5i+7j-3k
b.-5i+7j-3k
c.-5i+7j+3k
d.-5i-7j-3k
1
Expert's answer
2020-06-08T17:56:29-0400

α=i+2j+3k,β=2i+jk\alpha = i + 2j + 3k, \beta = 2i+j-k


* indicates the cross product

α\alpha x β\beta = (i+2j+3k)(i+2j+3k) * (2i+jk)(2i+j-k)

=2(ii)+(ij)(ik)+4(ji)+2(jj)2(jk)+6(ki)+3(kj)3(kk)= 2(i*i) + (i*j) - (i*k) + 4(j*i) + 2(j*j) -2(j*k) +6(k*i) + 3(k*j) - 3(k*k)


since ii=jj=kk=0,ij=k,jk=i,ki=j,ji=k,kj=i,i*i = j*j = k*k = 0, i*j = k, j*k = i, k*i = j, j*i = - k , k*j = -i,

ik=ji*k = -j


so α\alpha x β\beta =2(0)+k+j4k+2(0)2i+6j3i3(0)= 2(0) + k + j -4k +2(0) - 2i + 6j -3i -3(0)

=5i+7j3k= -5i + 7j - 3k


So option b is correct.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS