Question

i[abi]+j[abj]+k[abk]=?

Answer 1

ia.(b x i) + aj.(b x j) + ck.(b x k)

Answer 2

how do i further proceed if it is correct?

Answer 3

is this correct?

Answer 4

gives abi(k-j)-abj(k-i)+abk(j-i)

Answer 5

@ganeshie8

Answer 6

@hartnn ....

Answer 7

bxi = b3j -b2k
so
[ abi] = a2b3 -a3b2
so
i[abi] = (a2b3 - a3b2)i

Answer 8

note that we could get to this result by:
[abi] = [iab]
which is the i component of axb

Answer 9

so i[abi]
is the i component of axb in the i direction

Answer 10

so eventually:
i[abi]+j[abj]+k[abk] = i[iab] + j[jab] + k[kab] = axb

Answer 11

how is it equal to axb in the end?

Answer 12

we saw that
i[abi] = i[iab] = i ( i dot axb ) = i ( i component of axb)
so it is the i component of axb in i direction

Answer 13

i[abi]+j[abj]+k[abk]

i[iab]+j[jab]+k[kab]


\[

\hat{i} \left| \begin{array}{ccc}
\ 1 & 0 & 0 \\
\ a_1 & a_2 & a_3 \\
\ b_1 & b_2 & b_3 \\
\end{array} \right| +

\hat{j} \left| \begin{array}{ccc}
\ 0 & 1 & 0 \\
\ a_1 & a_2 & a_3 \\
\ b_1 & b_2 & b_3 \\
\end{array} \right| +

\hat{k} \left| \begin{array}{ccc}
\ 0 & 0 & 1 \\
\ a_1 & a_2 & a_3 \\
\ b_1 & b_2 & b_3 \\
\end{array} \right|

\]

\[
\left| \begin{array}{ccc}
\ \hat{i} & \hat{j} & \hat{k} \\
\ a_1 & a_2 & a_3 \\
\ b_1 & b_2 & b_3 \\
\end{array} \right|

\]

\(\vec{A} \times \vec{B}\)