Task. Given two vector A = 5 i \mathrm{A} = 5\mathrm{i} A = 5 i , B = 3 j \mathrm{B} = 3\mathrm{j} B = 3 j . What is the direction of the cross product [ B × A ] [B \times A] [ B × A ] ?
Solution. Since [ j × i ] = − k [j \times i] = -k [ j × i ] = − k , we obtain that [ B × A ] = − 15 k [B \times A] = -15k [ B × A ] = − 15 k .
We can produce direct calculation. We have that A = ( 5 , 0 , 0 ) = 5 i A = (5,0,0) = 5i A = ( 5 , 0 , 0 ) = 5 i , B = ( 0 , 3 , 0 ) = 3 j B = (0,3,0) = 3j B = ( 0 , 3 , 0 ) = 3 j . Then [ B × A ] [B \times A] [ B × A ] has the following coordinates:
[ B × A ] = ( ∣ 3 0 0 0 ∣ , ∣ 0 0 0 5 ∣ , ∣ 0 3 5 0 ∣ ) = ( 3 ∗ 0 − 0 ∗ 0 , 0 ∗ 5 − 0 ∗ 0 , 0 ∗ 0 − 3 ∗ 5 ) = ( 0 , 0 , − 15 ) = − 15 k . \begin{aligned}
[B \times A] &= \left( \left| \begin{array}{cc} 3 & 0 \\ 0 & 0 \end{array} \right|, \left| \begin{array}{cc} 0 & 0 \\ 0 & 5 \end{array} \right|, \left| \begin{array}{cc} 0 & 3 \\ 5 & 0 \end{array} \right| \right) \\
&= (3 * 0 - 0 * 0, 0 * 5 - 0 * 0, 0 * 0 - 3 * 5) \\
&= (0, 0, -15) = -15k.
\end{aligned} [ B × A ] = ( ∣ ∣ 3 0 0 0 ∣ ∣ , ∣ ∣ 0 0 0 5 ∣ ∣ , ∣ ∣ 0 5 3 0 ∣ ∣ ) = ( 3 ∗ 0 − 0 ∗ 0 , 0 ∗ 5 − 0 ∗ 0 , 0 ∗ 0 − 3 ∗ 5 ) = ( 0 , 0 , − 15 ) = − 15 k .
Answer. [ B × A ] = − 15 k [B \times A] = -15k [ B × A ] = − 15 k .