Question
$\mathrm{A}=+\mathrm{i} \wedge+\mathrm{j} \wedge-2 \mathrm{k} \wedge$ and $\mathrm{B} \overrightarrow{\mathrm{i}} \wedge-\mathrm{j} \wedge+\mathrm{k} \wedge$ Find the unit vector indirection of $\mathrm{A} \rightarrow \times \mathrm{B}^{\rightarrow}$(A) $[1 / \sqrt{(23)}](-\mathrm{i} \wedge-5 \mathrm{j} \wedge-2 \mathrm{k} \wedge)$(B) $[1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 \mathrm{j} \wedge-3 \mathrm{k} \wedge)$(C) $[1 / \sqrt{(29})](-i \wedge-5 j \wedge-3 k \wedge)$(D) $[1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 j \wedge-3 \mathrm{k} \wedge)$
Step 1
The cross product of two vectors A and B is given by: \[A \times B = (A_yB_z - A_zB_y)i + (A_zB_x - A_xB_z)j + (A_xB_y - A_yB_x)k\] where \(A_x, A_y, A_z\) and \(B_x, B_y, B_z\) are the components of vectors A and B respectively. Show more…
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