$$( \vec { i } \times 78 \vec { j } ) \cdot \vec { k } = \sqrt { 13 } \vec { i } \cdot ( \sqrt { 13 } \vec { j } \times 6 \vec { k } )$$

Question

Quizplus · Accepted Answer

Both sides of the equation evaluate to 78. On the left side, $\vec{i} \times 78\vec{j}$ results in $78\vec{k}$, and the dot product with $\vec{k}$ gives 78. On the right side, $\sqrt{13}\vec{j} \times 6\vec{k}$ results in $78\vec{i}$, and the dot product with $\sqrt{13}\vec{i}$ also gives 78.

(i⃗×78j⃗)⋅k⃗=13i⃗⋅(13j⃗×6k⃗)( \vec { i } \times 78 \vec { j } ) \cdot \vec { k } = \sqrt { 13 } \vec { i } \cdot ( \sqrt { 13 } \vec { j } \times 6 \vec { k } )(i×78j​)⋅k=13​i⋅(13​j​×6k)

$( \vec { i } \times 78 \vec { j } ) \cdot \vec { k } = \sqrt { 13 } \vec { i } \cdot ( \sqrt { 13 } \vec { j } \times 6 \vec { k } )$