# (2j×3i).k=? My text book says the answer is -6. But how?

Jun 24, 2018

See below

#### Explanation:

$2 \hat{j} \times 3 \hat{i} = 2 \times 3 \left(\hat{j} \times \hat{i}\right) = 6 \left(- \hat{i} \times \hat{j}\right) = - 6 \hat{k}$

Thus

$\left(2 \hat{j} \times 3 \hat{i}\right) \cdot \hat{k} = - 6 \setminus \hat{k} \cdot \hat{k} = - 6$

Here we have used the following properties

• $\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$
• $\hat{i} \times \hat{j} = \hat{k}$
• $\hat{k} \cdot \hat{k} = | \hat{k} {|}^{2} = 1$

You could of course also use the determinant form for the scalar triple product

$\left(2 \hat{j} \times 3 \hat{i}\right) \cdot \hat{k} = \hat{k} \cdot \left(2 \hat{j} \times 3 \hat{i}\right)$
$q \quad q \quad q \quad q \quad q \quad \quad = | \left(0 , 0 , 1\right) , \left(0 , 2 , 0\right) , \left(3 , 0 , 0\right) | = - 6$
but this will be an overkill for this problem.