If the vector b=3 j+4 k is written as the sum of a vector b_1, parallel to a= i+ j, and a …

Question

If the vector $\vec b=3\hat j+4\hat k$ is written as the sum of a vector $\vec b_1$, parallel to $\vec a=\hat i+\hat j$, and a vector $\vec b_2$, perpendicular to $\vec a$, then $\vec b_1	imes\vec b_2$ is equal to:

Accepted Answer

$\vec b_1 = \dfrac{\vec a\cdot\vec b}{|\vec a|^2}\vec a = \dfrac{3}{2}(\hat i+\hat j)$, so $\vec b_2 = \vec b-\vec b_1 = -\dfrac32\hat i+\dfrac32\hat j+4\hat k$

$$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\ \frac32&\frac32&0\ -\frac32&\frac32&4\end{vmatrix} = \hat i(6)-\hat j(6)+\hat k\left(\frac94+\frac94\right) = 6\hat i-6\hat j+\frac92\hat k$$

Solutions · 1