1. Given vectors

$\vec\alpha = 3\hat i + \hat j = (3,1,0), \qquad \vec\beta = 2\hat i - \hat j + 3\hat k = (2,-1,3)$

We are told

$\vec\beta = \vec\beta_1 - \vec\beta_2$

where

• $\vec\beta_1 \parallel \vec\alpha$
• $\vec\beta_2 \perp \vec\alpha$

We need to find $\vec\beta_1 \times \vec\beta_2$.

2. Rewrite the decomposition

From

$\vec\beta = \vec\beta_1 - \vec\beta_2$

we get

$\vec\beta_1 = \vec\beta + \vec\beta_2$

Since $\vec\beta_1$ is parallel to $\vec\alpha$ and $\vec\beta_2$ is perpendicular to $\vec\alpha$, this means $\vec\beta_1$ is the component of $\vec\beta$ along $\vec\alpha$, but with the sign adjusted due to the given form.

Let

$\vec\beta_1 = \lambda \vec\alpha = \lambda(3,1,0)$

Then

$\vec\beta_2 = \vec\beta_1 - \vec\beta$

because

$\vec\beta = \vec\beta_1 - \vec\beta_2 \implies \vec\beta_2 = \vec\beta_1 - \vec\beta$

Now use the condition $\vec\beta_2 \perp \vec\alpha$.

3. Apply perpendicularity condition

$\vec\beta_2 \cdot \vec\alpha = 0$

So,

$(\vec\beta_1 - \vec\beta)\cdot \vec\alpha = 0$

Substitute $\vec\beta_1 = \lambda \vec\alpha$:

$(\lambda \vec\alpha - \vec\beta)\cdot \vec\alpha = 0$

$\lambda (\vec\alpha\cdot\vec\alpha) - \vec\beta\cdot\vec\alpha = 0$

Now,

$\vec\alpha\cdot\vec\alpha = 3^2+1^2=10$

$\vec\beta\cdot\vec\alpha = (2)(3)+(-1)(1)+(3)(0)=6-1=5$

Thus,

$10\lambda - 5 = 0 \implies \lambda = \frac12$

Hence,

$\vec\beta_1 = \frac12(3\hat i+\hat j)=\frac32\hat i+\frac12\hat j$

4. Find $\vec\beta_2$

$\vec\beta_2 = \vec\beta_1 - \vec\beta$

$\vec\beta_2 = \left(\frac32,\frac12,0\right) - (2,-1,3)$

$\vec\beta_2 = \left(-\frac12,\frac32,-3\right)$

So,

$\vec\beta_2 = -\frac12\hat i + \frac32\hat j - 3\hat k$

5. Compute $\vec\beta_1 \times \vec\beta_2$