Dot Product of Vector r and Vector b

To find the value of $\vec{r} \cdot \vec{b}$, we begin with the given vector equation:

$\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$

Rearranging the terms gives:

$(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$

This implies that the vector $(\vec{r} - \vec{c})$ is parallel to $\vec{b}$. Thus, there exists a scalar $\lambda$ such that:

$\vec{r} - \vec{c} = \lambda \vec{b} \implies \vec{r} = \vec{c} + \lambda \vec{b}$

We are given the condition $\vec{r} \cdot \vec{a} = 0$. Substituting $\vec{r} = \vec{c} + \lambda \vec{b}$ into this condition:

$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$
$\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$

Now, using the given vectors:

• $\vec{a} = -\hat{i} - \hat{k}$
• $\vec{b} = -\hat{i} + \hat{j}$
• $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k}$

We calculate the dot products:

$\vec{c} \cdot \vec{a} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{i} - \hat{k}) = (1)(-1) + (2)(0) + (3)(-1) = -1 - 3 = -4$

$\vec{b} \cdot \vec{a} = (-\hat{i} + \hat{j}) \cdot (-\hat{i} - \hat{k}) = (-1)(-1) + (1)(0) + (0)(-1) = 1$

Substituting these values into the equation:

$-4 + \lambda (1) = 0 \implies \lambda = 4$

Now, we substitute $\lambda = 4$ into the expression for $\vec{r}$:

$\vec{r} = \vec{c} + 4\vec{b}$

Finally, we compute $\vec{r} \cdot \vec{b}$:

$\vec{r} \cdot \vec{b} = (\vec{c} + 4\vec{b}) \cdot \vec{b} = \vec{c} \cdot \vec{b} + 4|\vec{b}|^2$

Calculating $\vec{c} \cdot \vec{b}$ and $|\vec{b}|^2$:

$\vec{c} \cdot \vec{b} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{i} + \hat{j}) = (1)(-1) + (2)(1) + (3)(0) = 1$

$|\vec{b}|^2 = (-1)^2 + (1)^2 + (0)^2 = 2$

Thus,

$\vec{r} \cdot \vec{b} = 1 + 4(2) = 1 + 8 = 9$