Addition of Vectors

Given,

$$\vec a=2\hat i-\hat j+2\hat k$$ and $$\vec b=\hat i+2\hat j-2\hat k$$

First compute

$$\vec a\times\vec b$$

$$= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -1 & 2\\ 1 & 2 & -2 \end{vmatrix}$$

$$=\hat i(2-4)-\hat j(-4-2)+\hat k(4+1)$$

$$=-2\hat i+6\hat j+5\hat k$$

Now,

$$\vec a+2\vec b$$

$$=(2,-1,2)+2(1,2,-2)$$

$$=(4,3,-2)$$

Let

$$\vec p=\vec a\times\vec b=(-2,6,5)$$

and

$$\vec q=\vec a+2\vec b=(4,3,-2)$$

Using

$$|\vec p\times\vec q|^2=|\vec p|^2|\vec q|^2-(\vec p\cdot\vec q)^2,$$

first compute:

$$|\vec p|^2=(-2)^2+6^2+5^2$$

$$=4+36+25$$

$$=65$$

Also,

$$|\vec q|^2=4^2+3^2+(-2)^2$$

$$=16+9+4$$

$$=29$$

Now,

$$\vec p\cdot\vec q=(-2)(4)+6(3)+5(-2)$$

$$=-8+18-10$$

$$=0$$

Therefore,

$$|\vec p\times\vec q|^2=65\times29$$

$$=1885$$

$$\vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0} \implies (\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$$
This implies that the vector $$(\vec{r} - \vec{b})$$ is parallel to $$\vec{a}$$. Therefore, we can write:
$$\vec{r} - \vec{b} = \lambda \vec{a} \implies \vec{r} = \vec{b} + \lambda \vec{a}$$
Now, substitute this into the second condition ($$\vec{r} \cdot \vec{b} = 0$$):
$$(\vec{b} + \lambda \vec{a}) \cdot \vec{b} = 0$$
$$|\vec{b}|^2 + \lambda (\vec{a} \cdot \vec{b}) = 0$$
Let's compute the scalar values:
$$|\vec{b}|^2 = 1^2 + (-1)^2 + 1^2 = 3$$
$$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1$$
Substitute these back to solve for $$\lambda$$:
$$3 + \lambda(1) = 0 \implies \lambda = -3$$
Now, find vector $$\vec{r}$$:
$$\vec{r} = \vec{b} - 3\vec{a} = (\hat{i} - \hat{j} + \hat{k}) - 3(\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} - 4\hat{j} - 2\hat{k}$$
Finally, compute $$|\vec{r}|^2$$:
$$|\vec{r}|^2 = (-2)^2 + (-4)^2 + (-2)^2 = 4 + 16 + 4 = 24$$