Q24: Diagonals of the parallelogram on $\vec a = 3\hat i - 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat k$

Put $\vec a$ and $\vec b$ tail to tail along two sides of the parallelogram. Going along $\vec a$ and then along the side parallel to $\vec b$ reaches the far corner, so one diagonal is $\vec a + \vec b$ (triangle law).

The other diagonal joins the tips of $\vec b$ and $\vec a$⁠. Since $\vec b + (\vec a - \vec b) = \vec a$⁠, that diagonal is $\vec a - \vec b$⁠.

Add or subtract like components. $\vec b$ has no $\hat j$ part:

So the $\hat j$ component stays $-2$ in both the sum and the difference.

The length of a vector is its magnitude:

Here $4^2 + (-2)^2 + 4^2 = 36$ and $\sqrt{36} = 6$⁠.

$\vec a - \vec b = 2\hat i - 2\hat j + 0\hat k$⁠, so the squares are $2^2 = 4$⁠, $(-2)^2 = 4$⁠, $0^2 = 0$⁠.

$2^2 + (-2)^2 + 0^2 = 8$⁠, and $\sqrt8 = \sqrt{4\cdot 2} = 2\sqrt2$⁠.

A negative of a diagonal vector has the same length, so either sign gives the same answer.