Vectors in Space

Expand the left side coordinate-wise: $|\mathbf u - \mathbf v|^2 = \sum(u_i - v_i)^2 = |\mathbf u|^2 - 2\sum u_iv_i + |\mathbf v|^2$. Comparing the two expansions, the $|\mathbf u|^2$ and $|\mathbf v|^2$ terms cancel and $\sum u_iv_i = |\mathbf u||\mathbf v|\cos\theta$. $\blacksquare$

Now compute $|\mathbf u\times\mathbf v|^2$ from the component formula; the Lagrange identity gives $|\mathbf u\times\mathbf v|^2 = |\mathbf u|^2|\mathbf v|^2 - (\mathbf u\cdot\mathbf v)^2 = |\mathbf u|^2|\mathbf v|^2(1 - \cos^2\theta) = \left(|\mathbf u||\mathbf v|\sin\theta\right)^2$. Taking square roots matches the area exactly. $\blacksquare$

Rearranged, $|\mathbf u\cdot\mathbf v| \le |\mathbf u||\mathbf v|$, with equality exactly when $\mathbf u + t\mathbf v = \mathbf 0$ for some $t$ — parallel vectors. $\blacksquare$

The diagonal of the parallelogram IS the sum; the other diagonal, from head of $\mathbf v$ to head of $\mathbf u$, is the difference $\mathbf u - \mathbf v$. Every force diagram, velocity composition, and displacement chain in physics is this one picture reused.

Reading the picture: when $\theta$ is small the vectors reinforce and the dot product is large and positive; as $\theta$ passes $90^\circ$ the cosine — and the dot product — crosses zero and turns negative. So perpendicularity is the single cleanest test in the subject: compute one number, check if it is $0$. No angle ever needs to be found unless the problem asks for it.

The dashed red segment is what remains, $\mathbf u - \mathrm{proj}_{\mathbf v}\mathbf u$, and the small square marks the right angle that DEFINES the projection: the remainder is exactly perpendicular to $\mathbf v$. The remainder’s length is the distance from $\mathbf u$’s head to the line — which is why this one picture powers point-to-line distances, least squares, and Gram–Schmidt alike.

Now ask: how much do two vectors AGREE in direction? Multiply matching coordinates and add: $\mathbf u\cdot\mathbf v = u_1v_1 + u_2v_2 + u_3v_3$. Why is that a good measure? Because it is LINEAR in each slot ($(\mathbf a + \mathbf b)\cdot\mathbf v = \mathbf a\cdot\mathbf v + \mathbf b\cdot\mathbf v$, straight from distributing the products), it gives $\mathbf u\cdot\mathbf u = |\mathbf u|^2$ (agreement with yourself is total), and — the theorem proved on the proof slide — it equals $|\mathbf u||\mathbf v|\cos\theta$. So the sign of the dot product is the sign of $\cos\theta$: positive means the angle is acute (the vectors lean the same way), zero means perpendicular, negative means obtuse. Every fact in this topic flows from those three properties.

Step 4 — interpret: $\tfrac{63}{65}$ is close to $1$, so the angle is small ($\theta \approx 14.25^\circ$); both vectors point into the first quadrant, so a small positive angle is exactly what geometry predicts. Always end with this sanity check — a negative cosine for two first-quadrant vectors would signal an arithmetic slip.

Step 2 — perpendicularity check: $(0,0,5)\cdot(2,1,0) = 0$ ✓ and $(0,0,5)\cdot(1,3,0) = 0$ ✓, as the cross product must be orthogonal to both inputs.

Step 3 — magnitude check by the area formula: $|\mathbf u| = \sqrt5$, $|\mathbf v| = \sqrt{10}$, and $\cos\theta = \tfrac{2 + 3}{\sqrt5\sqrt{10}} = \tfrac{5}{\sqrt{50}} = \tfrac{1}{\sqrt2}$, so $\sin\theta = \tfrac{1}{\sqrt2}$ and $|\mathbf u||\mathbf v|\sin\theta = \sqrt{50}\cdot\tfrac{1}{\sqrt2} = 5$ — matching $|(0,0,5)| = 5$ exactly. Two independent routes to the same number is how you KNOW the computation is right.