Vectors in Space

Why the cosine appears: the law of cosines on the triangle with sides $\mathbf u$, $\mathbf v$ and $\mathbf u - \mathbf v$ gives $|\mathbf u - \mathbf v|^2 = |\mathbf u|^2 + |\mathbf v|^2 - 2|\mathbf u||\mathbf v|\cos\theta$, while multiplying out the left side coordinate by coordinate gives $|\mathbf u|^2 + |\mathbf v|^2 - 2(u_1v_1 + u_2v_2 + u_3v_3)$. The two expansions must agree, which leaves exactly $u_1v_1 + u_2v_2 + u_3v_3 = |\mathbf u||\mathbf v|\cos\theta$ (the full proof is below). The lengths are positive, so the sign of the dot product is the sign of $\cos\theta$. In the figure $\mathbf u = (3, 1)$ meets three partners: $(1, 2)$ gives $3 + 2 = 5$, an acute angle; $(-1, 3)$ gives $-3 + 3 = 0$, a right angle; $(-2, 1)$ gives $-6 + 1 = -5$, an obtuse angle.

Why the length is an area: the parallelogram on $\mathbf u$ and $\mathbf v$ has base $|\mathbf u|$ and height $|\mathbf v|\sin\theta$, and the cross product is built so that its length is that product. For plane vectors you can see it with no trigonometry. Take $\mathbf u = (4, 1, 0)$ and $\mathbf v = (1, 3, 0)$, so $\mathbf u\times\mathbf v = (0, 0, 4\cdot3 - 1\cdot1) = (0, 0, 11)$. The parallelogram sits inside a $5\times4$ box of area $20$; the pieces of the box outside it are two triangles of area $\tfrac12\cdot4\cdot1 = 2$, two of area $\tfrac12\cdot1\cdot3 = \tfrac32$ and two unit squares, and $20 - 4 - 3 - 2 = 11$. In letters, $(a+c)(b+d) - ab - cd - 2bc = ad - bc$.

Why the angle must be $60^\circ$: the third side of the triangle is $(1,1,0) - (0,1,1) = (1, 0, -1)$, also of length $\sqrt2$. All three sides are face diagonals of the unit cube, so the triangle is equilateral and each angle is $60^\circ$. The dot product reached the same answer in one line without ever drawing the cube.

Where the formula comes from: write $\mathbf u = c\,\mathbf v + \mathbf w$ and demand $\mathbf w\cdot\mathbf v = 0$. Dotting both sides with $\mathbf v$ removes $\mathbf w$ and leaves $\mathbf u\cdot\mathbf v = c\,(\mathbf v\cdot\mathbf v)$, so $c = \tfrac{\mathbf u\cdot\mathbf v}{\mathbf v\cdot\mathbf v}$ is forced: it is the only multiple of $\mathbf v$ that leaves a perpendicular remainder. The division by $\mathbf v\cdot\mathbf v$ also makes the answer ignore the length of $\mathbf v$: doubling $\mathbf v$ doubles the numerator, quadruples the denominator and doubles $\mathbf v$ itself, which cancels exactly.

The remainder is $\mathbf u - \mathrm{proj}_{\mathbf v}\mathbf u = \left(\tfrac12, -\tfrac12, 2\right)$. Check that it really is perpendicular to $\mathbf v$, since that is the property the formula was built to have: $\tfrac12 - \tfrac12 + 0 = 0$. The scalar component is $\tfrac{3}{\sqrt2} \approx 2.12$, shorter than $|\mathbf u| = 3$ as a shadow must be, and Pythagoras closes the loop: $|\mathrm{proj}|^2 + |\text{remainder}|^2 = \tfrac92 + \tfrac92 = 9 = |\mathbf u|^2$.

The along-ramp part is $-6\,\mathbf d = \left(-\tfrac{24}{5}, -\tfrac{18}{5}\right)$, and the rest, $\mathbf g - \left(-\tfrac{24}{5}, -\tfrac{18}{5}\right) = \left(\tfrac{24}{5}, -\tfrac{32}{5}\right)$, presses straight into the ramp. Its length is $\tfrac{\sqrt{576 + 1024}}{5} = \tfrac{40}{5} = 8$ N. The parts $6$, $8$, $10$ form a right triangle because the two parts are perpendicular: $\tfrac{24}{5}\cdot4 - \tfrac{32}{5}\cdot3 = 0$. Friction is proportional to the $8$ N pressing part, which is why the same block slides more easily on a steeper ramp.

Only the part of $\mathbf F$ along the motion does work, and that part times the distance is $\mathrm{comp}_{\mathbf d}\mathbf F\cdot|\mathbf d| = \mathbf F\cdot\mathbf d$. So $W = 3\cdot3 + (-1)\cdot2 + 2\cdot1 = 9 - 2 + 2 = 9$ J. The negative term is the $y$-part of the force pushing against the motion in $y$; the dot product books it as work taken away.

Why: $\mathbf v\times\mathbf w$ is perpendicular to the base parallelogram and its length is the base area. Dotting it with $\mathbf u$ multiplies that area by the component of $\mathbf u$ along the normal, which is the height of the solid, so the result is base times height. If $\mathbf u$ lies in the plane of $\mathbf v$ and $\mathbf w$, the height is zero and so is the volume. The sign records orientation: positive when $\mathbf u, \mathbf v, \mathbf w$ are arranged like the thumb, index and middle fingers of a right hand.

Then dot with $\mathbf u$ to multiply by the height: $\mathbf u\cdot(\mathbf v\times\mathbf w) = 1 + 12 + 0 = 13$, so the volume is $13$. Cross-check with the determinant expanded along its first row: $1(1\cdot1 - 3\cdot0) - 2(0\cdot1 - 3\cdot2) + 0 = 1 + 12 = 13$. Two routes, one number.

Zero volume, so yes: the three are coplanar, and indeed $(7, 8, 9) = 2(4, 5, 6) - (1, 2, 3)$. Change the last entry to $10$: the determinant becomes $1(50 - 48) - 2(40 - 42) + 3(32 - 35) = 2 + 4 - 9 = -3$, so $(1,2,3)$, $(4,5,6)$, $(7,8,10)$ span a solid of volume $3$. One changed entry turns a flat triple into a genuine solid, which is why this test is done with exact arithmetic, not by eye.

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 segment is what remains, $\mathbf u - \mathrm{proj}_{\mathbf v}\mathbf u = \left(-\tfrac{10}{17}, \tfrac{40}{17}\right)$, 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. Here the check is one line: $\left(-\tfrac{10}{17}\right)(4) + \tfrac{40}{17}(1) = 0$, and the remainder has length $\tfrac{10}{\sqrt{17}} \approx 2.43$, the distance from the point $(2, 3)$ to the line through $(4, 1)$.

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.