Exercises — Projection of vectors

WHAT we do: apply ${\text{comp}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$. WHY: $\vec{b}=(1,0)$ is the $x$-axis direction and already has length $1$, so the shadow length is the $x$-component. $\vec{a}\cdot \vec{b}=6(1)+8(0)=6,|\vec{b}|=\sqrt{1^2+0^2}=1.$ ${\text{comp}}_{\vec{b}}\vec{a}=\frac{6}{1}=\boxed{6}.$ WHAT IT LOOKS LIKE: the shadow of $(6,8)$ on the horizontal axis reaches out to $6$ — exactly its horizontal reach.

WHAT/WHY: use ${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}\vec{b}$ because we want a vector, not just a length. $\vec{a}\cdot \vec{b}=0(0)+5(2)=10,\vec{b}\cdot \vec{b}=0^2+2^2=4.$ ${\text{proj}}_{\vec{b}}\vec{a}=\frac{10}{4}(0,2)=\left(0,5\right)=\boxed{(0,5)}.$ WHAT IT LOOKS LIKE: $\vec{a}$ already points straight up the same line as $\vec{b}$, so its shadow is itself. The fraction $\frac{10}{4}=2.5$ times $(0,2)$ lands exactly on $(0,5)$. Perfect self-projection.

$\vec{a}\cdot \vec{b}=3(0)+4(1)=4,|\vec{b}|=1,{\text{comp}}_{\vec{b}}\vec{a}=\frac{4}{1}=\boxed{4}.$ The dot product picked out the vertical component $4$ because $\vec{b}$ is the $y$-axis.

$\vec{a}\cdot \vec{b}=4(3)+3(0)=12,\vec{b}\cdot \vec{b}=3^2+0^2=9.$ ${\text{proj}}_{\vec{b}}\vec{a}=\frac{12}{9}(3,0)=\left(4,0\right)=\boxed{(4,0)}.$ WHY $\vec{b}\cdot \vec{b}$ and not $|\vec{b}|$? The $\frac{1}{9}$ cancels the "extra length" of the long vector $(3,0)$. Notice the answer $(4,0)$ has length $4$ — the horizontal reach of $\vec{a}$ — not $12$. If we'd divided by $|\vec{b}|=3$ only once, we'd get $(4,0)\cdot 3=(12,0)$, three times too long.

$\vec{a}\cdot \vec{b}=2(3)+3(4)=6+12=18,|\vec{b}|=\sqrt{3^2+4^2}=\sqrt{25}=5.$ ${\text{comp}}_{\vec{b}}\vec{a}=\frac{18}{5}=\boxed{3.6}.$ WHAT IT LOOKS LIKE: the shadow of $\vec{a}$ on the slanted line through $(3,4)$ is $3.6$ units long — positive, so it falls on the same side $\vec{b}$ points.

$\vec{a}\cdot \vec{b}=1(-2)+1(0)=-2,|\vec{b}|=\sqrt{(-2{)}^2}=2.$ ${\text{comp}}_{\vec{b}}\vec{a}=\frac{-2}{2}=\boxed{-1}.$ WHY negative? $\vec{b}$ points left ($-x$), but $\vec{a}$ leans right and up — the angle between them exceeds $90^{\circ }$, so $\cos \theta <0$. The sign is data: the shadow lands on the back side of $\vec{b}$.

Parallel part: $\vec{a}\cdot \vec{b}=3,\vec{b}\cdot \vec{b}=1,{\text{proj}}_{\vec{b}}\vec{a}=\frac{3}{1}(1,0)=(3,0).$ Perpendicular part (what's left over — see Orthogonal decomposition): ${\vec{a}}_{\perp }=\vec{a}-{\text{proj}}_{\vec{b}}\vec{a}=(3,3)-(3,0)=(0,3).$ Check orthogonality: ${\vec{a}}_{\perp }\cdot \vec{b}=(0)(1)+(3)(0)=0$ ✓. WHAT IT LOOKS LIKE: $(3,3)$ = horizontal reach $(3,0)$ + vertical climb $(0,3)$, a clean right-angle split.

$\vec{a}\cdot \vec{b}=2(1)+1(2)+2(2)=2+2+4=8,\vec{b}\cdot \vec{b}=1+4+4=9.$ ${\text{proj}}_{\vec{b}}\vec{a}=\frac{8}{9}(1,2,2)=\left(\frac{8}{9},\frac{16}{9},\frac{16}{9}\right).$ ${\vec{a}}_{\perp }=\left(2-\frac{8}{9},1-\frac{16}{9},2-\frac{16}{9}\right)=\left(\frac{10}{9},-\frac{7}{9},\frac{2}{9}\right).$ Check: ${\vec{a}}_{\perp }\cdot \vec{b}=\frac{10}{9}(1)+(-\frac{7}{9})(2)+\frac{2}{9}(2)=\frac{10-14+4}{9}=\frac{0}{9}=0$ ✓.

WHY projection gives distance: the perpendicular part ${\vec{a}}_{\perp }$ is the straight-line gap from the tip to the line, and its length is that distance. $\vec{a}\cdot \vec{b}=5(1)+2(1)=7,\vec{b}\cdot \vec{b}=2,{\text{proj}}_{\vec{b}}\vec{a}=\frac{7}{2}(1,1)=(3.5,3.5).$ ${\vec{a}}_{\perp }=(5,2)-(3.5,3.5)=(1.5,-1.5).$ $\text{distance}=|{\vec{a}}_{\perp }|=\sqrt{1.5^2+(-1.5{)}^2}=\sqrt{4.5}=\frac{3}{\sqrt{2}}\approx \boxed{2.1213}.$

Remove the $\vec{u}$-component: $\vec{v}\cdot \vec{u}=2,\vec{u}\cdot \vec{u}=1,{\text{proj}}_{\vec{u}}\vec{v}=(2,0).$ $\vec{w}=\vec{v}-{\text{proj}}_{\vec{u}}\vec{v}=(2,3)-(2,0)=(0,3).$ Normalise ($|\vec{w}|=3$): $\hat{w}=\frac{(0,3)}{3}=\boxed{(0,1)}.$ WHAT IT LOOKS LIKE: $\vec{u}$ points along $x$; Gram-Schmidt strips $\vec{v}$'s horizontal share and keeps only the pure vertical, giving the orthonormal partner $(0,1)$.

WHY dot product: two vectors are perpendicular exactly when their projection onto each other is zero, i.e. $\vec{a}\cdot \vec{b}=0$. $\vec{a}\cdot \vec{b}=2t+4=0\Rightarrow t=\boxed{-2}.$ Check: $(-2,4)\cdot (2,1)=-4+4=0$ ✓. With $t=-2$, ${\text{proj}}_{\vec{b}}\vec{a}=\vec{0}$ — no shadow at all.

WHY dot product is work: $W=\vec{F}\cdot \vec{d}$ counts only the part of the force along the displacement — projection in disguise. $W=\vec{F}\cdot \vec{d}=6(4)+8(0)=24\text{J}.$ Wasted (perpendicular) part of the force: project $\vec{F}$ onto $\vec{d}$, then subtract. ${\text{proj}}_{\vec{d}}\vec{F}=\frac{24}{16}(4,0)=(6,0),{\vec{F}}_{\perp }=(6,8)-(6,0)=(0,8)\text{N}.$ The $(0,8)$ N pushes sideways and does no work because it is perpendicular to $\vec{d}$. $\boxed{W=24\text{J},{\vec{F}}_{\perp }=(0,8)}$.

Let $\vec{p}={\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}\vec{b}$. Write $k=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}$, so $\vec{p}=k\vec{b}$. Now project $\vec{p}$ again: ${\text{proj}}_{\vec{b}}\vec{p}=\frac{\vec{p}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}\vec{b}=\frac{(k\vec{b})\cdot \vec{b}}{\vec{b}\cdot \vec{b}}\vec{b}=\frac{k(\vec{b}\cdot \vec{b})}{\vec{b}\cdot \vec{b}}\vec{b}=k\vec{b}=\vec{p}.$ WHAT IT MEANS: $\vec{p}$ already lies on the line of $\vec{b}$, so its shadow is itself. A shadow of a shadow is the same shadow. $■$

$\vec{a}\cdot \vec{b}=5,\vec{b}\cdot \vec{b}=2,\vec{p}=\frac{5}{2}(1,1)=(2.5,2.5).$ Re-project: $\vec{p}\cdot \vec{b}=5,\frac{5}{2}(1,1)=(2.5,2.5)$ — unchanged ✓.

Scalar: ${\text{comp}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$. As $\vec{b}\to \vec{0}$, the numerator $\vec{a}\cdot \vec{b}\to 0$ but the denominator $|\vec{b}|\to 0$ too — a $\frac{0}{0}$ that has no limit (it depends on the direction $\vec{b}$ approaches from). Numerically, take $\vec{a}=(1,0)$: along $\vec{b}=(\epsilon ,0)$ you get $\frac{\epsilon }{\epsilon }=1$, but along $\vec{b}=(0,\epsilon )$ you get $\frac{0}{\epsilon }=0$. Two answers ⇒ undefined. Vector: ${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot \vec{b}}\vec{b}$. Here numerator scales like $\epsilon$, denominator like ${\epsilon }^2$, and the trailing $\vec{b}$ like $\epsilon$: overall $\frac{\epsilon }{{\epsilon }^2}\cdot \epsilon =O(1)$ but again direction-dependent — no single limit. WHY it must fail: projecting onto $\vec{0}$ asks "how much of $\vec{a}$ lies along a direction that doesn't exist." There is no line to cast a shadow on. Rule: projection requires $\vec{b}\ne \vec{0}$.

L1 plug in · L2 mind the denominator & sign · L3 decompose and measure distance with ${\vec{a}}_{\perp }$ · L4 fuse with Gram-Schmidt, perpendicularity, work · L5 prove idempotence and respect the $\vec{b}\ne \vec{0}$ rule.