Exercises — Dot product — formula, geometric meaning, work calculation

WHAT: The dot product always collapses two arrows into a single plain number — a scalar. Compute with the component form $\vec{A}\cdot \vec{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$: $\vec{A}\cdot \vec{B}=(1)(4)+(2)(5)+(3)(6)=4+10+18=32.$ No direction attached — just the number $32$. It is a scalar.

WHY this test: two non-zero vectors are perpendicular exactly when their dot product is $0$ (because $\cos 90^{\circ }=0$). $\vec{A}\cdot \vec{B}=(2)(3)+(3)(-2)=6-6=0.$ The dot product is $0$, so yes — they are perpendicular ($\theta =90^{\circ }$).

WHY: a vector dotted with itself gives $|\vec{A}{|}^2$ (angle with itself is $0^{\circ }$, $\cos 0=1$). $\vec{A}\cdot \vec{A}=(3)(3)+(4)(4)=9+16=25,|\vec{A}|=\sqrt{25}=5.$

Multiply like-with-like, then add: $\vec{A}\cdot \vec{B}=(5)(-3)+(-2)(4)+(1)(2)=-15-8+2=-21.$ WHAT the sign means: negative ⇒ the vectors mostly oppose each other (angle beyond $90^{\circ }$).

WHY $\cos \theta$: only the part of the force along the motion does work — that part is $|\vec{F}|\cos \theta$ (the shadow of $\vec{F}$ on $\vec{d}$, exactly the amber segment in the figure). $W=|\vec{F}||\vec{d}|\cos \theta =(20)(4)\cos 30^{\circ }=80\times 0.8660=69.28\text{ J}.$ Units: $\text{N⋅m}=\text{J}$.

WHY skip the angle: with components in hand, $W=\vec{F}\cdot \vec{d}$ is just multiply-and-add. $W=(6)(3)+(-8)(1)=18-8=10\text{ J}.$

WHY divide by magnitudes: rearranging the geometric form gives $\cos \theta =\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}$; the denominator strips out the lengths so only the alignment remains. $\vec{A}\cdot \vec{B}=(1)(2)+(2)(0)+(2)(-1)=2+0-2=0.$ $\cos \theta =\frac{0}{|\vec{A}||\vec{B}|}=0\Rightarrow \theta =90^{\circ }.$ The dot product is $0$, so they are perpendicular — no need to even compute the magnitudes here.

$\vec{A}\cdot \vec{B}=(-2)(3)+(1)(1)=-6+1=-5.$ Interpret: negative ⇒ $\cos \theta <0$ ⇒ $\theta$ is obtuse ($>90^{\circ }$). The vectors partly oppose one another. (Look again at the figure: if $\vec{A}$'s shadow fell on the opposite side of $\vec{B}$, the shadow length — and hence the dot product — turns negative.)

WHY this formula: the projection length is $|\vec{A}|\cos \theta =\frac{\vec{A}\cdot \vec{B}}{|\vec{B}|}$ (see Projection of a vector). Since $\vec{B}=(1,0)$ points along the $x$-axis, its length is $1$. $\vec{A}\cdot \vec{B}=(4)(1)+(3)(0)=4,|\vec{B}|=1.$ $\text{projection}=\frac{4}{1}=4.$ That is exactly the $x$-component of $\vec{A}$ — the shadow onto the $x$-axis is $A_x$.

WHY: perpendicular ⇔ dot product $=0$. Set it up and solve for $k$. $\vec{A}\cdot \vec{B}=(2)(3)+(k)(-2)+(1)(4)=6-2k+4=10-2k.$ $10-2k=0\Rightarrow k=5.$

WHY add: work is a scalar, and the dot product is distributive, so total work is the sum over stages: $W=\vec{F}\cdot {\vec{d}}_1+\vec{F}\cdot {\vec{d}}_2=\vec{F}\cdot ({\vec{d}}_1+{\vec{d}}_2)$. $\vec{F}\cdot {\vec{d}}_1=(3)(2)+(4)(0)=6,\vec{F}\cdot {\vec{d}}_2=(3)(0)+(4)(3)=12.$ $W=6+12=18\text{ J}.$ Check via the combined displacement ${\vec{d}}_1+{\vec{d}}_2=(2,3)$: $(3)(2)+(4)(3)=6+12=18$ ✓.

First build $\vec{R}=(1,0)+(0,1)=(1,1)$. $\vec{A}\cdot \vec{R}=(1)(1)+(0)(1)=1.$ $|\vec{A}|=1,|\vec{R}|=\sqrt{1^2+1^2}=\sqrt{2}.$ $\cos \theta =\frac{1}{1\cdot \sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =45^{\circ }.$ Makes sense: $\vec{R}$ is the diagonal of a unit square, tilted $45^{\circ }$ off $\vec{A}$.

Displacement in components: $\vec{d}=(5,0)\text{ m}$ (horizontal). Applied force: $W_{\text{app}}={\vec{F}}_{\text{app}}\cdot \vec{d}=(12)(5)+(5)(0)=60\text{ J}.$ Only the horizontal $12\text{ N}$ part does work; the vertical $5\text{ N}$ is perpendicular to $\vec{d}$ and contributes $0$ (its shadow on $\vec{d}$ has zero length). Friction: $W_{\text{fric}}=\vec{f}\cdot \vec{d}=(-4)(5)+(0)(0)=-20\text{ J}.$ Negative — friction removes energy, exactly as it should. Net work: $W_{\text{net}}=60+(-20)=40\text{ J}.$ By the Work-Energy Theorem, this $+40\text{ J}$ equals the block's gain in kinetic energy.

WHY we expect zero: in Circular motion the centripetal force always points to the centre, which is perpendicular to the velocity (and to the displacement, which follows $\vec{v}$). $W=\vec{F}\cdot \vec{d}=(5)(0)+(0)(0.03)=0\text{ J}.$ A perpendicular force does no work, so the speed (and kinetic energy) stays constant even though the direction keeps changing.

Set up $\cos 60^{\circ }=\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}$ with $\cos 60^{\circ }=\frac{1}{2}$. $\vec{A}\cdot \vec{B}=(2)(1)+(1)(c)+(-2)(1)=2+c-2=c.$ $|\vec{A}|=\sqrt{4+1+4}=3,|\vec{B}|=\sqrt{1+c^2+1}=\sqrt{c^2+2}.$ $\frac{c}{3\sqrt{c^2+2}}=\frac{1}{2}\Rightarrow 2c=3\sqrt{c^2+2}.$ Square both sides (valid since $c>0$ is required for a $60^{\circ }$ acute angle): $4c^2=9(c^2+2)\Rightarrow 4c^2=9c^2+18\Rightarrow -5c^2=18.$ This gives $c^2=-3.6$, which is impossible — no real $c$ works. WHAT this means: the geometry forbids a $60^{\circ }$ angle here. As $c$ grows, $\vec{B}$'s length grows faster than the dot product, so $\cos \theta$ can never reach $\frac{1}{2}$; the smallest attainable angle stays above $60^{\circ }$. Always sanity-check that a solution can exist before trusting algebra.