Dot and Cross Product Practice

These questions follow What Vectors Are, Adding and Scaling Vectors, The Dot Product, and The Cross Product — a warm-up on vectors themselves, then each product in turn. Sketch first wherever a sketch is possible; the picture audits the arithmetic.

Vector warm-up

1. Determine the magnitude and direction of the vector $(8,6)$.
2. Represent the vector with magnitude 8 and direction $30°$ anticlockwise from the positive $x$-axis in Cartesian form.
3. A plane on a heading of N 27° E has an airspeed of 375 km/h. The wind is blowing from the south at 62 km/h. Determine the plane’s actual direction of travel and its ground speed.

Answer 1

Magnitude: $\sqrt{8^2+6^2}=\sqrt{100}=10$. Direction: ${\tan }^{-1}\left(\frac{6}{8}\right)\approx 36.9°$ above the positive $x$-axis. A 3–4–5 triangle scaled by 2 — worth recognising on sight.

Answer 2

$(8\cos 30°,8\sin 30°)=(4\sqrt{3},4)\approx (6.93,4)$. Audit: $\sqrt{48+16}=\sqrt{64}=8$. ✓

Answer 3

Components, taking east as $x$ and north as $y$. The heading N 27° E leans $27°$ east of north: $(375\sin 27°,375\cos 27°)\approx (170.3,334.1)$. Wind from the south blows toward the north: $(0,62)$. Sum: $(170.3,396.1)$. Ground speed: $\sqrt{170.3^2+396.1^2}\approx 431$ km/h. Direction: ${\tan }^{-1}\left(\frac{170.3}{396.1}\right)\approx 23.3°$ east of north — about N 23° E. The tailwind adds speed and swings the travel direction closer to north than the heading; both changes should feel right before the arithmetic confirms them.

The dot product

4. For $\vec{a}=(3,-1,2)$ and $\vec{b}=(1,4,-2)$, compute $\vec{a}\cdot \vec{b}$ and determine the angle between the vectors.
5. Determine $k$ so that $(2,k,1)$ and $(3,-2,4)$ are orthogonal.
6. Determine the projection of $\vec{a}=(6,2)$ onto $\vec{b}=(3,4)$ — the scalar amount and the vector.

Answer 4

Dot product: $3(1)+(-1)(4)+2(-2)=3-4-4=-5$. Magnitudes: $|\vec{a}|=\sqrt{14}$, $|\vec{b}|=\sqrt{21}$. So $\cos \theta =\frac{-5}{\sqrt{14}\sqrt{21}}=\frac{-5}{7\sqrt{6}}\approx -0.292,\theta \approx 107°$ The negative dot product promised an obtuse angle before the inverse cosine said which one.

Answer 5

Orthogonal means dot product zero: $2(3)+k(-2)+1(4)=10-2k=0$, so $k=5$. Check: $(2,5,1)\cdot (3,-2,4)=6-10+4=0$. ✓

Answer 6

$\vec{a}\cdot \vec{b}=18+8=26$ and $|\vec{b}|=5$. Scalar projection: $\frac{26}{5}=5.2$ — how much of $\vec{a}$ points along $\vec{b}$. Vector projection: that amount in $\vec{b}$‘s unit direction, $\frac{26}{25}(3,4)=(3.12,4.16)$. This is the wagon-handle computation: only the along-the-road part of the pull does work.

The cross product

7. For $\vec{a}=(1,0,1)$ and $\vec{b}=(0,1,-1)$, compute $\vec{a}\times \vec{b}$, and verify the result is orthogonal to both original vectors.
8. Determine the area of the parallelogram with sides $\vec{a}=(2,1,0)$ and $\vec{b}=(1,3,0)$.
9. You are loosening a stubborn bolt with a wrench. Using $|\vec{r}\times \vec{F}|=|\vec{r}||\vec{F}|\sin \theta$, explain how to maximise the torque — and why a longer handle helps as much as a stronger pull.

Answer 7

$\vec{a}\times \vec{b}=(0(-1)-1(1),1(0)-1(-1),1(1)-0(0))=(-1,1,1)$ Verification by dot product: $(1,0,1)\cdot (-1,1,1)=-1+0+1=0$ and $(0,1,-1)\cdot (-1,1,1)=0+1-1=0$. Both zero — the answer stands. Run this check every time; it is free.

Answer 8

$\vec{a}\times \vec{b}=(1(0)-0(3),0(1)-2(0),2(3)-1(1))=(0,0,5)$ Area $=|(0,0,5)|=5$ square units. Both sides lie in the $xy$-plane, so the cross product points straight up the $z$-axis — perpendicular to the plane containing the parallelogram, exactly as it must.

Answer 9

Torque is largest when every factor is as large as possible: maximise $|\vec{r}|$ by gripping at the end of the handle, maximise $|\vec{F}|$ by pulling hard, and maximise $\sin \theta$ by pulling at $90°$ to the handle, where $\sin \theta =1$. The three factors multiply, so doubling the handle length doubles the torque exactly as doubling the force does — which is why a length of pipe slipped over a wrench handle is the mechanic’s cheat, and why pushing along the handle ($\theta =0$) turns nothing at all.