Study Companion: S02 — Vectors and Their Operations

Forces, positions, moments and velocities are all vectors. Statics adds them, splits them into parts, measures the angles between them and multiplies them. The operations are few. Each has a picture that makes it obvious and a component formula that makes it fast, and the two always agree. This lesson builds both forms for each operation, in an order where each one solves a problem the previous one could not.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 1/3 ("Working with Vectors") and Appendix C/7 ("Vector Operations"). Marghitu, Dupac and Madsen, Statics with MATLAB, sections 1.1–1.11 and its Examples 1.1–1.3. Marghitu computes everything in MATLAB; those computations appear as a labelled second view. Notation follows Meriam: unit vectors \mathbf i, \mathbf j, \mathbf k (Marghitu prints them dotless) and direction cosines l, m, n (Marghitu also writes \cos\alpha, \cos\beta, \cos\gamma).

How to use this companion

Each nugget is one operation. It opens with a claim. A picture shows why the operation works, a number checks it, and its equation follows, labelled with what kind of statement it is. After that come a worked example with units, a real use, the most common wrong answer, and a check question. A bridge says why the next operation is needed. The Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

Two tugs pull a ship out of harbour. One pulls with 30\,\mathrm{kN} straight ahead, the other with 20\,\mathrm{kN} at 60^\circ to the first. A single tug replacing both would need to pull how hard, and in which direction? "50 kN straight ahead" is wrong on both counts. The answer, 43.6\,\mathrm{kN} at 23.4^\circ, comes from Nugget 01 and is confirmed by Nugget 02. By Nugget 04 the same tools tell you how hard a guy wire presses its mast into the ground.

Nugget 01: Arrows Add Head to Tail, Not Size to Size

Claim. Two vectors add by the parallelogram law. The size of the sum is not the sum of the sizes.

The picture. Draw the tugs' pulls to scale from one point, O:

Complete the parallelogram with \mathbf F_1 and \mathbf F_2 as its sides. The diagonal from O is the single force with the same effect, the resultant \mathbf R. The right-hand side of the parallelogram is \mathbf F_2 slid to start at the tip of \mathbf F_1. So the same diagonal comes from placing the vectors head to tail (the triangle law), in either order.

Check with numbers. In the triangle with sides 30 and 20, the angle opposite \mathbf R is 180^\circ - 60^\circ = 120^\circ. The law of cosines gives R^2 = 30^2 + 20^2 - 2(30)(20)\cos120^\circ = 900 + 400 + 600 = 1900, so R = 43.59\,\mathrm{kN}. The law of sines gives the direction, \sin\alpha = 20\sin120^\circ/43.59 = 0.3974, so \alpha = 23.41^\circ from \mathbf F_1. The arrows measured on the drawing agree. The sizes added would give 50, which is wrong.

The idea. A vector has a magnitude and a direction. Marghitu splits the direction into an orientation (the line it lies along) and a sense (which way along that line). The rules every vector obeys follow from the picture:

• \mathbf P + \mathbf Q = \mathbf Q + \mathbf P (commutative: either side of the parallelogram first);
• \mathbf P + (\mathbf Q + \mathbf R) = (\mathbf P + \mathbf Q) + \mathbf R (associative: a chain of head-to-tail arrows ends at the same point however you group it);
• -\mathbf V has the size of \mathbf V and the opposite sense; subtraction is adding the negative: \mathbf P - \mathbf Q = \mathbf P + (-\mathbf Q);
• a scalar multiple s\mathbf V has magnitude |s|V, lies along \mathbf V, and has the same sense as \mathbf V if s > 0, the opposite sense if s < 0. Scalars distribute: (s_1 + s_2)\mathbf V = s_1\mathbf V + s_2\mathbf V, s(\mathbf V_1 + \mathbf V_2) = s\mathbf V_1 + s\mathbf V_2, and s_1(s_2\mathbf V) = (s_1s_2)\mathbf V;
• the zero vector \mathbf 0 has zero size and no direction. It is not the number 0.

The equation (definition: vector addition is defined by this geometry; that forces add this way is the physical law of Stevinus, confirmed by experiment):

where \phi is the angle between \mathbf F_1 and \mathbf F_2 placed tail to tail. Limiting cases: \phi = 0 gives R = F_1 + F_2 (pulling together); \phi = 180^\circ gives R = |F_1 - F_2| (pulling against each other).

Worked number. The difference \mathbf D = \mathbf F_2 - \mathbf F_1 is the other diagonal of the parallelogram, from the tip of \mathbf F_1 to the tip of \mathbf F_2. Its size, by the law of cosines with the 60^\circ angle between them: D^2 = 30^2 + 20^2 - 2(30)(20)\cos60^\circ = 700, so D = 26.46\,\mathrm{kN}.

Real example. Two tugs, two ropes on a stuck car, two people carrying a bag by one handle each: in every case the useful pull is less than the sum, and it shrinks as the angle between them opens.

Plausible wrong answer. "R = 30 + 20 = 50\,\mathrm{kN}." That holds only for \phi = 0. Meriam stresses the distinction in notation: \mathbf V_1 + \mathbf V_2 (bold) is vector addition, and the parallelogram shows V \neq V_1 + V_2.

Check. Two equal 10\,\mathrm N forces act at 120^\circ to each other. What is their resultant? (R^2 = 100 + 100 + 200\cos120^\circ = 100, so R = 10\,\mathrm N, along the bisector.)

Narration spine. Draw the two tug arrows from one point. Sweep in the parallelogram's other two sides. Draw the diagonal and label it. Slide \mathbf F_2 to the tip of \mathbf F_1 to show the triangle law gives the same diagonal. Finally, lay the two arrows end to end in a straight line beside it (50 kN) and compare with the diagonal (43.6 kN).

Bridge. Drawing and the law of cosines work for two vectors in a plane. Five forces in three dimensions would need a chain of triangles. We want a method that turns vector addition into ordinary addition of numbers.

Nugget 02: Components Are Adding Run Backwards

Claim. Any vector can be split into parts along chosen directions, and for perpendicular directions the parts simply add as numbers.

The picture. Run the parallelogram backwards: start with \mathbf R and ask which vectors along two chosen directions add up to it.

On the right, the chosen directions are the two tugs' directions (1 along x, 2 at 60^\circ). The parts come out as 30 and 20 kN, the tugs themselves. On the left the directions are perpendicular, and the parts are the rectangular components R_x and R_y. Each is just the shadow of \mathbf R on its axis.

Check with numbers. Resolve each tug into x and y parts: \mathbf F_1 = 30\mathbf i\ \mathrm{kN} and \mathbf F_2 = 20\cos60^\circ\,\mathbf i + 20\sin60^\circ\,\mathbf j = 10\mathbf i + 17.32\mathbf j\ \mathrm{kN}. Add like numbers: \mathbf R = 40\mathbf i + 17.32\mathbf j\ \mathrm{kN}. Then R = \sqrt{40^2 + 17.32^2} = \sqrt{1900} = 43.59\,\mathrm{kN} and \theta = \tan^{-1}(17.32/40) = 23.41^\circ. That is exactly Nugget 01's answer, found with no triangle at all.

The idea. Marghitu (1.3–1.4) puts the reason precisely. A linear combination of vectors is s_1\mathbf v_1 + \dots + s_n\mathbf v_n. A set of non-zero vectors is linearly independent if none of them is a linear combination of the others. In space at most three vectors can be independent (that is what three-dimensional means), and in a plane at most two. Three independent vectors form a basis: every vector is a combination of them, and in exactly one way, so its components are unique. Any two non-parallel directions in a plane work, perpendicular or not (the right-hand picture). The rectangular (Cartesian) basis \mathbf i, \mathbf j, \mathbf k of mutually perpendicular unit vectors makes the components independent of each other. That is why they add separately. A vector equation \mathbf v = \mathbf 0 is therefore three scalar equations, v_x = 0, v_y = 0, v_z = 0, the form in which equilibrium will be solved.

The equation (mathematical identity):

Worked number. Marghitu's Example 1.1 adds three plane vectors: \mathbf v_1 = 0.600\mathbf i + 0.800\mathbf j, \mathbf v_2 = -0.866\mathbf i + 0.500\mathbf j and \mathbf v_3 = -1.299\mathbf i - 0.750\mathbf j (units). Summing the columns, \mathbf v = -1.565\mathbf i + 0.550\mathbf j, with v = \sqrt{1.565^2 + 0.550^2} = 1.659. Its angle: \tan^{-1}(0.550/{-1.565}) on a calculator gives -19.36^\circ. But v_x < 0 and v_y > 0 put the vector in the second quadrant, so the true angle is 180^\circ - 19.36^\circ = 160.64^\circ. The book prints -19.363^\circ. That is the quadrant slip below.

Real example. A surveyor adds a traverse of legs (120 m north-east, then 80 m south, and so on) by summing north and east components. A chain of triangles would compound the angle errors.

Plausible wrong answer. "\theta = \tan^{-1}(V_y/V_x), read straight off the calculator." The calculator cannot tell (-1.565, 0.550) from (1.565, -0.550). Use the signs, or atan2(Vy, Vx).

Check. Resolve \mathbf R = 43.59\,\mathrm{kN} at 23.41^\circ back into components. (43.59\cos23.41^\circ = 40.0, 43.59\sin23.41^\circ = 17.3\,\mathrm{kN}.)

Second view (Marghitu, MATLAB). A vector is a row of its components, and addition is element by element:

F1 = [30 0 0];  F2 = 20*[cosd(60) sind(60) 0];   % kN
R = F1 + F2           % [40 17.32 0]
norm(R)               % 43.59
atan2d(R(2), R(1))    % 23.41 deg; atan2d keeps the quadrant

Narration spine. Start from Nugget 01's diagonal. Drop its shadows onto the x- and y-axes, then lift the two shadows into arrows that rebuild it head to tail. Repeat with the two tug directions and watch 30 and 20 come back. Add the tugs' components in a two-column table. Finally rotate a vector into the second quadrant while a calculator readout of \tan^{-1} jumps to the wrong quadrant.

Bridge. In a plane one angle gives a direction. In three dimensions a force along a guy wire is fixed by two points, not by an angle. We need a direction that we can build from coordinates.

Nugget 03: A Unit Vector Is a Direction You Can Multiply

Claim. Dividing a vector by its length leaves a pure direction, the unit vector. Any force is its magnitude times the unit vector along its line.

The picture. A radio mast OA stands 12\,\mathrm m tall. A guy wire runs from its top A to an anchor B on the ground, 5\,\mathrm m along x and -4\,\mathrm m along y. The wire pulls the mast top with 2.5\,\mathrm{kN} towards B.

The arrow from A to B is the position vector of B relative to A, found by subtracting coordinates. It has the right direction but the wrong size (metres, not kilonewtons). Divide it by its own length and the size becomes 1 and the units cancel. What is left is a direction, which we then scale by 2.5\,\mathrm{kN}.

Check with numbers. \mathbf r_{AB} = (5 - 0)\mathbf i + (-4 - 0)\mathbf j + (0 - 12)\mathbf k = 5\mathbf i - 4\mathbf j - 12\mathbf k\ \mathrm m. Its length is \sqrt{25 + 16 + 144} = \sqrt{185} = 13.60\,\mathrm m. The unit vector is \mathbf n = 0.3676\mathbf i - 0.2941\mathbf j - 0.8823\mathbf k, and its length is exactly \sqrt{(25 + 16 + 144)/185} = 1. The force is \mathbf T = 2.5\,\mathbf n = 0.919\mathbf i - 0.735\mathbf j - 2.206\mathbf k\ \mathrm{kN}.

The idea. In 3D, \mathbf V = V_x\mathbf i + V_y\mathbf j + V_z\mathbf k with V = \sqrt{V_x^2 + V_y^2 + V_z^2} (Pythagoras twice: once across the base, once up the height). Its unit vector is \mathbf n = \mathbf V/V, dimensionless. Its components are the cosines of the angles \theta_x, \theta_y, \theta_z that \mathbf V makes with the axes, the direction cosines l = \cos\theta_x, m = \cos\theta_y, n = \cos\theta_z. (Meriam uses m for a direction cosine and for mass, and n for a cosine and for \mathbf n; the context always decides which.) Each angle runs from 0^\circ to 180^\circ and is found by sliding the vectors tail to tail (Marghitu 1.5). Because \mathbf n has length 1, the three cosines are not independent.

The equation (definition and the mathematical identity it implies):

The position vector of P relative to O is \mathbf r_{OP} = x_P\mathbf i + y_P\mathbf j + z_P\mathbf k. The vector from P to M is \mathbf r_{PM} = \mathbf r_M - \mathbf r_P, and its length is the distance d = \sqrt{(x_M - x_P)^2 + (y_M - y_P)^2 + (z_M - z_P)^2}.

Worked number. The wire's direction angles are \theta_x = \cos^{-1}0.3676 = 68.43^\circ, \theta_y = \cos^{-1}(-0.2941) = 107.10^\circ and \theta_z = \cos^{-1}(-0.8823) = 151.92^\circ. Check: 0.3676^2 + 0.2941^2 + 0.8823^2 = 1.000.

Real example (one you can see). Every radio mast, tent pole and sailing-boat mast is held by guy wires to anchors at known points. The pull of each wire on the mast is found exactly this way, from two sets of coordinates.

Plausible wrong answer. "I can choose any three direction angles." Marghitu's Example 1.3 takes \theta_x = 30^\circ and \theta_y = \theta_z = 60^\circ. But \cos^230^\circ + 2\cos^260^\circ = 0.75 + 0.5 = 1.25 \neq 1, so no vector has those three angles. The example's vector comes out with magnitude 150\sqrt{1.25} = 167.7\,\mathrm m instead of the stated 150\,\mathrm m. Choose two angles, and the third is fixed by l^2 + m^2 + n^2 = 1 (up to its sign).

Check. Can a unit vector have components (0.6, 0.6, 0.6)? (0.36 \times 3 = 1.08 \neq 1. No.)

Second view (Marghitu Example 1.2, MATLAB). A 500\,\mathrm{kN} force acts from A(20, 10, 30)\,\mathrm m towards M, which lies in the xy-plane with OM = 50\,\mathrm m at 45^\circ to x. So \mathbf r_M = (35.36, 35.36, 0)\,\mathrm m, \mathbf r_{AM} = (15.36, 25.36, -30)\,\mathrm m, r_{AM} = 42.17\,\mathrm m and \mathbf u = (0.364, 0.601, -0.711). The force is \mathbf F = 182.0\mathbf i + 300.6\mathbf j - 355.7\mathbf k\ \mathrm{kN}, at 68.65^\circ, 53.04^\circ and 135.34^\circ to the axes. (The book prints \mathbf u with "(m)"; a unit vector has no unit.)

rA = [20 10 30];  rM = 50*[cosd(45) sind(45) 0];   % m
u  = (rM - rA)/norm(rM - rA);                      % dimensionless
F  = 500*u                                         % kN
acosd(u)                                           % direction angles, deg

Narration spine. Show the mast and wire in 3D. Draw the arrow from A to B in metres, then shrink it smoothly to length 1 and label it \mathbf n. Stretch it back to 2.5\,\mathrm{kN} in force units. Drop its three shadows onto the axes and mark their angles. A readout of l^2 + m^2 + n^2 stays at 1 as the anchor moves.

Bridge. We now have the wire's force as a vector. The question an engineer asks next is a single number: how much of that force pushes down the mast? That needs a way to measure one vector along another.

Nugget 04: The Dot Product Measures How Much of One Vector Lies Along Another

Claim. \mathbf P\cdot\mathbf Q = PQ\cos\theta is the length of \mathbf P times the shadow of \mathbf Q on it. In components it is P_xQ_x + P_yQ_y + P_zQ_z.

The picture.

Drop a perpendicular from the tip of \mathbf Q onto the line of \mathbf P. The shadow has length Q\cos\theta: positive when \theta < 90^\circ, zero at 90^\circ, negative beyond. The dot product multiplies that shadow by P. Projecting \mathbf P onto \mathbf Q instead gives P\cos\theta \times Q, the same number, so the order does not matter.

Check with numbers. To find how hard the guy wire of Nugget 03 presses down the mast, project \mathbf T onto the downward unit vector -\mathbf k: \mathbf T\cdot(-\mathbf k) = (0.919)(0) + (-0.735)(0) + (-2.206)(-1) = 2.206\,\mathrm{kN}. By the picture: T\cos\phi, where \phi is the angle between the wire and the downward vertical. \cos\phi = -n = 0.8823, and 2.5 \times 0.8823 = 2.206\,\mathrm{kN}. The two agree.

The idea. Why do the two forms agree? Use the rules the picture guarantees. The dot product is commutative and distributes over addition, and for unit vectors \mathbf i\cdot\mathbf i = \mathbf j\cdot\mathbf j = \mathbf k\cdot\mathbf k = 1 (\theta = 0) and \mathbf i\cdot\mathbf j = \mathbf j\cdot\mathbf k = \mathbf k\cdot\mathbf i = 0 (\theta = 90^\circ). Expand (P_x\mathbf i + P_y\mathbf j + P_z\mathbf k)\cdot(Q_x\mathbf i + Q_y\mathbf j + Q_z\mathbf k) into nine terms. Six vanish, and three survive: P_xQ_x + P_yQ_y + P_zQ_z. In particular \mathbf i\cdot\mathbf V = V_x: dotting with a unit vector reads off a component, so \mathbf V = (\mathbf i\cdot\mathbf V)\mathbf i + (\mathbf j\cdot\mathbf V)\mathbf j + (\mathbf k\cdot\mathbf V)\mathbf k (Marghitu 1.7).

The equation (definition, and the identity derived above):

Rules: \mathbf P\cdot\mathbf Q = \mathbf Q\cdot\mathbf P; \mathbf P\cdot(\mathbf Q + \mathbf R) = \mathbf P\cdot\mathbf Q + \mathbf P\cdot\mathbf R; (s\mathbf P)\cdot\mathbf Q = s(\mathbf P\cdot\mathbf Q); \mathbf P\cdot\mathbf P = P^2. Two non-zero vectors are perpendicular exactly when \mathbf P\cdot\mathbf Q = 0. The result is a scalar. The units are the product of the two vectors' units, so \mathrm N\cdot\mathrm m for a force dotted with a displacement.

Worked number. The component of a vector \mathbf F along a line with unit vector \mathbf e is F_e = \mathbf F\cdot\mathbf e, and the vector part is (\mathbf F\cdot\mathbf e)\mathbf e. A suitcase pulled 20\,\mathrm m along the floor by a 60\,\mathrm N handle force at 35^\circ above the floor receives \mathbf F\cdot\mathbf d = 60 \times 20 \times \cos35^\circ = 983\,\mathrm J of work. Only the part of the pull along the floor does any work.

Real example. The guy wire's 2.206\,\mathrm{kN} down the mast is the compression the mast's foundation must carry, per wire. Three wires mean 6.6\,\mathrm{kN} before the mast's own weight is counted.

Plausible wrong answer. "\mathbf P\cdot\mathbf Q = 0, so one of them is zero." For vectors, a zero dot product usually means perpendicular: \mathbf i\cdot\mathbf j = 0, and neither is zero.

Check. Find the angle between \mathbf a = \mathbf i + \mathbf j and \mathbf b = \mathbf j + \mathbf k. (\mathbf a\cdot\mathbf b = 1, ab = \sqrt2\sqrt2 = 2, \cos\theta = 0.5, \theta = 60^\circ.)

Second view (Marghitu Example 1.3). For two position vectors \mathbf r_A = (129.90, 75.00, 75.00)\,\mathrm m and \mathbf r_B = (136.60, -36.60, 141.42)\,\mathrm m, dot(rA,rB)/(norm(rA)*norm(rB)) gives \cos\theta, so \theta = 40.23^\circ. The projection of \mathbf r_A on \mathbf r_B, dot(rA, rB/norm(rB)), is 128.03\,\mathrm m.

Narration spine. Draw \mathbf P and \mathbf Q from one point and drop \mathbf Q's shadow onto \mathbf P. Swing \mathbf Q through 90^\circ: the shadow shrinks to zero and then turns negative, while a readout of \mathbf P\cdot\mathbf Q follows it. Expand the component product as a 3\times3 grid in which the six off-diagonal cells fade out. End on the guy wire, its shadow on the mast reading 2.206 kN.

Bridge. The dot product kept the part of one vector along another. Turning a bolt needs the other part, the one across the lever, and an answer that points along the bolt. That is a new kind of product.

Nugget 05: The Cross Product Builds the Perpendicular

Claim. \mathbf P\times\mathbf Q is a vector perpendicular to both. Its length is the area of the parallelogram they span, and its sense follows the right-hand screw rule.

The picture.

The parallelogram on \mathbf P and \mathbf Q has base P and height Q\sin\theta, so its area is PQ\sin\theta. The cross product is the vector with that length, standing perpendicular to the plane of the parallelogram. Curl the fingers of your right hand from \mathbf P towards \mathbf Q (through the smaller angle), and your thumb gives its sense. A right-handed screw turned that way advances along the same line. Reversing the order reverses the turn, so \mathbf Q\times\mathbf P = -\mathbf P\times\mathbf Q.

Check with numbers. A spanner 0.25\,\mathrm m long lies along \mathbf r = 0.25\mathbf i + 0.05\mathbf j\ \mathrm m, and a hand pushes its end with \mathbf F = -20\mathbf i + 80\mathbf j\ \mathrm N. Picture: r = 0.2550\,\mathrm m, F = 82.46\,\mathrm N. From \cos\theta = \mathbf r\cdot\mathbf F/(rF) = (-5 + 4)/21.02 = -0.0476, \theta = 92.73^\circ and \sin\theta = 0.99887. The area is 0.2550 \times 82.46 \times 0.99887 = 21.0\,\mathrm{N\cdot m}. Components (below): \mathbf r\times\mathbf F = (0.25 \times 80 - 0.05 \times (-20))\,\mathbf k = 21.0\,\mathbf k\ \mathrm{N\cdot m}. The two agree, and +\mathbf k is out of the page: the bolt turns counter-clockwise seen from above.

The idea. From the definition, for a right-handed set of axes (\mathbf i\times\mathbf j = \mathbf k):

Going round the cycle i \to j \to k \to i gives +, and going against it gives -. The cross product distributes over addition, \mathbf P\times(\mathbf Q + \mathbf R) = \mathbf P\times\mathbf Q + \mathbf P\times\mathbf R (Marghitu calls this, for n terms, Varignon's theorem; in Meriam that name belongs to the moment theorem of S04). So expanding the two component sums term by term and applying the table gives the component formula. A set with \mathbf i\times\mathbf j = -\mathbf k is left-handed, and every formula here flips sign in it. This book always uses right-handed axes.

The equation (definition, and its component identity):

Parallel vectors (\theta = 0 or 180^\circ) have \mathbf P\times\mathbf Q = \mathbf 0. (s\mathbf P)\times\mathbf Q = s(\mathbf P\times\mathbf Q). Units: the product of the factors' units, \mathrm m\cdot\mathrm N = \mathrm{N\cdot m} for the spanner.

Worked number. Marghitu's Example 1.3: \mathbf r_A\times\mathbf r_B for \mathbf r_A = (129.90, 75, 75) and \mathbf r_B = (136.60, -36.60, 141.42)\,\mathrm m. The \mathbf i term is 75 \times 141.42 - 75 \times (-36.60) = 13\,352. The \mathbf j term is 75 \times 136.60 - 129.90 \times 141.42 = -8125. The \mathbf k term is 129.90 \times (-36.60) - 75 \times 136.60 = -15\,000. So \mathbf r_A\times\mathbf r_B = (1.335\times10^4, -8.13\times10^3, -1.50\times10^4)\,\mathrm{m^2}.

Real example (one you can try). Ordinary bolts have right-hand threads. Look down on a bolt head and turn it counter-clockwise: the bolt rises out of the hole, along \mathbf r\times\mathbf F, exactly as the right-hand rule predicts. "Righty-tighty" is the cross product's sign convention.

Plausible wrong answer. "\mathbf r\times\mathbf F = \mathbf F\times\mathbf r." They are opposite. Swap the order in a moment calculation and every rotation in the problem reverses.

Check. Find \mathbf k\times\mathbf i and \mathbf i\times\mathbf k. (\mathbf j and -\mathbf j.)

Narration spine. Draw \mathbf P and \mathbf Q, sweep out their parallelogram, and lift a perpendicular arrow whose length tracks the shaded area. Rotate \mathbf Q towards \mathbf P until the area and the arrow vanish together. Show the right hand curling from \mathbf P to \mathbf Q, then reverse the order and flip the arrow. Expand the determinant term by term beside the spanner.

Bridge. One dot product gives a number and one cross product gives a vector. Combine them and two compact results follow. One measures volume, and it will give the moment of a force about an axis (S08). The other rearranges a double cross product.

Nugget 06: Triple Products Measure Volume and Untangle Double Crosses

Claim. \mathbf a\cdot(\mathbf b\times\mathbf c) is the volume of the box on \mathbf a, \mathbf b and \mathbf c. And \mathbf a\times(\mathbf b\times\mathbf c) = (\mathbf a\cdot\mathbf c)\mathbf b - (\mathbf a\cdot\mathbf b)\mathbf c.

The picture.

\mathbf b\times\mathbf c (or here \mathbf a\times\mathbf b) is a vector whose length is the area of one face and which points perpendicular to that face. Dotting the third edge with it multiplies that area by the third edge's height above the face, which is the volume. If the three vectors lie in one plane the box is flat, and the volume is zero.

Check with numbers. \mathbf a = 2\mathbf i, \mathbf b = 3\mathbf j and \mathbf c = \mathbf i + \mathbf j + 4\mathbf k (m). Base: \mathbf a\times\mathbf b = 6\mathbf k\ \mathrm{m^2}. Height of \mathbf c above the base: \mathbf c\cdot\mathbf k = 4\,\mathrm m. Volume: (\mathbf a\times\mathbf b)\cdot\mathbf c = 6 \times 4 = 24\,\mathrm{m^3}. The determinant gives the same: \begin{vmatrix}2&0&0\\0&3&0\\1&1&4\end{vmatrix} = 2(12 - 0) = 24.

The idea. The scalar triple product [\mathbf a,\mathbf b,\mathbf c] = \mathbf a\cdot(\mathbf b\times\mathbf c) is a signed volume. It is positive when \mathbf a, \mathbf b, \mathbf c form a right-handed set. Every face gives the same box, so the dot and the cross can be swapped, \mathbf a\cdot\mathbf b\times\mathbf c = \mathbf a\times\mathbf b\cdot\mathbf c. Cycling the vectors leaves it unchanged, and swapping two flips the sign. No parentheses are needed, because \mathbf a\times(\mathbf b\cdot\mathbf c) would cross a vector with a scalar, which means nothing. The vector triple product is different: the parentheses are essential, because the cross product is not associative.

The equation (mathematical identities):

Meriam writes the second identity as (\mathbf P\times\mathbf Q)\times\mathbf R = (\mathbf R\cdot\mathbf P)\mathbf Q - (\mathbf R\cdot\mathbf Q)\mathbf P. It is the same identity. Step the sources skip: (\mathbf P\times\mathbf Q)\times\mathbf R = -\mathbf R\times(\mathbf P\times\mathbf Q), and Marghitu's form with \mathbf a = \mathbf R, \mathbf b = \mathbf P, \mathbf c = \mathbf Q gives -[(\mathbf R\cdot\mathbf Q)\mathbf P - (\mathbf R\cdot\mathbf P)\mathbf Q], which is Meriam's.

Worked number. \mathbf a = \mathbf i + 2\mathbf j, \mathbf b = \mathbf j + 3\mathbf k, \mathbf c = 2\mathbf i + \mathbf k. Directly: \mathbf b\times\mathbf c = \mathbf i + 6\mathbf j - 2\mathbf k, and \mathbf a\times(\mathbf b\times\mathbf c) = -4\mathbf i + 2\mathbf j + 4\mathbf k. By the identity: \mathbf a\cdot\mathbf c = 2 and \mathbf a\cdot\mathbf b = 2, so 2\mathbf b - 2\mathbf c = (0, 2, 6) - (4, 0, 2) = -4\mathbf i + 2\mathbf j + 4\mathbf k. They agree.

Real example. The moment of a force about a hinge line (a door's hinges, a shaft's axis) is \mathbf e\cdot(\mathbf r\times\mathbf F), a scalar triple product. S08 builds it.

Plausible wrong answer. "Cross products can be regrouped like ordinary products." Try \mathbf i\times(\mathbf i\times\mathbf j) = \mathbf i\times\mathbf k = -\mathbf j, but (\mathbf i\times\mathbf i)\times\mathbf j = \mathbf 0\times\mathbf j = \mathbf 0. With the worked numbers above, (\mathbf a\times\mathbf b)\times\mathbf c = -3\mathbf i - 4\mathbf j + 6\mathbf k, not -4\mathbf i + 2\mathbf j + 4\mathbf k.

Check. Are \mathbf a = (1, 2, 3), \mathbf b = (4, 5, 6) and \mathbf c = (7, 8, 9) coplanar? (Self-check Q12.)

Second view (Marghitu Example 1.3). For the projection \mathbf r_C = (136.60, -36.60, 0)\,\mathrm m of \mathbf r_B onto the xy-plane, det([rC; rB; rA]) and dot(rC, cross(rB, rA)) both give -2.12\times10^{6}\,\mathrm{m^3}. The sign is negative because \mathbf r_C, \mathbf r_B, \mathbf r_A form a left-handed set.

Narration spine. Build the box from \mathbf a and \mathbf b along the floor and \mathbf c slanting up. Raise \mathbf a\times\mathbf b as an arrow off the base and project \mathbf c onto it to show the height. Shear \mathbf c sideways: the volume readout does not change. Tilt \mathbf c into the floor and it falls to zero. Then show \mathbf i\times(\mathbf i\times\mathbf j) against (\mathbf i\times\mathbf i)\times\mathbf j side by side.

Bridge. So far every vector has stood still. A crank turns and a cable's tension varies along its length. We need vectors that change, and the rules for differentiating and integrating them.

Nugget 07: A Changing Vector Is Differentiated Component by Component

Claim. In a fixed set of axes, a vector's derivative is the vector of its components' derivatives. The product rules carry over, as long as the order of a cross product is kept.

The picture. A crank pin moves on a circle of radius 0.1\,\mathrm m at \omega = 5\,\mathrm{rad/s}.

The pin's position \mathbf r keeps the same length while it turns, so its rate of change cannot have any part along \mathbf r. If it did, the length would grow or shrink. The velocity is tangent to the circle.

Check with numbers. \mathbf r(t) = 0.1(\cos\omega t\,\mathbf i + \sin\omega t\,\mathbf j)\ \mathrm m. Differentiate each component: d\mathbf r/dt = 0.1\omega(-\sin\omega t\,\mathbf i + \cos\omega t\,\mathbf j) = 0.5(-\sin\omega t\,\mathbf i + \cos\omega t\,\mathbf j)\ \mathrm{m/s}. Its size is 0.5\,\mathrm{m/s} at every instant. Its dot product with \mathbf r is 0.05(-\cos\omega t\sin\omega t + \sin\omega t\cos\omega t) = 0. So it is perpendicular, as the picture said.

The idea. The derivative of a vector is defined exactly as for a scalar, d\mathbf a/dt = \lim_{\Delta t\to0}[\mathbf a(t + \Delta t) - \mathbf a(t)]/\Delta t. In general, d\mathbf v/dt = \dot v_x\mathbf i + v_x\,d\mathbf i/dt + \dots, and when the axes \mathbf i, \mathbf j, \mathbf k do not change (always in this book), the d\mathbf i/dt terms vanish. The general proof of the picture's result: d(\mathbf r\cdot\mathbf r)/dt = 2\,\mathbf r\cdot\dot{\mathbf r}. If r is constant the left side is zero, so \mathbf r\cdot\dot{\mathbf r} = 0. Integration works the same way, component by component. Over a volume element d\tau = dx\,dy\,dz, \int\mathbf V\,d\tau = \mathbf i\int V_x\,d\tau + \mathbf j\int V_y\,d\tau + \mathbf k\int V_z\,d\tau. Marghitu adds the gradient of a scalar function V(x, y, z), \nabla V = \frac{\partial V}{\partial x}\mathbf i + \frac{\partial V}{\partial y}\mathbf j + \frac{\partial V}{\partial z}\mathbf k, from its total differential dV = \frac{\partial V}{\partial x}dx + \frac{\partial V}{\partial y}dy + \frac{\partial V}{\partial z}dz. He also adds the curl of a vector field, \nabla\times\mathbf F, the determinant with rows \mathbf i\ \mathbf j\ \mathbf k, \partial_x\ \partial_y\ \partial_z, F_x\ F_y\ F_z. Statics uses the gradient once, in S31, where a force is minus the gradient of a potential energy.

The equation (mathematical identities; Meriam C/7.9, Marghitu (1.19)–(1.21)):

Worked number. A linear spring of rate k = 2000\,\mathrm{N/m} stretched by x along \mathbf i stores V = \frac12kx^2. At x = 0.03\,\mathrm m, V = 0.9\,\mathrm J and \nabla V = kx\,\mathbf i = 60\mathbf i\ \mathrm N. The spring pulls back with -\nabla V = -60\mathbf i\ \mathrm N, which is Hooke's law, recovered from the energy. A uniform load of 400\,\mathrm{N/m} pressing down along a 3\,\mathrm m beam integrates component by component to \int_0^3 (-400\,\mathbf j)\,dx = -1200\,\mathbf j\ \mathrm N.

Real example. The book's recurring hardware, a spring-operated switch: a crank turns a shaft while a spring pulls on the crank pin. The pin's position vector, its velocity and the spring's force are all vectors that change with the crank angle. Every later lesson about that mechanism differentiates them this way.

Plausible wrong answer. "d(\mathbf P\times\mathbf Q)/dt = \dot{\mathbf Q}\times\mathbf P + \dots." Swapping the order changes the sign. The rule keeps each factor in its original place.

Check. If \mathbf u(t) is always a unit vector, what can you say about \dot{\mathbf u}? (It is perpendicular to \mathbf u, by the same argument as the crank.)

Second view (Marghitu, MATLAB). Symbolic vectors are differentiated with diff, and the product rules can be checked symbolically:

syms t real
a = [cos(5*t) sin(5*t) 0]*0.1;   % crank pin, m
v = diff(a, t)                   % [-0.5*sin(5t), 0.5*cos(5t), 0]
simplify(dot(a, v))              % 0

Narration spine. Turn the crank and draw the position arrow. Attach the velocity arrow at its tip, always at a right angle to it, while a readout of \mathbf r\cdot\dot{\mathbf r} stays at zero. Then switch to a spring: plot V = \frac12kx^2 as a curve, draw its slope at x = 0.03\,\mathrm m, and flip it into the spring's -60\,\mathrm N pull.

Synthesis

The tugs' puzzle needed only Nuggets 01 and 02: one resultant, 43.6\,\mathrm{kN} at 23.4^\circ, found by drawing and confirmed by components. The deeper result is that every vector operation has two forms that agree. Addition is a parallelogram and a column sum. The dot product is a shadow and P_xQ_x + P_yQ_y + P_zQ_z. The cross product is an area with a right-hand direction and a determinant. The triple product is a volume and another determinant. The pictures tell you what an operation means and when it is zero. The components let you compute it in three dimensions without drawing anything. The rest of statics is these operations applied to forces: sums of forces, moments \mathbf r\times\mathbf F, components along members, and moments about axes.

• Central idea: vectors are combined by geometry, and components turn that geometry into arithmetic on three numbers at a time.
• Key relationship: \mathbf V = V\mathbf n with l^2 + m^2 + n^2 = 1; \mathbf P\cdot\mathbf Q = PQ\cos\theta; |\mathbf P\times\mathbf Q| = PQ\sin\theta, perpendicular to both by the right-hand rule.
• Most important assumption: a fixed, right-handed set of mutually perpendicular axes. In a left-handed set every cross product changes sign.
• Open question: the cross product of a position with a force measures turning. How large is the turning effect of a force about a point, and why does it depend only on the perpendicular distance to its line of action? (S04.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Marghitu's programs; the Python is a line-for-line translation, and both print the same numbers.

Python (numpy, sympy). S02_vectors.py adds vectors by components and keeps the quadrant with arctan2, builds Example 1.2's force from two points, evaluates Example 1.3's angle, projection, cross and box products, checks the double-cross identity and differentiates the crank-pin vector with sympy.

Part 1: adding by components
  F1 + F2 = [40.   17.32  0.  ] kN, |R| = 43.59 kN, at 23.41 deg
  |F2 - F1| = 26.46 kN
  Ex 1.1: v = [-1.565  0.55 ], |v| = 1.659, arctan -19.36 deg, arctan2 160.64 deg
Part 2: Example 1.2, a force from A towards M
  r_AM = [ 15.36  25.36 -30.  ] m, |r_AM| = 42.17 m, u = [ 0.364  0.601 -0.711]
  F = [ 182.   300.6 -355.7] kN, angles [ 68.65  53.04 135.34] deg
Part 3: Example 1.3, two position vectors
  angle 40.23 deg, projection of rA on rB 128.03 m
  rA x rB = [ 13351.5  -8125.5 -14999.3] m^2
  det[rC; rB; rA] = -2.121e+06, rC . (rB x rA) = -2.121e+06 m^3
Part 4: identities and derivatives
  a x (b x c) = [-4  2  4], (a.c) b - (a.b) c = [-4  2  4]
  crank pin v = [-sin(5*t)/2, cos(5*t)/2, 0], a . v = 0
  spring: dV/dx = k*x, at k = 2000, x = 0.03: V = 0.90 J, force -60 N

The arctan and arctan2 pair is the quadrant slip of Nugget 02 in one line: the same two components give -19.36^\circ and 160.64^\circ.

MATLAB (Octave-compatible). S02_vectors.m does the same with norm, atan2d, acosd, dot, cross and det.

F1 + F2 = [40.00 17.32 0.00] kN, |R| = 43.59 kN at 23.41 deg
|F2 - F1| = 26.46 kN
Ex 1.1: v = [-1.565 0.550], |v| = 1.659, atand -19.36 deg, atan2d 160.64 deg
Ex 1.2: |rAM| = 42.17 m, u = [0.364 0.601 -0.711]
  F = [182.0 300.6 -355.7] kN, angles [68.65 53.04 135.34] deg
Ex 1.3: angle 40.23 deg, projection 128.03 m
  rA x rB = [13351.5 -8125.5 -14999.3] m^2
  det = -2.121e+06, rC.(rB x rA) = -2.121e+06 m^3
a x (b x c) = [-4 2 4], (a.c)b - (a.b)c = [-4 2 4]

Traps

• Adding magnitudes. 30\,\mathrm{kN} + 20\,\mathrm{kN} at 60^\circ is 43.6\,\mathrm{kN}, not 50. Add vectors, not sizes. (From Nugget 01.)
• The calculator's quadrant. \tan^{-1}(0.550/{-1.565}) returns -19.36^\circ, but the vector points at 160.64^\circ. Check the signs of the components, or use atan2. (From Nugget 02.)
• Free choice of direction angles. l^2 + m^2 + n^2 = 1 fixes the third angle once two are chosen; 30^\circ, 60^\circ, 60^\circ is impossible. (From Nugget 03.)
• Using the position vector as the force. \mathbf r_{AB} is in metres. Divide by its length first, then multiply by the force's magnitude. (From Nugget 03.)
• Zero dot product means a zero vector. It means perpendicular: \mathbf i\cdot\mathbf j = 0. (From Nugget 04.)
• Swapping the cross product's order. \mathbf Q\times\mathbf P = -\mathbf P\times\mathbf Q; a moment written \mathbf F\times\mathbf r has the wrong sign. (From Nugget 05.)
• Left-handed axes. With \mathbf i\times\mathbf j = -\mathbf k every cross product flips. Draw x, y, z so that the right hand curls from x to y with the thumb on z. (From Nugget 05.)
• Regrouping a double cross product. \mathbf i\times(\mathbf i\times\mathbf j) = -\mathbf j but (\mathbf i\times\mathbf i)\times\mathbf j = \mathbf 0. (From Nugget 06.)
• Differentiating a cross product out of order. Keep each factor in place: \dot{\mathbf P}\times\mathbf Q + \mathbf P\times\dot{\mathbf Q}. (From Nugget 07.)
• Equal vectors, equal effects. Two equal forces at different points can act differently on a body; equality of vectors is not physical equivalence. (From Nugget 01.)

Self-check

1. \mathbf V_1 has magnitude 8 units and rises 3 for every 4 along +x; \mathbf V_2 has magnitude 15 units and points 60^\circ above the negative x-axis. Find the magnitude and direction of \mathbf S = \mathbf V_1 + \mathbf V_2, by components, and check the magnitude with the law of cosines.
2. For the same vectors, find \mathbf D = \mathbf V_2 - \mathbf V_1.
3. A resultant force is \mathbf R = -2.4\mathbf i - 1.8\mathbf j\ \mathrm{kN}. Find its magnitude and its angle measured counter-clockwise from +x.
4. A force is \mathbf F = 150\mathbf i - 120\mathbf j + 200\mathbf k\ \mathrm N. Find its direction angles and check the cosines.
5. For \mathbf v_1 = 3\mathbf i - 2\mathbf j - 6\mathbf k, \mathbf v_2 = 4\mathbf i + 5\mathbf k and \mathbf v_3 = \mathbf i + 7\mathbf j + 2\mathbf k, find (\mathbf v_1\times\mathbf v_2)\times\mathbf v_3 and (\mathbf v_1\times\mathbf v_2)\cdot\mathbf v_3.
6. Find the angle between \mathbf v_1 = 6\mathbf i - 7\mathbf j + 2\mathbf k and \mathbf v_2 = 3\mathbf i + 4\mathbf j + 9\mathbf k, and find \mathbf v_1\times\mathbf v_2.
7. For \mathbf v_1 = 2\mathbf i + 4\mathbf j + 6\mathbf k, \mathbf v_2 = \mathbf i + 3\mathbf j + 4\mathbf k and \mathbf v_3 = -3\mathbf i - 2\mathbf k, find the vector triple product \mathbf v_1\times(\mathbf v_2\times\mathbf v_3) directly and by the identity.
8. A 9\,\mathrm m pole stands at the origin along z. A cable from its top to an anchor at (-6, 2, 0)\,\mathrm m carries 3.3\,\mathrm{kN}. Write the cable's force on the pole top as a vector and give its direction angles.
9. Three unit vectors satisfy \mathbf e_1\times\mathbf e_2 = -\mathbf e_3. Is (\mathbf e_1, \mathbf e_2, \mathbf e_3) a right-handed set? If \mathbf e_1 = \mathbf i and \mathbf e_2 = \mathbf k, what is \mathbf e_3?
10. A force \mathbf F = 400\mathbf i - 300\mathbf j + 1200\mathbf k\ \mathrm N acts on a strut whose direction is \mathbf e = \frac13(2\mathbf i + \mathbf j - 2\mathbf k). Find the component of \mathbf F along the strut, its vector part along the strut, and the size of the part perpendicular to it.
11. For \mathbf P(t) = t\,\mathbf i + t^2\mathbf j and \mathbf Q(t) = 3\mathbf i + t\,\mathbf k, find d(\mathbf P\times\mathbf Q)/dt at t = 2 by differentiating the cross product, then by the product rule.
12. Are \mathbf a = \mathbf i + 2\mathbf j + 3\mathbf k, \mathbf b = 4\mathbf i + 5\mathbf j + 6\mathbf k and \mathbf c = 7\mathbf i + 8\mathbf j + 9\mathbf k coplanar? What if c_z = 10 instead?

Fully worked solutions

1. \mathbf V_1 lies along the slope 3{:}4, whose hypotenuse is 5: \mathbf V_1 = 8(\frac45\mathbf i + \frac35\mathbf j) = 6.4\mathbf i + 4.8\mathbf j. \mathbf V_2 = 15(-\cos60^\circ\,\mathbf i + \sin60^\circ\,\mathbf j) = -7.5\mathbf i + 12.99\mathbf j. \mathbf S = -1.1\mathbf i + 17.79\mathbf j. S = \sqrt{1.1^2 + 17.79^2} = \sqrt{1.21 + 316.5} = 17.82 units. S_x < 0 and S_y > 0, so second quadrant: \theta = 180^\circ - \tan^{-1}(17.79/1.1) = 180^\circ - 86.46^\circ = 93.54^\circ from +x. Law of cosines: \mathbf V_1 is at 36.87^\circ and \mathbf V_2 at 120^\circ, 83.13^\circ apart, so S^2 = 8^2 + 15^2 + 2(8)(15)\cos83.13^\circ = 289 + 28.7 = 317.7 and S = 17.82. The two methods agree.

2. \mathbf D = (-7.5 - 6.4)\mathbf i + (12.99 - 4.8)\mathbf j = -13.9\mathbf i + 8.19\mathbf j. D = \sqrt{193.2 + 67.1} = 16.13 units. Second quadrant: \theta = 180^\circ - \tan^{-1}(8.19/13.9) = 180^\circ - 30.51^\circ = 149.49^\circ.

3. R = \sqrt{2.4^2 + 1.8^2} = \sqrt{9} = 3.0\,\mathrm{kN}. \tan^{-1}(-1.8/{-2.4}) = \tan^{-1}0.75 = 36.87^\circ on a calculator, but both components are negative, so the vector is in the third quadrant: \theta = 180^\circ + 36.87^\circ = 216.87^\circ.

4. F = \sqrt{150^2 + 120^2 + 200^2} = \sqrt{22\,500 + 14\,400 + 40\,000} = \sqrt{76\,900} = 277.3\,\mathrm N. l = 150/277.3 = 0.5409, \theta_x = 57.25^\circ; m = -120/277.3 = -0.4327, \theta_y = 115.64^\circ; n = 200/277.3 = 0.7212, \theta_z = 43.84^\circ. Check: 0.2926 + 0.1873 + 0.5202 = 1.000.

5. \mathbf v_1\times\mathbf v_2: \mathbf i: (-2)(5) - (-6)(0) = -10; \mathbf j: (-6)(4) - (3)(5) = -39; \mathbf k: (3)(0) - (-2)(4) = 8. So \mathbf v_1\times\mathbf v_2 = -10\mathbf i - 39\mathbf j + 8\mathbf k. Cross with \mathbf v_3 = (1, 7, 2): \mathbf i: (-39)(2) - (8)(7) = -134; \mathbf j: (8)(1) - (-10)(2) = 28; \mathbf k: (-10)(7) - (-39)(1) = -31. (\mathbf v_1\times\mathbf v_2)\times\mathbf v_3 = -134\mathbf i + 28\mathbf j - 31\mathbf k. Dot: (-10)(1) + (-39)(7) + (8)(2) = -10 - 273 + 16 = -267. (The determinant of the rows \mathbf v_1, \mathbf v_2, \mathbf v_3 is also -267.)

6. \mathbf v_1\cdot\mathbf v_2 = 18 - 28 + 18 = 8. v_1 = \sqrt{36 + 49 + 4} = \sqrt{89} = 9.434; v_2 = \sqrt{9 + 16 + 81} = \sqrt{106} = 10.296. \cos\theta = 8/(9.434 \times 10.296) = 0.0824, \theta = 85.28^\circ. \mathbf v_1\times\mathbf v_2: \mathbf i: (-7)(9) - (2)(4) = -71; \mathbf j: (2)(3) - (6)(9) = -48; \mathbf k: (6)(4) - (-7)(3) = 45. So -71\mathbf i - 48\mathbf j + 45\mathbf k.

7. Directly: \mathbf v_2\times\mathbf v_3: \mathbf i: (3)(-2) - (4)(0) = -6; \mathbf j: (4)(-3) - (1)(-2) = -10; \mathbf k: (1)(0) - (3)(-3) = 9. Then \mathbf v_1\times(-6, -10, 9): \mathbf i: (4)(9) - (6)(-10) = 96; \mathbf j: (6)(-6) - (2)(9) = -54; \mathbf k: (2)(-10) - (4)(-6) = 4. Result 96\mathbf i - 54\mathbf j + 4\mathbf k. By the identity: \mathbf v_1\cdot\mathbf v_3 = -6 - 12 = -18, \mathbf v_1\cdot\mathbf v_2 = 2 + 12 + 24 = 38; -18(1, 3, 4) - 38(-3, 0, -2) = (-18 + 114, -54, -72 + 76) = (96, -54, 4). Same.

8. From the top (0, 0, 9) to the anchor: \mathbf r = -6\mathbf i + 2\mathbf j - 9\mathbf k\ \mathrm m, r = \sqrt{36 + 4 + 81} = 11\,\mathrm m. \mathbf n = (-0.5455, 0.1818, -0.8182). \mathbf F = 3.3\,\mathbf n = -1.8\mathbf i + 0.6\mathbf j - 2.7\mathbf k\ \mathrm{kN}. Angles: \theta_x = \cos^{-1}(-0.5455) = 123.06^\circ, \theta_y = \cos^{-1}0.1818 = 79.52^\circ, \theta_z = \cos^{-1}(-0.8182) = 144.90^\circ.

9. No: a right-handed set has \mathbf e_1\times\mathbf e_2 = +\mathbf e_3, so this one is left-handed. With \mathbf e_1 = \mathbf i and \mathbf e_2 = \mathbf k: \mathbf i\times\mathbf k = -\mathbf j = -\mathbf e_3, so \mathbf e_3 = \mathbf j. (The order x, z, y is left-handed.)

10. Check \mathbf e is a unit vector: \frac19(4 + 1 + 4) = 1. F_e = \mathbf F\cdot\mathbf e = \frac13(800 - 300 - 2400) = -633.3\,\mathrm N: 633.3\,\mathrm N directed against \mathbf e. Vector part: -633.3\,\mathbf e = -422.2\mathbf i - 211.1\mathbf j + 422.2\mathbf k\ \mathrm N. F = \sqrt{160\,000 + 90\,000 + 1\,440\,000} = 1300\,\mathrm N, so the perpendicular part has size \sqrt{1300^2 - 633.3^2} = 1135.3\,\mathrm N.

11. \mathbf P\times\mathbf Q: \mathbf i: (t^2)(t) - (0)(0) = t^3; \mathbf j: (0)(3) - (t)(t) = -t^2; \mathbf k: (t)(0) - (t^2)(3) = -3t^2. Derivative: 3t^2\mathbf i - 2t\mathbf j - 6t\mathbf k, at t = 2: 12\mathbf i - 4\mathbf j - 12\mathbf k. Product rule at t = 2: \dot{\mathbf P} = \mathbf i + 4\mathbf j, \mathbf Q = 3\mathbf i + 2\mathbf k, \dot{\mathbf P}\times\mathbf Q = 8\mathbf i - 2\mathbf j - 12\mathbf k; \mathbf P = 2\mathbf i + 4\mathbf j, \dot{\mathbf Q} = \mathbf k, \mathbf P\times\dot{\mathbf Q} = 4\mathbf i - 2\mathbf j. Sum: 12\mathbf i - 4\mathbf j - 12\mathbf k. Same.

12. \mathbf a\cdot(\mathbf b\times\mathbf c) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0: the box is flat, so the vectors are coplanar. (Indeed \mathbf c = 2\mathbf b - \mathbf a.) With c_z = 10: 1(50 - 48) - 2(40 - 42) + 3(32 - 35) = 2 + 4 - 9 = -3, so they are not coplanar. The box has volume 3, and the set is left-handed.