Vector geometry and products

Keep the point, displacement and length distinct

A third coordinate records movement in a third perpendicular direction. Work in an orthonormal Cartesian frame: the coordinate axes are perpendicular and use the same unit scale. You can then add components and use Pythagoras just as in two dimensions.

For A = (-1,2,0) and B = (3,-2,2), subtract the starting point from the endpoint:

This displacement says how to move from A to B. Its magnitude is the distance between the points:

The reverse displacement is vector BA = (-4,4,-2). It has the same length, but the opposite direction. Translating both points changes their position vectors a and b by the same amount, so their difference stays unchanged.

Vector addition joins displacements head to tail. Multiplying a vector by a positive scalar scales its length; a negative scalar also reverses its direction. Multiplying by zero gives the zero vector, which has no direction.

The same rules apply to two components. For u = (3,-2) and v = (-1,5),

Keep the coordinate order and apply the operation to every component. In the difference, subtracting the second component 5 gives -2-5 = -7; it does not give the sum 3.

A unit vector has magnitude one. Divide a nonzero vector by its magnitude to retain its direction. Here the unit vector from A towards B is

Check that its squared magnitude is 4/9 + 4/9 + 1/9 = 1. Normalising each component separately would destroy the vector’s direction.

Derive a ratio from a displacement

Write a point on the line through A and B as

The same scalar t multiplies every component. When 0 < t < 1, P lies inside the segment AB. The endpoint cases are t = 0 and t = 1; values outside this interval place P beyond an endpoint.

If P divides AB internally with AP:PB = 1:2, it is one third of the way from A to B. Thus t = 1/3, and

The distances are AP = 2 and PB = 4, consistent with the ratio and AB = 6. More generally, for positive m,n and internal ratio AP:PB = m:n,

The coefficient of A involves n. Deriving it from the fraction of the displacement avoids swapping the weights.

For a distinct point C, collinearity means that c-a is a scalar multiple of b-a. Every component must give one consistent scalar, including a zero-component condition. Matching just one coordinate is insufficient.

Check your understanding 1 Check the coordinate that does not change

Let A = (1,2,3), B = (1,4,5) and C = (2,8,9). Are the three points collinear? Check every coordinate without dividing by a zero component.

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They are not collinear. The displacements are vector AB = (0,2,2) and vector AC = (1,6,6). The last two components suggest a factor of three, but three times the first component of AB is still zero. Every point on line AB has x-coordinate one, while C has x-coordinate two.

Changing C to (1,8,9) would make vector AC = 3 vector AB in every coordinate. The zero component is a condition to check, not a number to divide by.

Check your understanding 2 Locate the point beyond B

Keep A = (-1,2,0) and B = (3,-2,2). Q lies beyond B on the line AB, with AQ:QB = 3:1. Find Q and explain why using the internal ratio formula would put it in the wrong place.

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Write q = a + t(b-a) with t > 1. The lengths satisfy AQ:QB = t:(t-1) = 3:1. Hence t = 3(t-1), so t = 3/2 and Q = (5,-4,3).

Check AQ = 9 and QB = 3. The internal formula would use t = 3/4, placing a point inside AB. The location “beyond B” changes the distance relationship; it is not a reason to change the signs of individual coordinates by guesswork.

For an external point beyond A, the parameter is negative. If AR:RB = 1:3, then the two lengths are -t|vector AB| and (1-t)|vector AB|. Thus

This gives R = (-3,4,-1), with AR = 3 and RB = 9. For positive lengths in an external ratio AP:PB = m:n, the same reasoning gives t = m/(m-n). When m > n, P lies beyond B; when m < n, it lies beyond A. Equal positive ratios give the midpoint internally, but no finite external point when A and B are distinct.

Use the scalar product to measure alignment

For vectors u and v,

For nonzero vectors, it also equals |u||v| cos θ, where 0 ≤ θ ≤ 180°. A positive dot product gives an angle below 90°; a negative one gives an angle above 90°; zero gives perpendicular directions. A zero vector cannot determine an angle, even though its dot product with every vector is zero.

For u = (2,-1,2) and v = (1,2,0), the dot product is 2-2 + 0 = 0. Both vectors are nonzero, so they are perpendicular.

The scalar product is commutative and distributes over addition. In particular, u · u = |u|². It is a scalar, so you cannot read its value as three vector components.

Separate a signed component from a length

Suppose you want the component of a = (2,1,2) along d = (0,0,-5). First form the unit direction n hat = d/|d| = (0,0,-1).

The signed scalar component is

The negative sign says that the component points opposite to the chosen unit direction. The corresponding vector component is

Its length is 2, never -2. Subtract the parallel vector to get the perpendicular remainder:

Its dot product with d is zero and its magnitude is square root of 5. The check |a|² = 2² + (square root of 5)² = 9 confirms the perpendicular decomposition.

For any nonzero direction d, combining normalisation and multiplication gives

Using a · d alone as a component would give -10 in this example: it also contains the factor |d| = 5. A direction vector need not be a unit vector.

Check your understanding 3 Reverse the reference direction

Replace d = (0,0,-5) by -d in the example. Which of the signed scalar component, parallel vector and perpendicular magnitude change? Explain without repeating the whole calculation.

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The unit direction changes sign, so the signed scalar component changes from -2 to 2. The parallel vector stays (0,0,2): both factors in (a · n hat)n hat reverse sign. The perpendicular vector and its magnitude square root of 5 also stay unchanged.

A signed coordinate depends on which orientation you call positive. The geometrical components do not change when you reverse only the reference direction.

The vector product gives a perpendicular direction

In a right-handed coordinate frame, the components of u × v are

For the perpendicular vectors u = (2,-1,2) and v = (1,2,0), these give

Check perpendicularity to both inputs: -8-2 + 10 = 0 and -4 + 4 + 0 = 0. There are two possible perpendicular orientations; the order of the inputs fixes which one you get. The right-hand rule points your thumb along u × v when your fingers curl from u towards v through the smaller angle. Reversing the order negates the result.

The magnitude is |u||v| sin θ, the area of the parallelogram spanned by the vectors. Here it is square root of (16 + 4 + 25) = 3 square root of 5. That also equals 3 × square root of 5, as expected for perpendicular sides. The triangle with those two sides from one vertex has half that area.

The vector product distributes over addition, but is not commutative: u × v = -v × u. Also u × u = 0. For two nonzero vectors, a zero cross product means parallel directions, including opposite directions. It does not mean the vectors are perpendicular.

With a unit direction n hat, |a × n hat| is the magnitude of the component perpendicular to that direction. Compare it with |a · n hat|, the magnitude of the parallel component. The absolute value removes orientation; it does not turn a scalar product into a vector product.

Check your understanding 4 Dot and cross answer different questions

For a = (1,0,0) and b = (1,1,0), find a·b, a×b and the angle between the vectors.

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The dot product is 1, the cross product (0,0,1), and cos θ = 1/square root of 2, giving θ = 45°. The scalar product is a number; the vector product is a directed vector. They are not interchangeable outputs.

Check your understanding 5 Work from the geometry

In a right-handed orthonormal frame, O is the origin, A = (1,0,1) and B = (0,2,0). Find the area of triangle OAB. Decide whether replacing B by B' = (0,-2,0) changes that area or the direction of vector OA × vector OB. Justify both conclusions.

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The two sides from O are (1,0,1) and (0,2,0). Their vector product is (-2,0,2), with magnitude 2 square root of 2. The triangle’s area is therefore square root of 2 square units. Alternatively, the sides are perpendicular by their dot product, so half base times height gives 1/2 × square root of 2 × 2.

Replacing B by B′ negates the second side vector. It reverses the cross product to (2,0,-2) while preserving its magnitude and the triangle’s area. Check both the factor one half and the direction: an area alone cannot show which orientation the product has.

Match the quantity to the question

A displacement has length units. A scalar component along a unit direction also has length units; a cross product of two length vectors has area units. Before calculating, decide whether the question asks for a position, a vector, a signed scalar, a nonnegative length or an area. Then check the dimensions and any zero-vector exception.