Three vectors are coplanar if and only if:

Their scalar triple product is zero

Their dot products are all zero

Their cross products are all zero

Vector Algebra

Vectors are the most economical topic in the paper, because two products do almost all the work and they answer different questions. The dot product returns a number and measures how much two vectors point the same way, so it is the tool for angles, projections and perpendicularity. The cross product returns a vector perpendicular to both and measures the area they span, so it is the tool for normals, areas and parallelism. Deciding which of the two a question wants — before writing anything — is most of the skill, and it also makes the geometry of the next chapter almost mechanical.

The Two Products

Dot product against cross product
Feature	Dot product a·b	Cross product a × b
Result	A scalar	A vector perpendicular to both
Formula	\|a\|\|b\| cos θ	\|a\|\|b\| sin θ n̂
Zero when	The vectors are perpendicular	The vectors are parallel
Commutative?	Yes, a·b = b·a	No — a × b = −(b × a)
Geometric meaning	Projection of one on the other	Area of the parallelogram they span
Typical use	Angle, work done, perpendicularity test	Normal to a plane, torque, area

🧠 Read the question for "angle" or "perpendicular" versus "area" or "normal"

The choice of product is almost always dictated by a single word. Angle, projection, work, perpendicular all call for the dot product, because each is about how much of one vector lies along another. Area, normal, torque, parallel, coplanar all call for the cross product or the scalar triple product, because each is about the space two or three vectors span. So "find the angle between the diagonals of a parallelogram" is a dot-product question and "find the area of a triangle with given vertices" is a cross-product one — half the vector questions in the paper are decided at the reading stage rather than the calculating stage.

Two derived quantities recur. The projection of a on b is (a·b)/|b|, a scalar, while the projection vector is that value times the unit vector along b — a distinction questions test directly. And the area of a triangle with two sides a and b is ½|a × b|, so the area of a triangle with vertices A, B, C is ½|AB⃗ × AC⃗|, which is usually faster than any coordinate formula.

The Triple Products

The scalar triple product [a b c] = a·(b × c) is a single number equal to the determinant of the three components, and its absolute value is the volume of the parallelepiped the three vectors span. So the vectors are coplanar exactly when [a b c] = 0, since a flat box has no volume — this is the standard coplanarity test and it is far quicker than any alternative. The product is unchanged by a cyclic rotation of the three vectors and changes sign under a swap.

🎯 Why the scalar triple product is a volume, and why that makes coplanarity easy

b × c is a vector whose magnitude is the area of the base parallelogram and whose direction is perpendicular to that base. Dotting a with it picks out the component of a along that perpendicular — which is exactly the height of the box. Area times height is volume, so [a b c] is the volume, signed according to the orientation. The payoff is immediate: three vectors lie in one plane if and only if the box they define is flat, so setting the determinant to zero tests coplanarity in one line. The same reasoning tells you four points are coplanar when the three vectors joining one of them to the others have zero triple product.

⚠️ The cross product is neither commutative nor associative

Two habits carried over from ordinary multiplication fail here. Order matters: a × b = −(b × a), so swapping the factors flips the direction of the result, which is why the order of vertices matters when you compute a normal. And brackets matter: a × (b × c) is generally not equal to (a × b) × c. The vector triple product expands as a × (b × c) = (a·c)b − (a·b)c, and the pattern worth remembering is that the result lies in the plane of b and c, with the middle vector taking the positive sign. Dropping the brackets or reordering the factors is the single most common source of a wrong sign in this chapter.

Two applications close the chapter. The section formula in vector form gives the point dividing AB in the ratio m : n internally as (nA⃗ + mB⃗)/(m + n), and externally with a minus sign — note that the coefficients cross over, which is the usual slip. And three points are collinear when the vectors joining them are parallel, that is when their cross product vanishes, which is the vector counterpart of the determinant test from coordinate geometry.

Key takeaways

The dot product gives a scalar and vanishes for perpendicular vectors; the cross product gives a vector and vanishes for parallel ones.
Angle, projection, work and perpendicularity mean dot; area, normal, torque and coplanarity mean cross.
Projection of a on b is the scalar (a·b)/|b|; the projection vector multiplies that by b̂.
The area of a triangle on sides a and b is ½|a × b|.
[a b c] is the volume of the parallelepiped, so the vectors are coplanar exactly when it is zero.
a × b = −(b × a), so the order of the factors decides the direction of the result.
a × (b × c) = (a·c)b − (a·b)c, and the result lies in the plane of b and c.
The internal section point is (nA + mB)/(m + n) — the coefficients cross over.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Two non-zero vectors are perpendicular if:
1. Their cross product is zero
2. Their dot product is zero
3. Their sum is zero
4. Their magnitudes are equal
Show answer
Answer: B — Their dot product is zero
Since a·b = |a||b|cos θ, the dot product vanishes exactly when cos θ = 0, that is at 90°. A zero cross product means sin θ = 0 and so signals parallel vectors instead — the two conditions are complementary, and mixing them up reverses the answer to a great many questions.
Three vectors are coplanar if and only if:
1. Their scalar triple product is zero
2. Their dot products are all zero
3. Their cross products are all zero
4. Their magnitudes are equal
Show answer
Answer: A — Their scalar triple product is zero
The scalar triple product equals the volume of the parallelepiped the three vectors span, and three vectors lie in one plane precisely when that box is flat and the volume is zero. All cross products vanishing would mean all three are mutually parallel, which is a much stronger condition and not what coplanarity requires.
The area of a triangle with adjacent sides represented by a and b is:
1. |a·b|
2. |a × b|
3. ½|a × b|
4. ½|a·b|
Show answer
Answer: C — ½|a × b|
The magnitude of the cross product is the area of the parallelogram on the two vectors, and a triangle is half of that parallelogram, so the area is ½|a × b|. Option two is the parallelogram area, offered because forgetting the factor of a half is the commonest slip; the dot product gives a scalar with no area interpretation at all.
a × (b × c) equals:
1. (a × b) × c
2. (a·c)b − (a·b)c
3. (a·b)c − (a·c)b
4. (a × b)·c
Show answer
Answer: B — (a·c)b − (a·b)c
The vector triple product expands with the middle vector of the bracket taking the positive sign: a × (b × c) = (a·c)b − (a·b)c, and the result therefore lies in the plane of b and c. The cross product is not associative, so the first option is generally a different vector, and the last option is a scalar and cannot equal a vector.
The projection of a on b is:
1. (a·b)/|a|
2. (a·b)/|b|
3. |a × b|/|b|
4. a·b
Show answer
Answer: B — (a·b)/|b|
The projection measures how much of a lies along b, so it is |a|cos θ, and writing cos θ = (a·b)/(|a||b|) leaves (a·b)/|b|. Dividing by |a| instead gives the projection of b on a, which is a different quantity — the divisor is always the magnitude of the vector you are projecting onto.
For the cross product, a × b equals:
1. b × a
2. −(b × a)
3. |a||b|cos θ
4. Zero always
Show answer
Answer: B — −(b × a)
The cross product is anti-commutative, so reversing the order reverses the direction of the resulting vector while keeping its magnitude. This is why the order of vertices matters when computing a normal to a plane, and why an area computed as ½|a × b| uses a modulus — the magnitude is what carries the geometric meaning, not the sign.
The point dividing AB internally in the ratio m : n is:
1. (mA⃗ + nB⃗)/(m + n)
2. (nA⃗ + mB⃗)/(m + n)
3. (A⃗ + B⃗)/2
4. (mA⃗ − nB⃗)/(m − n)
Show answer
Answer: B — (nA⃗ + mB⃗)/(m + n)
The coefficients cross over: the ratio m : n puts the larger weight on the farther endpoint, giving (nA⃗ + mB⃗)/(m + n). A quick sanity check is to set m = n, which should give the midpoint — and it does. The last option with the minus signs is the external division formula, and the third is the special case of a midpoint.
The volume of the parallelepiped formed by a, b and c is:
1. |a·(b × c)|
2. |a × (b × c)|
3. |a||b||c|
4. ½|a·(b × c)|
Show answer
Answer: A — |a·(b × c)|
b × c has magnitude equal to the base area and direction normal to the base, so dotting with a extracts the height and the product is the volume. The vector triple product in the second option is a vector, not a volume, and the half in the last option belongs to a tetrahedron, whose volume is one sixth of the parallelepiped rather than one half.