How do I find the area of a triangle using vectors?

Take two sides from the same vertex as vectors, find their cross product and halve its magnitude: area = ½|AB × AC|.

Are equations of lines and planes tested in DSE M2?

No. The M2 curriculum covers parallelism, orthogonality, division of a line segment, projection, angle and area. Distances to a plane are found with a normal vector from the cross product and a projection, not with a plane equation.

Why do students lose marks on vector long questions?

Mostly because the 3D relationship in the question is hard to picture, so the wrong pair of vectors is chosen. Drawing the figure, or opening it in a 3D graph you can rotate, makes it clear which vectors lie in the plane and which one is the normal.

Applications of Vectors (DSE M2): Projection, Area and Angle

DSE M2 uses vectors for five jobs: testing parallel and perpendicular lines, dividing a line segment, projecting one vector onto another, finding the angle between two vectors, and finding the area of a triangle or parallelogram with ½|AB × AC|.

Solve Applications of Vectors questions with AI

Triangle ABC from Example 1 with the normal AB × AC = 3i + 4j + 5k drawn at A. Try it yourself →

Applications of Vectors: Formulas

Projection of b onto a((a · b) / |a|²) a

Length of the projection|a · b| / |a|

Angle between vectorscos θ = (a · b) / (|a||b|)

Area of triangle ABC½ |AB × AC|

Area of parallelogram|a × b|

Section formula (AP : PB = m : n)OP = (na + mb) / (m + n)

Parallel / perpendiculara × b = 0 (parallel); a · b = 0 (perpendicular)

Distance from a Point to a Plane Through Three Points

Equations of planes are not in the M2 syllabus, so the examinable method uses a normal vector and a projection:

Take two vectors in the plane, e.g. AB and AC.
A normal to the plane is n = AB × AC.
Join the point P to any point of the plane, e.g. AP.
The distance is the length of the projection of AP onto n: |AP · n| / |n|.

The same idea gives the distance from a point to a line: subtract the projection along the line from the vector to the point, then take the magnitude.

Worked Examples — Applications of Vectors

Example 1: Area of a triangle from three points

The points A(1, 0, 2), B(3, 1, 0) and C(0, 2, 1) are given. Find the area of triangle ABC.

AB = 2i + j − 2k, AC = −i + 2j − k
AB × AC = [(1)(−1) − (−2)(2)]i − [(2)(−1) − (−2)(−1)]j + [(2)(2) − (1)(−1)]k = 3i + 4j + 5k
|AB × AC| = √(9 + 16 + 25) = √50 = 5√2
Area = ½ × 5√2 = 5√2 / 2

Answer: 5√2 / 2 square units (≈ 3.54)

Example 2: Projection of one vector onto another

Let a = 3i + 4k and b = i + 2j + 2k be the position vectors of A and B. Find the projection of b onto a, and hence the distance from B to the line OA.

a · b = 3 + 0 + 8 = 11, |a|² = 25
Projection = (11/25)(3i + 4k) = (33/25)i + (44/25)k
Length of the projection = 11/5, and |b| = 3
By Pythagoras, distance = √(3² − (11/5)²) = √(104/25) = 2√26 / 5

Answer: Projection (33/25)i + (44/25)k; distance 2√26 / 5 (≈ 2.04)

💡 Exam Traps & Secrets

⚠️ Projection onto a, so divide by |a|²The vector projection of b onto a is a multiple of a: ((a · b)/|a|²)a. Dividing by |b|², or multiplying by b, projects the wrong way round.

⚠️ Triangle area needs the ½|AB × AC| is the area of the parallelogram. The triangle is half of it. Both vectors must also start from the same vertex.

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Area with cross product)Find the area of the triangle with vertices A(1, 2, 3), B(2, 0, 1) and C(0, 1, 4).

(Projection)Find the projection of 2i + 3j − k onto i + 2j + 2k.

(Point-to-plane distance)Find the distance from P(1, 1, 5) to the plane through A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3) using vectors.