The scalar product turns vectors into a tool for geometry. The foot of the perpendicular F from a point P to a line r = a + λd is found by using two facts: F is on the line, and the vector PF is perpendicular to d, so PF · d = 0. The shortest distance from P to the line is |PF|. Areas of triangles and parallelograms use ½|AB||AC| sin θ, with cos θ from the scalar product.
Vector geometry
Foot of the perpendicular F: F = a + λd and PF · d = 0
Shortest distance from P to the line = |PF|
Triangle area = ½|AB||AC| sin θ
Parallelogram area = |AB||AD| sin θ
sin θ = √(1 − cos²θ)
The foot of the perpendicular
The foot of the perpendicular F from a point P to a line is the point on the line closest to P. Write F as a general point on the line, a + λd. Then the vector PF = F − P must be perpendicular to the direction d, so PF · d = 0. That gives a linear equation in λ; substitute λ back to find F.
The shortest distance from a point to a line
The shortest distance from P to a line is the perpendicular distance, which is the length |PF|. So find the foot F first, then work out the magnitude of PF. Questions often ask for this as an exact surd.
The area of a triangle
Using two sides from one vertex, area = ½|AB||AC| sin θ, where θ is the angle between AB and AC. Find cos θ from the scalar product, then sin θ = √(1 − cos²θ), which keeps the answer exact. If AB · AC = 0 the angle is 90° and the area is simply half the product of the lengths.
The area of a parallelogram
A parallelogram is two congruent triangles, so its area is |AB||AD| sin θ, using two adjacent sides from the same vertex. As before, find cos θ from the scalar product and convert to sin θ.
Shapes and proofs with vectors
To find a missing vertex of a parallelogram ABCD, use the fact that opposite sides are equal vectors: the vector DC equals the vector AB, so c = d + (b − a). To prove that two lines or sides are perpendicular, show that the relevant scalar product is 0. Questions often chain these: find a foot, then a distance, then an area using the right angle at the foot.
Worked examples
Example 1. Find the foot of the perpendicular from P(4, 4, 3) to the line r = (i − k) + λ(2i + j + 2k).
- General point: F = (1 + 2λ, λ, −1 + 2λ).
- PF = F − P = (2λ − 3, λ − 4, 2λ − 4).
- PF · (2, 1, 2) = 4λ − 6 + λ − 4 + 4λ − 8 = 9λ − 18 = 0, so λ = 2.
Answer: F = (5, 2, 3).
Marks: M1 for PF · d = 0, A1 for λ = 2, A1 for F.
Example 2. Find the shortest distance from P(4, 1, 1) to the line r = (i + 2k) + λ(2i + j − 2k).
- F = (1 + 2λ, λ, 2 − 2λ), so PF = (2λ − 3, λ − 1, 1 − 2λ).
- PF · (2, 1, −2) = 9λ − 9 = 0, so λ = 1 and F = (3, 1, 0).
- PF = (−1, 0, −1), so |PF| = √2.
Answer: The shortest distance is √2.
Marks: M1 for F, M1 for |PF|, A1 for √2.
Example 3. Find the area of triangle ABC for A(1, 2, 1), B(4, 3, 2) and C(2, 0, 3).
- AB = (3, 1, 1) and AC = (1, −2, 2).
- AB · AC = 3 − 2 + 2 = 3. |AB| = √11 and |AC| = 3.
- cos θ = 1/√11, so sin θ = √(10/11).
- Area = ½ × √11 × 3 × √(10/11) = (3/2)√10.
Answer: (3/2)√10.
Marks: M1 for cos θ, M1 for the area formula, A1 for (3/2)√10.
Example 4. Find the area of the parallelogram OABC with OA = 2i + j + 2k and OC = i + 2j − k.
- OA · OC = 2 + 2 − 2 = 2. |OA| = 3 and |OC| = √6.
- cos θ = 2/(3√6), so sin θ = √(50/54).
- Area = 3√6 × √(50/54) = √50.
Answer: 5√2.
Marks: M1 for cos θ, M1 for the area formula, A1 for 5√2.
Example 5. ABCD is a parallelogram with A(1, 1, 2), B(4, 2, 3) and D(2, 4, 1). Find C.
- AB = b − a = (3, 1, 1).
- Opposite sides are equal vectors, so c = d + AB = (2, 4, 1) + (3, 1, 1).
Answer: C = (5, 5, 2).
Marks: M1 for c = d + AB, A1 for C.
Example 6. The line l is r = (2i + j + k) + λ(i + 2j + 2k), with A(2, 1, 1) on l, and P is (6, 3, 6). Find the foot F, the distance PF, and the area of triangle APF.
- F = (2 + λ, 1 + 2λ, 1 + 2λ), so PF = (λ − 4, 2λ − 2, 2λ − 5).
- PF · (1, 2, 2) = 9λ − 18 = 0, so λ = 2 and F = (4, 5, 5).
- PF = (−2, 2, −1), so PF = 3. AF = (2, 4, 4), so AF = 6.
- The angle at F is 90°, so the area is ½ × 6 × 3.
Answer: F = (4, 5, 5), PF = 3, area 9.
Marks: M1 for F, A1 for PF = 3, M1 for the right-angled area, A1 for 9.
Common mistakes
- Using a fixed point instead of the general point. Write F as a + λd before applying the perpendicular condition.
- Dotting PF with the wrong vector. PF must be dotted with the line's direction vector d.
- Using sin θ directly from a calculator decimal. Use sin θ = √(1 − cos²θ) to keep an exact answer.
- Forgetting the ½ for a triangle. Triangle area = ½|AB||AC| sin θ; a parallelogram has no ½.
- Missing the right angle at the foot. Triangle APF has a right angle at F, which simplifies its area.
How vector geometry is examined
Foot-of-perpendicular and area questions close the Pure 4 vectors chapter (WMA14) in Edexcel International A-Level; in Edexcel A-Level these methods are taught in Further Maths rather than in single Maths (9MA0). Questions commonly:
- find the foot of the perpendicular from a point to a line
- find the shortest distance from a point to a line, exactly
- find the area of a triangle or parallelogram from its vertices
- find a missing vertex, or prove a property of a shape
Full multi-step questions with mark schemes are among the Pure 4 vectors exam questions. The vectors worksheet for Pure 4 and the vectors revision engine give more practice.
These methods rely on the scalar product and on vector equations of lines.
Frequently asked questions
What is the method for the foot of the perpendicular from a point to a line?
Write a general point F on the line using λ, form the vector from P to F, set its scalar product with the direction vector to 0, solve for λ, and substitute back.
How do you find the shortest distance from a point to a line?
Find the foot of the perpendicular F, then the distance is |PF|.
How do you find the area of a triangle with vectors?
Area = ½|AB||AC| sin θ, where cos θ comes from the scalar product of AB and AC.