# The scalar product · Grade 12 · 4 units

> The scalar product, Grade 12 · 4 units: The angle between vectors and more. Explanation, worked examples and free practice following the Israeli curriculum.

In The scalar product (Grade 12 · 4 units) you practise 6 skills: The angle between vectors · What the scalar product is · The scalar product in coordinates · Perpendicular vectors · A vector perpendicular to a plane · Lengths and areas.

The angle between two vectors is the angle between them when they leave the same point. If one vector starts at the end of the other, move it so that the two starts meet, then measure. The angle is between 0 and 180 degrees. The scalar product of two vectors is the product of their magnitudes and the cosine of the angle between them. The result is a number, not a vector. It is positive when the angle is acute, negative when the angle is obtuse, and zero when the vectors are perpendicular. A vector times itself is its magnitude squared.

The scalar product is commutative, a number multiplying one of the vectors comes out of the product, and it distributes over addition of vectors. These properties are taken without proof. They give the expansions: the square of the sum of two vectors is the sum of the squared magnitudes plus twice the scalar product; for the difference, subtract twice the scalar product instead, and the sum times the difference is the difference of the squared magnitudes.

In coordinates, the scalar product is the sum of the products of matching components, and the magnitude of a vector is the square root of the sum of its squared components. Divide the product by the product of the magnitudes to get the cosine of the angle between the vectors.

Two vectors that are not the zero vector are perpendicular exactly when their scalar product is zero. A vector is perpendicular to a plane when it is perpendicular to two vectors of the plane that are not collinear. The product also gives the lengths of segments, and the areas of triangles and parallelograms.

**Tip:** Check the sign: a negative scalar product means the angle between the vectors is obtuse.

## What you learn

### The angle between vectors

- Example: In the cube ABCDA′B′C′D′ , what is the angle between the vectors DC⃗ and DD′⃗ ? (A box with its corners marked by letters)
- Answer: 90 °
- Solution: Both vectors are edges of the cube that leave the same corner, and they are perpendicular. So the angle between them is 90° .

### What the scalar product is

- Example: The magnitudes |u⃗| = 3 and |v⃗| = 10 are given, and the angle between the vectors is 0° . What is the scalar product u⃗ ⋅ v⃗ ? (Two arrows leaving the same point, and the angle between them)
- Answer: 30
- Solution: Multiply the magnitudes by the cosine of the angle, cos 0° = 1 : 3 × 10 × 1 = 30 .

### The scalar product in coordinates

- Example: What is the scalar product (−5, 6, −5) ⋅ (0, 5, 3) ? (Three axes in space leaving the origin)
- Answer: 15
- Solution: Multiply each component by the matching one and add the three products: 0 + 30 − 15 , which is 15 .

### Perpendicular vectors

- Example: Are the vectors u⃗ = (2, −3, −6) and v⃗ = (6, −2, 3) perpendicular? (Three axes in space leaving the origin)
- Options: No, because their scalar product is not zero · No, because their magnitudes are different · Yes, because they are collinear · Yes, because their scalar product is zero
- Answer: Yes, because their scalar product is zero
- Solution: The scalar product is 12 + 6 − 18 = 0 , so the vectors are perpendicular.

### A vector perpendicular to a plane

- Example: In the right pyramid SABCD the base is a square, and the point O is the centre of the base. Which pair of products proves that SO⃗ is perpendicular to the plane of the base? (A pyramid with its corners marked by letters)
- Options: SO⃗ ⋅ AB⃗ = 0 and SO⃗ ⋅ CD⃗ = 0 · SO⃗ ⋅ AC⃗ = 0 alone · SO⃗ ⋅ SA⃗ = 0 and SO⃗ ⋅ SB⃗ = 0 · SO⃗ ⋅ AB⃗ = 0 and SO⃗ ⋅ AD⃗ = 0
- Answer: SO⃗ ⋅ AB⃗ = 0 and SO⃗ ⋅ AD⃗ = 0
- Solution: The vectors AB⃗ and AD⃗ lie in the plane of the base and are not collinear, so SO⃗ ⋅ AB⃗ = 0 and SO⃗ ⋅ AD⃗ = 0 prove it. One vector is not enough, and neither are two collinear vectors.

### Lengths and areas

- Example: The points A(2, 0, −2) and B(−1, 1, −4) are given. What is the length of the segment AB ? (Three axes in space, and two points joined by a segment)
- Answer: √14
- Solution: The vector is AB⃗ = (−3, 1, −2) , and the sum of its squared components is 14 . So AB = √14 .

## Not on the site yet

- Proofs with the scalar product

The exercises are made by rules and the answers are worked out in code. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 4 units (grades 10 to 12).

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