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Scalar products and angles

The scalar product relates coordinates to lengths and angles.

For real vectors a and b, a·b is the sum of the products of corresponding components. It also equals |a||b|cos θ for the angle between nonzero vectors. If either vector is zero, the angle is undefined.

A zero scalar product establishes perpendicular nonzero vectors. For a line-to-plane angle, a plane normal is often useful, but its angle to the line is complementary to the line-to-plane angle. Make clear which angle the question asks for.

Worked example

Find the angle between (1,0) and (1,1).

  1. The scalar product is 1.
  2. The lengths are 1 and √2.
  3. cos θ = 1/√2.

Answer: θ = π/4, or 45°

Perpendicularity identifies the closest point

A line r = a + td has a closest point to P when the displacement a + td − P is perpendicular to d. Set their scalar product to zero. This gives one linear equation for t when d is nonzero.

To intersect a line with a plane n·r = c, substitute the line equation. If n·d is nonzero there is one intersection. If it is zero, the line is parallel to the plane: check n·a to distinguish a line in the plane from one with no intersection.

The closest point to the origin on r = (1, 2, 0) + t(1, 1, 0)

  1. Perpendicularity gives (1 + t) + (2 + t) = 0, hence t = −3/2.
  2. The closest point is (−1/2, 1/2, 0).
  3. Its squared distance is 1/4 + 1/4 = 1/2. The distance is √2/2.
  4. The displacement has zero scalar product with (1, 1, 0), confirming perpendicularity.
Can you divide by n·d when it is zero?

No. Check the constant terms to decide whether there are no intersections or infinitely many.

Take the idea further

Translate the target point and repeat the projection. The direction stays the same, but the displacement must be measured from the new target.

More advanced methods

Practise scalar products and angles

Course mapping

These specification references show where the topic occurs. The questions cover only some parts of each topic.

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