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How do you find angles between lines and planes and the shortest distances in three dimensions?

Angles between two lines, between a line and a plane, and between two planes, and the shortest distance from a point to a line or plane and between two skew lines.

A focused answer to the OCR A-Level Further Mathematics A content on distances and angles in three dimensions, covering the angle between two lines, between a line and a plane and between two planes, and the shortest distance from a point to a line or plane and between two skew lines, using the scalar and vector products.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Angle between two lines
  3. Angle between a line and a plane
  4. Angle between two planes
  5. Shortest distance from a point to a plane
  6. Shortest distance between skew lines
  7. Try this

What this dot point is asking

OCR wants you to find the angle between two lines (using their directions), between a line and a plane (using the direction and the normal), and between two planes (using their normals), and to find the shortest perpendicular distances: from a point to a line, from a point to a plane, and between two skew lines, drawing on the scalar and vector products.

Angle between two lines

The angle between two lines is the angle between their direction vectors, found from the scalar product. Taking the modulus of the dot product returns the acute angle, which is what is usually wanted.

Angle between a line and a plane

This is the subtle one. The natural calculation uses the line's direction and the plane's normal, but that gives the angle to the normal, so the angle to the plane is its complement, which is why a sine appears.

Angle between two planes

Two planes meet at an angle equal to the angle between their normals, found from the scalar product exactly as for two lines, taking the modulus for the acute angle.

Shortest distance from a point to a plane

The shortest distance from a point to a plane is measured along the normal. The standard formula substitutes the point into the plane expression and divides by the magnitude of the normal.

Shortest distance between skew lines

Two skew lines have a unique common perpendicular, and its length is the shortest distance. It is found by projecting the displacement between a point on each line onto the common normal direction d1×d2\mathbf{d}_1\times\mathbf{d}_2.

Distances and angles complete the vectors strand, combining the scalar product (angles and the point-plane distance) with the vector product (the skew-line distance and the plane normal).

Try this

Q1. Find the acute angle between lines with directions i+j\mathbf{i} + \mathbf{j} and i−j\mathbf{i} - \mathbf{j}. [2 marks]

  • Cue. cos⁡θ=∣1−1∣22=0\cos\theta = \dfrac{|1 - 1|}{\sqrt{2}\sqrt{2}} = 0, so θ=90∘\theta = 90^\circ.

Q2. Find the distance from the origin to the plane 2x+2y+z=92x + 2y + z = 9. [2 marks]

  • Cue. ∣0+0+0−9∣4+4+1=93=3\dfrac{|0 + 0 + 0 - 9|}{\sqrt{4 + 4 + 1}} = \dfrac{9}{3} = 3.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksFind the acute angle between the line with direction d=2i+j−2k\mathbf{d} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} and the plane with normal n=i+2j+2k\mathbf{n} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}, to the nearest degree.
Show worked answer →

The angle θ\theta between a line and a plane uses sin⁡θ=∣d⋅n∣∣d∣∣n∣\sin\theta = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} (M1).

Scalar product (M1): d⋅n=(2)(1)+(1)(2)+(−2)(2)=2+2−4=0\mathbf{d}\cdot\mathbf{n} = (2)(1) + (1)(2) + (-2)(2) = 2 + 2 - 4 = 0 (A1).

Since the scalar product is zero, sin⁡θ=0\sin\theta = 0, so θ=0∘\theta = 0^\circ: the line is parallel to the plane (it makes a zero angle with it) (A1, A1).

Markers reward the sin⁡θ\sin\theta formula (using the normal), the scalar product, and the interpretation that a zero value means the line is parallel to the plane.

OCR 20226 marksFind the shortest (perpendicular) distance from the point P(4,1,−2)P(4, 1, -2) to the plane 2x−y+2z=62x - y + 2z = 6.
Show worked answer →

Use the formula for the distance from a point to a plane ax+by+cz=dax + by + cz = d (M1): distance =∣ax0+by0+cz0−d∣a2+b2+c2= \dfrac{|a x_0 + b y_0 + c z_0 - d|}{\sqrt{a^2 + b^2 + c^2}}.

Substitute P(4,1,−2)P(4, 1, -2) and the plane (M1, A1): numerator =∣2(4)−(1)+2(−2)−6∣=∣8−1−4−6∣=∣−3∣=3= |2(4) - (1) + 2(-2) - 6| = |8 - 1 - 4 - 6| = |-3| = 3.

Denominator (A1): 22+(−1)2+22=9=3\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3.

Distance (A1, A1): 33=1\dfrac{3}{3} = 1.

Markers reward the correct formula, substituting the point and plane, the magnitude of the normal, and the final distance 11.

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