Pick your class, follow the steps in the lab, and check what you see against the working under the graph.
Class 11 — Coordinates in three dimensions, distance and section formula
Concept: a point in space has three coordinates (x, y, z). The distance between A and B is √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²), the midpoint averages the coordinates, and the section formula splits a segment in a ratio m : n. The three coordinate planes cut space into eight octants.
- Open Vectors in space and choose Distance & section formula.
- Change A and B and read the distance: try A = (0, 0, 0) and B = (2, 3, 6) to get 7.
- Set the ratio 1 : 1 and check that P is the midpoint M.
- Set m : n = 2 : 1, then tick divide externally and see P move outside the segment.
- Move a point through a coordinate plane and read which octant it is in.
Takeaway: the same distance and section formulas as in the plane, with one more coordinate.
Class 12 — Vectors: magnitude, direction and addition
Concept: a vector has magnitude and direction. The magnitude of xi + yj + zk is √(x² + y² + z²), the unit vector is a ÷ |a|, and the direction cosines l, m, n (the cosines of the angles with the axes) satisfy l² + m² + n² = 1. Vectors add component by component, by the triangle law or the parallelogram law.
- Open Vectors in the plane and drag a and b: the sum is the diagonal of the parallelogram.
- Tick also show a − b and press opposite vectors.
- Choose Scalar & unit vector and change k: |k·a| = |k||a|.
- Open Vectors in space → Components & direction and press 3i + 4j + 12k: |a| = 13.
- Read the direction cosines and check l² + m² + n² = 1.
Takeaway: a vector is its magnitude times its unit vector, and the direction cosines are the coordinates of that unit vector.
Class 12 — The dot product and projections
Concept: a · b = |a||b| cos θ = x₁x₂ + y₁y₂ + z₁z₂. It is a scalar, it is 0 for perpendicular vectors, and the projection of a on b is (a · b) ÷ |b|.
- Open Vectors in the plane → Dot product & projection and press perpendicular (a · b = 0).
- Press 45° apart and read the angle from cos θ = a · b ÷ (|a||b|).
- Drag b around a and watch the projection (teal) shrink to 0 and turn negative.
- Open Vectors in space → Dot product & angle and press perpendicular.
- Check the sign: a · b positive means an acute angle.
Takeaway: the dot product measures how much two vectors point the same way.
Class 12 — The cross product and the scalar triple product
Concept: a × b is perpendicular to both a and b, with length |a||b| sin θ, which is the area of the parallelogram on a and b. The scalar triple product [a b c] = a · (b × c) is the volume of the parallelepiped on a, b, c, and is 0 when the three vectors are coplanar.
- Open Vectors in space → Cross product and press i × j = k.
- Turn the scene to check that a × b is perpendicular to the parallelogram.
- Read the parallelogram area and the triangle area (half of it).
- Tick add a third vector c and press unit cube: the volume is 27 (for 3 × 3 × 3).
- Press parallel (a × b = 0) and see the area collapse to 0.
Takeaway: the cross product gives a perpendicular vector and an area; the triple product gives a volume.
Class 12 — The section formula for vectors
Concept: the point P dividing AB internally in the ratio m : n has position vector (m·b + n·a) ÷ (m + n); externally it is (m·b − n·a) ÷ (m − n). The midpoint is the case m = n.
- Open Vectors in the plane → Section formula and drag A and B.
- Set m = n and check that P is the midpoint.
- Set m : n = 3 : 1 and see P close to B.
- Tick divide externally and see P outside AB.
- Read AP : PB in the cards: it equals m ÷ n.
Takeaway: P = (mB + nA) ÷ (m + n): each end is weighted by the opposite part of the ratio.
Class 12 — Lines in space: equations, angle and shortest distance
Concept: a line is r = a + λb (a point and a direction), or in Cartesian form (x − x₁) ÷ a = (y − y₁) ÷ b = (z − z₁) ÷ c. Two lines in space either meet, are parallel, or are skew. The angle between them is the angle between their directions, and the shortest distance between skew lines is |(a₂ − a₁) · (b₁ × b₂)| ÷ |b₁ × b₂|.
- Open Lines & planes → Two lines and press skew lines: the shortest distance is 1 (the red dashed segment).
- Press intersecting at (1, 2, 3) and read the point of intersection.
- Press parallel lines and then the same line.
- Edit line 2 with the sliders until the distance becomes 0: the lines now meet.
- Press perpendicular and meeting and read the 90° angle.
Takeaway: lines meet when the shortest distance is 0; skew lines have a positive distance and non-parallel directions.
Class 12 — Planes: equation and distance from a point
Concept: the plane through a point with normal n is n · r = d, or ax + by + cz = d. The distance from a point Q to the plane is |n · Q − d| ÷ |n|, and the foot of the perpendicular lies along the normal.
- Open Point & plane and press x + y + z = 3 and Q(3, 2, 1).
- Read the distance |6 − 3| ÷ √3 and the foot F of the perpendicular.
- Press a point on the plane: the distance is 0.
- Change the normal and the value of d and see the plane tilt and move.
- Read the intercepts on the axes.
Takeaway: distance = |n · Q − d| ÷ |n|, along the normal.
Class 12 — Angles between a line and a plane, and between two planes
Concept: the angle between a line and a plane is 90° minus the angle between the line and the normal: sin θ = |n · b| ÷ (|n||b|). The angle between two planes is the angle between their normals: cos θ = |n₁ · n₂| ÷ (|n₁||n₂|). Two planes meet in a line with direction n₁ × n₂.
- Open Line & plane and press line meets the plane: read the angle and the point where it meets.
- Press line parallel to the plane and line inside the plane: n · b = 0 in both.
- Open Two planes and press planes at 45°.
- Press perpendicular planes and parallel planes.
- Read the line of intersection and check that its direction is perpendicular to both normals.
Takeaway: line and plane use the sine of the angle with the normal's complement; plane and plane use the normals' cosine.
Class 12 — The vector triple product and Lagrange's identity
Concept: a × (b × c) is a vector that lies in the plane of b and c, and a × (b × c) = b(a · c) − c(a · b). The scalar triple product [a b c] = a · (b × c) is the volume of the box on a, b, c.
- Open Vectors in space → Triple products and press the first preset: a = (1, 2, 3), b = (2, −1, 1), c = (0, 1, 2). Then b × c = (−3, −4, 2) and a × (b × c) = (16, −11, 2).
- Check with the rule: a · c = 8 and a · b = 3, so b(a · c) − c(a · b) = 8(2, −1, 1) − 3(0, 1, 2) = (16, −11, 2). The two vectors are equal (the lab shows ✓).
- Not associative: (a × b) × c = (15, −10, 5) is a different vector. Its rule is b(a · c) − a(b · c) = 8(2, −1, 1) − 1(1, 2, 3) = (15, −10, 5). The first lies in the plane of b and c, the second in the plane of a and b.
- Scalar triple product: [a b c] = −5 here. It does not change when you rotate the order (a, b, c → b, c, a → c, a, b) and changes sign when two vectors swap. The tetrahedron on a, b, c has volume 5/6. Press the third preset (a = (2, 1, 3), b = (1, 1, 1), c = (3, 2, 4)): [a b c] = 0, so the vectors are coplanar.
- Lagrange's identity: |a × b|² = |a|²|b|² − (a · b)². For a = (1, 2, 3), b = (2, −1, 1): a × b = (5, 5, −5), 75 = 14 × 6 − 3² = 84 − 9.
Takeaway: “BAC − CAB”: a × (b × c) = b(a · c) − c(a · b). Coplanar vectors have [a b c] = 0.
Class 12 — Point and line, plane through three points, family of planes, common perpendicular
Concept: the standard 3D questions each reduce to a cross product or a dot product. Open Lines & planes.
- Distance of a point from a line (view Point & line): the line r = (−2, −1, 0) + λ(1, 2, 2) and Q(2, 3, 1). Then (Q − a) × b = (4, 4, 1) × (1, 2, 2) = (6, −7, 4), whose length is √101 = 10.05, and |b| = 3, so the distance is 10.05 ÷ 3 = 3.35. The foot of the perpendicular is at λ = (Q − a) · b ÷ |b|² = 14/9, F = (−0.44, 2.11, 3.11), and the image of Q is 2F − Q = (−2.89, 1.22, 5.22).
- Plane through three points (view Three points): A(1, 0, 0), B(0, 2, 0), C(0, 0, 3). n = AB × AC = (−1, 2, 0) × (−1, 0, 3) = (6, 3, 2), so the plane is 6x + 3y + 2z = 6, in intercept form x/1 + y/2 + z/3 = 1. The area of the triangle is ½|n| = 3.5. For D(1, 1, 1) the distance is 5/7 = 0.714, [AB AC AD] = 5 and the tetrahedron has volume 5/6. Press the second preset to put D on the plane: the distance is 0 and the four points are coplanar.
- Family of planes (view Family of planes): P₁: x + y + z = 3 and P₂: x − y = 1. Every plane through their line of intersection is (x + y + z − 3) + λ(x − y − 1) = 0. Through the point (2, 0, 0): −1 + λ(1) = 0, λ = 1, the plane 2x + z = 4. Through the origin: −3 + λ(−1) = 0, λ = −3, the plane −2x + 4y + z = 0. You do not need to find the line first.
- Shortest distance between skew lines (view Two lines, the opening example): r = (−3, −2, −1) + λ(2, 1, 1) and r = (1, −3, 2) + μ(0, 1, 1). b₁ × b₂ = (0, −2, 2), (a₂ − a₁) · (b₁ × b₂) = (4, −1, 3) · (0, −2, 2) = 8, so d = 8 ÷ √8 = 2.828. The lab now also gives the feet of the common perpendicular, (1, 0, 1) and (1, −2, 3), whose direction is b₁ × b₂.
- Image of a point in a plane (view Point & plane): the foot of Q(3, 2, 1) on x + y + z = 3 is (2, 1, 0), so the image is 2F − Q = (1, 0, −1).
Takeaway: distance from a point to a line = |(Q − a) × b| ÷ |b|; plane through three points has normal AB × AC; family of planes P₁ + λP₂ = 0; skew-line distance = |(a₂ − a₁) · (b₁ × b₂)| ÷ |b₁ × b₂|.