Applications: Planes and the Vector Product

The vector product of two edges of a triangle is a vector perpendicular to the triangle, and its magnitude is the area of the parallelogram those two edges make. The triangle is half of that parallelogram.

1. Take the two edges from corner A. AB = 4 − 00 − 04 − 1 = 403 and AC = 0 − 02 − 07 − 1 = 026.
2. Work out the vector product one component at a time. The first component is 0 × 6 − 3 × 2 = −6, the second is 3 × 0 − 4 × 6 = −24, and the third is 4 × 2 − 0 × 0 = 8.
3. (a) AB × AC = −6−248. Check: its dot product with AB is −24 + 0 + 24 = 0, and with AC it is 0 − 48 + 48 = 0, so it is perpendicular to the sail, as a vector product must be.
4. The magnitude of the vector product is the area of the parallelogram with sides AB and AC: √(−6)² + (−24)² + 8² = √36 + 576 + 64 = √676 = 26 square meters.
5. (b) The sail is half of that parallelogram, so its area is 12 × 26 = 13 square meters, and the fabric costs 13 × 15 = $195.

answer(a) −6−248; (b) 13 square meters, and the fabric costs $195

techniqueParallelogram Area from the Vector Product · The Vector Product

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A force that acts away from a pivot has a turning effect, its moment. As a vector, the moment is the vector product of the position of the point where the force acts with the force itself. Its direction is the axis about which the force tries to turn the sign, and its magnitude is the size of the turning effect.

1. The moment is r × F, with r = 301 first and F = 02010 second. The order matters: F × r is the opposite vector.
2. Work out the components. The first is 0 × 10 − 1 × 20 = −20, the second is 1 × 0 − 3 × 10 = −30, and the third is 3 × 20 − 0 × 0 = 60.
3. (a) The moment is −20−3060 N m. Check: r · (r × F) = −60 + 0 + 60 = 0 and F · (r × F) = 0 − 600 + 600 = 0, so the moment is perpendicular to both the bracket and the force.
4. The size of the turning effect is the magnitude of the moment: √(−20)² + (−30)² + 60² = √400 + 900 + 3600 = √4900 = 70 N m.
5. (b) 70 N m is less than 80 N m, so the bolt is strong enough, with 80 − 70 = 10 N m to spare.

answer(a) −20−3060 N m; (b) yes: the moment has magnitude 70 N m, which leaves 10 N m to spare

techniqueThe Vector Product

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Three points that are not in a line fix a plane. Its vector equation starts at one point and moves any amount along two directions in the plane. Its Cartesian equation needs a normal, a vector perpendicular to both directions, which is what the vector product gives. A fourth point lies in the plane exactly when it satisfies the equation.

1. Take two edges from corner A: AB = −240 and AC = 02−1. They are not parallel, so A, B and C are not in a line.
2. (a) Start at A and move λ along AB and μ along AC: r = 402 + λ−240 + μ02−1.
3. A normal is perpendicular to both edges, so take their vector product. The first component is 4 × (−1) − 0 × 2 = −4, the second is 0 × 0 − (−2) × (−1) = −2, and the third is (−2) × 2 − 4 × 0 = −4. So AB × AC = −4−2−4 = −2212, and n = 212 is a simpler normal along the same line.
4. The Cartesian equation is 2x + y + 2z = d, and A lies on it, so d = 8 + 0 + 4 = 12. The plane is 2x + y + 2z = 12. Check: B gives 4 + 4 + 4 = 12 and C gives 8 + 2 + 2 = 12.
5. (b) Substitute D: 2 × 2 + 6 + 2 × 1 = 4 + 6 + 2 = 12. D satisfies the equation, so all four corners lie in the plane 2x + y + 2z = 12 and the panel is flat. Check: CD = −240 = AB, so the panel is a parallelogram.

answer(a) r = 402 + λ−240 + μ02−1; (b) 2x + y + 2z = 12, and D lies on it, so the panel is flat

techniqueThe Cartesian Equation of a Plane · The Vector Equation of a Plane · The Vector Product

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The drone's path is a line, and the point where it reaches the hillside is on the line and in the plane at the same time. So substitute the line's coordinates, which depend on λ, into the plane's equation: that gives one equation in λ.

1. Read the coordinates off the line: after λ seconds the drone is at x = 2 + λ, y = 2λ and z = 7 − 2λ.
2. The drone reaches the hillside when these coordinates satisfy the plane's equation: 2(2 + λ) − 2λ + 2(7 − 2λ) = 10.
3. Expand the brackets: 4 + 2λ − 2λ + 14 − 4λ = 10, so 18 − 4λ = 10, 4λ = 8 and λ = 2.
4. (a) At λ = 2 the drone is at (2 + 2, 2 × 2, 7 − 4) = (4, 4, 3). Check: 2 × 4 − 4 + 2 × 3 = 8 − 4 + 6 = 10, so the point is on the hillside.
5. (b) The flight lasts 2 s. The drone's speed is the magnitude of its velocity, √1² + 2² + (−2)² = √9 = 3 m/s, so it flies 3 × 2 = 6 m. Check: from P to (4, 4, 3) is 24−4, of length √4 + 16 + 16 = √36 = 6 m.

answer(a) (4, 4, 3); (b) the flight lasts 2 s and the drone flies 6 m

techniqueThe Intersection of a Line and a Plane

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The shortest route from a point to a plane is along the normal, because any other route is the hypotenuse of a right-angled triangle with the normal route as one side. The distance is the value of ax + by + cz − d at the point, divided by the length of the normal. The foot of that route is where the line along the normal meets the plane.

1. Read the normal from the coefficients: n = −1−22, of length √1 + 4 + 4 = 3.
2. Substitute R into the left side of the equation: −3 − 2 × 3 + 2 × 1 = −7. The plane needs 2, so R is 2 − (−7) = 9 short of it.
3. (a) The distance is |−7 − 2|3 = 93 = 3 m, so the cable is 3 m long. Check by projection: A(0, 0, 1) is on the ceiling, since 0 − 0 + 2 = 2, and AR = 330. Its component along the unit normal is −3 − 6 + 03 = −3, a distance of 3 m.
4. The cable runs along the normal: r = 331 + λ−1−22. Substitute into the plane: −(3 − λ) − 2(3 − 2λ) + 2(1 + 2λ) = 2, which is −7 + 9λ = 2, so λ = 1.
5. (b) At λ = 1 the cable meets the ceiling at (3 − 1, 3 − 2, 1 + 2) = (2, 1, 3). Check: −2 − 2 + 6 = 2, and the cable is 1 × 3 = 3 m long, as in (a).

answer(a) 3 m; (b) at (2, 1, 3)

techniqueThe Distance from a Point to a Plane · The Intersection of a Line and a Plane

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The dot product gives the angle between a line and the NORMAL of a plane. The normal is at 90^° to the plane, so the angle between the line and the plane itself is 90^° minus that. In one step: sinθ = |d · n||d||n|.

1. Read the normals from the equations. The roof x + z = 3 has normal n = 101, and the ground z = 0 has normal 001.
2. For the roof: d · n = 0 × 1 + 1 × 0 + (−1) × 1 = −1, and |d| = |n| = √2.
3. The angle α between the ray and the normal has cosα = |−1|√2 × √2 = 12, so α = 60^°.
4. (a) The normal is at 90^° to the roof, so the ray meets the roof at 90^° − 60^° = 30^°. In one step, sinθ = 12 gives θ = 30^°.
5. (b) For the ground: d · 001 = −1, so sinθ = |−1|√2 × 1 = 1√2 and θ = 45^°: the sun is 45^° above the horizon. Check: the ray drops 1 m for every 1 m it travels across, and tan 45^° = 1.

answer(a) 30^°; (b) 45^°

techniqueThe Angle Between a Line and a Plane

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Two planes meet along a line, and the angle between them is the angle between their normals: turning one plane onto the other turns its normal by the same amount. The dot product of the normals gives its cosine. The horizontal is the plane z = 0, whose normal points straight up.

1. Read the normals from the coefficients: the roof has n₁ = 304 and the panel has n₂ = 0−34. Each has length √9 + 16 = 5.
2. Their dot product is n₁ · n₂ = 3 × 0 + 0 × (−3) + 4 × 4 = 16.
3. (a) cosθ = 165 × 5 = 0.64, so θ = cos⁻¹ 0.64 ≈ 50.2^°. The cosine is positive, so this angle is already acute, and it is the angle between the panel and the roof.
4. For the horizontal, use the normal 001: its dot product with n₂ is 4, and its length is 1.
5. (b) cosθ = 45 × 1 = 0.8, so the panel is tilted at cos⁻¹ 0.8 ≈ 36.9^° to the horizontal. Check: on the panel z = 3 + 34y, so it rises 3 m for every 4 m north, and tan⁻¹34 ≈ 36.9^°.

answer(a) 50.2^°; (b) 36.9^°

techniqueThe Angle Between Two Planes

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The hip is the line where two planes meet. It lies in both planes, so its direction is perpendicular to both normals, and the vector product of the normals gives that direction. One point on both planes then fixes the line.

1. The normals are n₁ = 102 and n₂ = 012. The hip lies in both faces, so its direction is perpendicular to both of them.
2. Take the vector product. The first component is 0 × 2 − 2 × 1 = −2, the second is 2 × 0 − 1 × 2 = −2, and the third is 1 × 1 − 0 × 0 = 1. So n₁ × n₂ = −2−21, and going down the roof the direction is 22−1.
3. P is on both faces, since 0 + 2 × 5 = 10 in each equation. So the hip is r = 005 + λ22−1.
4. (a) The coordinates are x = 2λ, y = 2λ and z = 5 − λ. Make λ the subject of each and set the three equal: x2 = y2 = z − 5−1.
5. The eaves are at z = 3, so 5 − λ = 3 and λ = 2. The hip ends at (4, 4, 3). Check: 4 + 2 × 3 = 10 in both equations.
6. (b) The rafter runs from (0, 0, 5) to (4, 4, 3), so its length is √4² + 4² + (−2)² = √36 = 6 m. Check: it is 2 steps of the direction vector, whose length is √4 + 4 + 1 = 3, and 2 × 3 = 6.

answer(a) x2 = y2 = z − 5−1; (b) 6 m, from P down to the corner of the eaves at (4, 4, 3)

techniqueThe Intersection of Two Planes · The Vector Product · The Cartesian Form of a Line

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Two lines in space either are parallel, meet at one point, or are skew: not parallel and never meeting. Setting their coordinates equal gives three equations in only two unknowns, λ and μ. Two of the equations fix λ and μ, and the third either agrees, so the lines meet, or does not, so they are skew.

1. No two directions are parallel: 120 has no multiple equal to 101 or to 021, because their zero components are in different places.
2. If A and B met, then 1 + λ = μ, 2λ = 4 and 2 = μ. The second equation gives λ = 2 and the third gives μ = 2.
3. (a) The first equation would then say 1 + 2 = 2, which is false. No values of λ and μ satisfy all three, so A and B never meet, and since they are not parallel they are skew. Check: where they cross as seen from above, at x = 3 and y = 4, pipe A is 2 m up and pipe B is 3 m up, so B passes 1 m above A.
4. If A and C met, then 1 + λ = 2, 2λ = 2μ and 2 = 1 + μ. The first gives λ = 1, the third gives μ = 1, and the second agrees: 2 = 2.
5. (b) All three equations hold, so A and C meet, at λ = 1 on A: (1 + 1, 2, 2) = (2, 2, 2). Check on C, with μ = 1: (2, 0 + 2, 1 + 1) = (2, 2, 2), the same point, where the joint must go.

answer(a) the directions are not parallel, and the three equations for a meeting point have no solution, so A and B are skew, with B passing 1 m above A; (b) A and C meet at (2, 2, 2)

techniqueIntersecting and Skew Lines

In the Cartesian form each fraction is the same number, the parameter λ, so setting each one equal to λ gives the coordinates of every point on the wire. The denominators are the components of the direction vector, and the numbers subtracted on top are the coordinates of a point on the line.

1. Set each fraction equal to λ: x = 6λ, y = 2λ and z = 6 − 3λ. In vector form the wire is r = 006 + λ62−3, starting from the platform at λ = 0.
2. The anchor is on the ground, where z = 0: 6 − 3λ = 0, so λ = 2, and the anchor is at (12, 4, 0).
3. (a) The direction vector has length √36 + 4 + 9 = √49 = 7, and the wire runs from λ = 0 to λ = 2, so it is 2 × 7 = 14 m long. Check: √12² + 4² + 6² = √196 = 14.
4. Above the post, x = 6 gives λ = 1, and then y = 2 × 1 = 2, which is the post's y as well, so the wire passes directly over the post.
5. (b) At λ = 1 the wire is at height z = 6 − 3 = 3 m. The top of the post is at 2 m, so the wire clears it by 3 − 2 = 1 m.

answer(a) the anchor is at (12, 4, 0) and the wire is 14 m long; (b) 1 m

techniqueThe Cartesian Form of a Line · The Intersection of a Line and a Plane

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