Planes and the Vector Product

The Vector Product #

A third vector, perpendicular to both.

The vector product of two vectors is a new vector at right angles to both

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The answer is a vector, not a number: each place crosses the other two and subtracts.

Both arrows lie flat with no depth, and the answer is pure depth: (0, 0, 5).

Dot the answer with a, then with b. Both come out zero: perpendicular to each.

Swap the order and every sign flips: b × a points exactly opposite to a × b.

A vector crossed with itself is zero: two copies of one arrow span nothing.

It distributes over addition: a × (b + c) = a × b + a × c, checked here.

Expanding Products of Vectors #

Both products open brackets; the vector product keeps the order.

Both products distribute over addition, so a product of sums expands like brackets, keeping the order for ×

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The dot product distributes over addition too: a · (b + c) = a · b + a · c, checked here.

Expand (a + b) · (a + b) like brackets of numbers. a · a is $|a{|}^2$|a|², and b · a = a · b, so the middle terms add.

So |a + b| comes from the two lengths and a · b alone, with no components needed.

With the vector product the order stays as written: b × a = −(a × b) and a × a = 0, so the middle terms add.

Parallelogram Area from the Vector Product #

Its length is the sheet the two arrows span.

The length of a vector product is the area of the parallelogram its two vectors span

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Two vectors from one corner span a parallelogram: base |a|, height $|b|\sin \theta$|b| sin θ.

Base times perpendicular height is $|a||b|\sin \theta$|a| |b| sin θ — the length of a × b exactly.

The diagonal cuts it into two matching triangles, each one half of the area.

A vector product of length 12 gives a parallelogram of 12 and a triangle of 6.

The Vector Equation of a Plane #

A point and two directions inside it.

A point and two directions inside a plane name every point of it as $r=a+\lambda b+\mu c$r = a + λb + μc

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A plane is a flat sheet, endless, with two different directions running inside.

One counter walks a line. A second direction and counter fill a whole plane.

$r=a+\lambda b+\mu c$r = a + λb + μc: reach the plane at a, then walk $\lambda$λ of b and $\mu$μ of c.

Pick the two numbers and out comes a point. Every point has its own pair.

The two directions must not be multiples, or the sweep collapses back to a line.

The Cartesian Equation of a Plane #

One normal, and out falls ax + by + cz = d.

Every point of a plane has the same dot product with the plane’s normal

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A normal n stands at right angles to every direction that lies inside the plane.

Dot the plane equation with n and both counters vanish: r · n = a · n.

Call that fixed number d. Then r · n = d holds at every point of the plane.

Write r as (x, y, z) and the dot product opens out into 2x + 3y + z = 7.

It runs backwards too: the coefficients of x, y and z are the normal itself.

The Angle Between a Line and a Plane #

Sine, not cosine — the normal is not the plane.

The angle between a line and a plane is the complement of its angle to the normal

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A normal stands at right angles to its plane, so the two angles always add to 90°.

Complementary angles swap cosine for sine, so the same quotient gives $\sin \theta$sin θ.

The common mistake is answering with $\cos \theta$cos θ — that is the angle to the normal.

A line of direction (1, 2, 2) meets the floor: $\sin \theta =\frac{2}{3}$sin θ = 2/3, so $\theta$θ is near 42°.

Reverse the line and the dot product turns negative, but the angle does not.

The Angle Between Two Planes #

Two normals carry the whole answer.

The angle between two planes is the angle between their normals

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Two planes tilt apart by exactly the angle their two normals tilt apart.

Read a normal off each equation: the coefficients already are the vector.

Dot the two normals, then divide by both lengths: $\cos \theta$cos θ comes to 8 ÷ 9.

Multiples mean parallel planes; a zero dot product means they meet at right angles.

Flip one normal and the cosine turns negative; its size gives the acute angle.

The Intersection of a Line and a Plane #

Substitute the line, then solve for one t.

Substituting a line into a plane equation leaves one equation in t

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Every point of the line carries one t. Just one of them lands on the plane.

Put the line’s x, y and z into the plane and it collapses to one equation.

Solve for t, then put it back into the line to find the meeting point.

When the t terms cancel, d · n is zero and the line runs along the plane.

The Distance from a Point to a Plane #

Drop a perpendicular along the normal.

The shortest distance from a point to a plane is measured along the normal

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Any other route to the plane is a hypotenuse, so the perpendicular is shortest.

Take the gap from a point of the plane up to P, then measure its part along n.

Put P into the equation, subtract the 6, then divide by 3: the distance is 3.

Walk back from P along the normal by that distance and you land at the foot.

The origin gives a negative reading, and its size is still the true distance.

The Distance from a Point to a Line #

Drop a perpendicular to a line, and measure it with the vector product.

The distance from a point to a line is |AP$|\sin \theta$| sin θ, which is the length of AP × d divided by |d|

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The distance from P to the line is the perpendicular PF. Any other route to the line is a hypotenuse, so it is longer.

AP is the hypotenuse of a right triangle with PF as the leg opposite $\theta$θ, the angle between AP and d, so PF = |AP$|\sin \theta$| sin θ.

The vector product already holds |AP$|\sin \theta$| sin θ, multiplied by |d|, so divide its length by |d|.

For the line through (1, 0, 0) along (2, 1, 2) and P = (1, 3, 3), |AP × d| = 9 and |d| = 3, so P is 3 away.

The foot is F = A + t d with t = AP $\cdot d÷|d{|}^2$· d ÷ |d|². Here t = 1, so F = (3, 1, 2), and PF · d = 0 checks the right angle.

The Cartesian Form of a Line #

Make t the subject and the parameter vanishes.

Making t the subject of each component turns r = a + t d into two equations with no parameter

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Read r = a + t d row by row: three equations sharing one t.

Make t the subject of each row. All three expressions equal the same number.

So set them equal. That is the Cartesian form, with the parameter gone.

Read it back the same way: numerators give the point, denominators the direction.

A zero in the direction freezes that coordinate, so it is written on its own.

Intersecting and Skew Lines #

Three equations, two unknowns, and one test.

Two lines in space meet only if the t and s that fit two components also fit the third

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In space, two lines can miss without being parallel. AB and CG never meet: they are skew.

Two lines meet only if one point sits on both, so set the two positions equal.

Equating components gives three equations for two unknowns — one more than needed.

Solve any two of them. The x and y rows give t = 2 and s = −2.

The spare equation is the test. It holds, so the lines really do meet at (3, 4, 2).

Move one line and the test fails: no point is on both, and they are skew.

The Intersection of Two Planes #

The crease runs perpendicular to both normals.

Two non-parallel planes meet along a line whose direction is perpendicular to both normals

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Two planes that are not parallel share a whole line, never a single point.

The line is perpendicular to both normals, so the cross product gives its direction.

For one point, fix a coordinate — set z = 0 and solve the two equations left.

A point and a direction make a line: $r=\left(-2,8,0\right)+\lambda \left(1,-2,1\right)$r = (−2, 8, 0) + λ(1, −2, 1).

Parallel normals share no line: the planes are the same one twice, or they never meet.