Represent and intersect spatial lines and planes

Build on the vector and system foundations

Use spatial components, independence and ordered normals together with elimination and parameterised algebra. A line is one position plus one nonzero direction; a plane is one position plus two independent directions

A line through two distinct points

For position vectors a,b, use r=a+t(b-a), t real. At t=0 the point is A; at t=1 it is B. The unrestricted parameter describes the whole line; 0≤t≤1 describes only the segment. If A=B, that formula gives one point and does not determine a unique line

Work a spatial line equation

For A=(1,-2,3), B=(3,-1,1), the direction B-A=(2,1,-2). Thus x=1+2t, y=-2+t, z=3-2t. At t=2 the point is (5,0,-1). The direction is not a position vector

Line

\(\displaystyle x=1+2t,\quad y=-2+t,\quad z=3-2t\)

What is the coefficient of t in y?

The two-dimensional case uses the same rule

For A=(3,-1) and B=(3,1), r=(3,-1)+t(0,2) gives the vertical line x=3. A zero x-component in its direction is allowed because the direction vector itself is nonzero. No finite slope y=mx+c describes this vertical line

Eliminate a parameter without dividing by zero

For x=2, y=1+t, z=3-2t, use t=y-1 to obtain 2y+z=5 together with x=2. These two independent Cartesian equations describe the line. A symmetric form is valid only for nonzero denominators; retain fixed-coordinate equations separately

Line

\(\displaystyle x=2,\quad y=1+t,\quad z=3-2t\)

Which Cartesian description retains the complete line?

Fix a parameter convention when comparing answers

For y=1+2(x-3), z=4-(x-3), choose t=x. Then r=(0,-5,7)+t(1,2,-1). Choosing t=x-3 instead gives r=(3,1,4)+t(1,2,-1), the same set with a different parameter origin

Equation and parameter

\(\displaystyle z=4-(x-3),\quad t=x\)

Find the constant z-coordinate

A plane normal supplies one scalar equation

With a nonzero normal n and a point A of position a, n·(r-a)=0 describes the plane. Equivalently n·r=n·a=d. Its Cartesian coefficients are the normal components. Scaling both n and d by the same nonzero number leaves the plane unchanged

Use two edges with a common starting point

For A=(1,-1,2), B=(3,0,2), C=(0,1,3), form AB=(2,1,0) and AC=(-1,2,1). Their ordered product is (1,-2,5). It is nonzero, so the points are noncollinear. The normal is perpendicular to both edges

Continue the point-normal example

The preceding ordered normal is n=(1,-2,5). Through A=(1,-1,2), the constant is n·a=1+2+10=13. The plane is x-2y+5z=13. Both original edge vectors are perpendicular to n, so all three original points lie in it

Plane data

\(\displaystyle \mathbf n=(1,-2,5),\quad A=(1,-1,2),\quad\mathbf n\cdot\mathbf r=d\)

Find d

Two independent plane directions give two parameters

For that plane, r=(1,-1,2)+s(2,1,0)+t(-1,2,1) describes every point. The directions have cross product (1,-2,5)≠0. Parallel directions would describe a line, and zero directions cannot supply the missing dimension

Convert a Cartesian plane to a parameterisation

For 2x-y+z=4, choose x=s,y=t. Then z=4-2s+t, so r=(0,0,4)+s(1,0,-2)+t(0,1,1). Substitute back to check that the constant is 4 and both parameter coefficients cancel

Plane and parameters

\(\displaystyle 2x-y+z=4,\quad x=s,\quad y=t\)

What is the coefficient of s in z?

A line meets a plane by satisfying one parameter equation

Substitute r=a+td into n·r=c. This gives (n·d)t=c-n·a. If n·d≠0, the unique t gives one intersection. If n·d=0, compare the right side: nonzero means parallel and disjoint; zero means the entire line lies in the plane

Work an exact line-plane intersection

For r=(1,0,2)+t(1,2,-1) and x+y+z=6, substitution gives 3+2t=6. Hence t=3/2 and P=(5/2,3,1/2). All three coordinates use the same t, and their sum is 6.

Line and plane

\(\displaystyle \mathbf r=(1,0,2)+t(1,2,-1),\quad x+y+z=6\)

Give the exact intersection parameter t

Separate a parallel line from a contained line

For r=(1,0,2)+t(1,-1,0), the coordinate sum is always 3. The line is contained in x+y+z=3 but is parallel and disjoint from x+y+z=4. A direction perpendicular to the normal establishes parallelism; the position decides containment

Use independent parameters for two lines

Set a+su=b+tv and solve for s,t. With nonparallel directions, two coordinates give at most one candidate pair; check the third. If no pair satisfies all three, the nonparallel spatial lines are skew. Equal s and t is not required unless both parameters were explicitly defined as the same time

Work two intersecting lines with different parameter values

For r=(1,0,2)+s(1,1,0) and r=(3,1,1)+t(0,1,1), the x equation gives s=2 and the z equation gives t=1. The y coordinates then both equal 2. The common point is (3,2,2), although the parameter values differ

Second line and common point

\(\displaystyle \mathbf r=(3,1,1)+t(0,1,1),\quad P=(3,2,2)\)

What is the parameter t at the common point?

Check the coordinate that was not used

The lines r=(s,s,0) and r=(1,t,1) have a candidate s=t=1 from x and y, but z=0 and z=1 disagree. They are skew. If the second line were r=(1,t,0), that same pair would give an intersection. A flat projection can hide the z discrepancy

Parallel directions need one more point test

If v=ku for a nonzero k, the lines are either parallel distinct or coincident. Check whether b-a is a multiple of u. For r=(1,0,2)+su and r=(1,0,2)+3u+t(-2u), the same unrestricted line is described. Restricted segments would need their parameter intervals checked too

Two planes with independent normals meet in a line

For n1·r=d1 and n2·r=d2, a common direction is n1×n2 when it is nonzero. Find one point satisfying both equations, then combine the point and direction. If the normals are parallel, compare the constants with the same multiplier to distinguish coincident and parallel distinct planes

Work a plane-pair intersection

From x+y+z=3 and x-y+z=1, subtraction gives y=1 and then x+z=2. Set z=t: r=(2,1,0)+t(-1,0,1). Substitution satisfies both original equations for every t.

Planes

\(\displaystyle x+y+z=3,\quad x-y+z=1\)

What does the marked point in this z=0 section represent?

Planes and parameter

\(\displaystyle x+y+z=3,\quad x-y+z=1,\quad z=t\)

Find the coefficient of t in x

Three equations can leave a point, line, plane or no solution

In three variables, count independent constraints and test consistency. Three independent equations give one point. Two independent consistent constraints give a line; one gives a plane. A row 0=nonzero makes the common set empty. A zero determinant alone does not distinguish infinite solutions from none

Work the geometric alternatives

The planes x+y+z=3, x-y+z=1 and z=1 meet at (1,1,1). Replacing the third by 2x+2z=4 leaves the same line as the first two because it is their sum. The three planes x=0, y=0, x+y=1 have pairwise intersections but no common point. Three consistent nonzero multiples of one equation describe one plane

Equations

\(\displaystyle x+y+z=3,\quad x-y+z=1,\quad 2x+2z=4\)

What is the dimension of the common solution set?

Angles come from directions and normals

For two planes, the smaller angle equals the smaller angle between their nonzero normals. For a line and a plane, use the line's direction and the plane's normal. The angle to the plane is complementary to the smaller angle to its normal

Use exact squared ratios without rounding

For u=(1,1,0) and n=(0,1,1), the absolute dot product is 1 and both norms are √2. If these are plane normals, cosθ=1/2 and θ=60°. If u is a line direction and n a plane normal, sinφ=1/2 and φ=30°. In either requested squared-ratio calculation the answer is 1/4.

Vectors

\(\displaystyle \mathbf u=(1,1,0),\quad\mathbf n=(0,1,1)\)

Squared ratio

\(\displaystyle \frac{(\mathbf u\cdot\mathbf n)^2}{\lVert\mathbf u\rVert^2\lVert\mathbf n\rVert^2}\)

Enter the exact squared ratio

Prove equivalent line parameterisations, ungraded

Let d≠0. Prove that r=a+td and r=b+su describe the same unrestricted line when b=a+αd and u=βd with β≠0. Explain how restricted parameter intervals alter the conclusion

Model: exhibit the parameter correspondence

Substitution gives b+su=a+(α+βs)d. As s ranges over all real numbers, t=α+βs also ranges over all real numbers because β≠0. Conversely s=(t-α)/β recovers every point of the first line. For an interval in s, map both endpoints and account for reversal when β<0; the resulting segment need not equal the other restricted segment

Derive equivalent plane descriptions, ungraded

For independent u,v and a point a, show that r=a+su+tv is equivalent to n·(r-a)=0 with n=u×v. Explain why a nonzero normal and independent directions are required

Model: prove both inclusions

Because n is perpendicular to u and v, every parameterised point satisfies the scalar equation. Independence makes n nonzero; u,v,n form a basis because a combination αu+βv+γn=0 dotted with n gives γ=0, then α=β=0. Expand any r-a in that basis. The scalar plane equation forces its n coefficient to vanish, leaving su+tv. Dependent directions would have n=0 and the scalar equation 0=0 would describe all space, not the intended plane

Classify a spatial line pair completely, ungraded

For r=(s,s,0) and r=(1,t,k), classify the pair for all real k and state any intersection point and parameters. Explain why equating only x and y is insufficient

Model: keep the third-coordinate consistency condition

The directions (1,1,0) and (0,1,0) are independent. The first two equations require s=t=1. The third is 0=k. Thus k=0 gives the unique intersection (1,1,0); every k≠0 gives skew lines. The lines are never parallel or coincident. A two-dimensional projection would hide the failing condition

Interpret three plane equations geometrically, ungraded

Compare the systems x=0,y=0,z=2; x=0,y=0,x+y=0; x=0,y=0,x+y=1; and x=0,2x=0,-3x=0. Give each complete solution set and justify its dimension or inconsistency without relying on a sketch

Model: independent rows, redundant rows and contradictions

The first system gives the point (0,0,2). The second gives the z-axis {(0,0,t):t real}, because the third equation is redundant. The third gives no solution: its first two equations force x+y=0. The fourth gives the whole plane x=0 with two free coordinates. These are respectively dimension 0, dimension 1, empty, and dimension 2.

Construct and audit a plane-intersection diagram, ungraded

For x+y+z=3 and x-y+z=1, derive the full intersection line and draw labelled spatial and z=0 section sketches. Use a graphing tool if available, record what you actually entered, and explain what the fixed section can and cannot show

Model: verify the line before interpreting the view

The line is r=(2,1,0)+t(-1,0,1). It satisfies both equations identically. The z=0 traces meet at (2,1,0), only the t=0 point of that line. A perspective sketch may make nonintersecting spatial lines appear to cross; exact equation checks settle intersection. Reading this supplied diagram/model is not evidence that the learner constructed a spatial sketch or operated a tool

Assessment boundary

Complete equivalence proofs, unrestricted algebraic classifications, independent spatial sketches and tool use remain ungraded