Module 33: Vectors, lines, planes, and surfaces in space

8 exercises, each with a complete solution. 1 figure. 1 note from the engine.

On this page

A. Prerequisites and observable goals
B. The central idea: direction and magnitude become one object
C. Vector operations and spatial equations
D. Why products measure geometry
E. Recognize the spatial object
F. Worked examples
Computing bridge: Coordinates belong to a chosen basis
G. Independent practice
H. Complete solutions: Module 33
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Coordinate geometry, square roots, trigonometry, and systems of linear equations.

After

Use vector arithmetic and products, construct lines and planes, compute geometric distances and areas, and identify basic quadric surfaces from traces.

B. The central idea: direction and magnitude become one object#

A point names a location; a vector names a displacement or another directed quantity. The vector from $P=(p_1,p_2,p_3)$ to $Q=(q_1,q_2,q_3)$ is $Q-P=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle$ . A vector can be translated without changing its components; a point cannot. Scalars, vectors, and equations of sets are different mathematical objects and require different operations.

A vector is a displacement. Shown in the plane: from

P=(1,1)

Q=(4,3)

the displacement is

Q-P=\langle3,2\rangle

3

across and

2

up. The grey arrow starts somewhere else and is the same vector, because it has the same components. A point cannot be moved this way.

C. Vector operations and spatial equations#

For $\vv a=\langle a_1,a_2,a_3\rangle$ , length is $\|\vv a\|=\sqrt{a_1^2+a_2^2+a_3^2}$ . A nonzero vector’s unit direction is $\vv a/\|\vv a\|$ . The dot product is the scalar

\vv a\cdot\vv b=a_1b_1+a_2b_2+a_3b_3=\|\vv a\|\|\vv b\|\cos\theta.

Nonzero vectors are perpendicular exactly when their dot product is zero. Projection onto a nonzero $\vv b$ is $\operatorname{proj}_{\vv b}\vv a=(\vv a\cdot\vv b)\vv b/\|\vv b\|^2$ .

The three-dimensional cross product is the vector

\vv a\times\vv b=\langle a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1\rangle.

It is perpendicular to both factors, has magnitude $\|\vv a\|\|\vv b\||\sin\theta|$ , and follows right-hand orientation. Swapping the factors reverses its sign. Its magnitude is parallelogram area; half is triangle area. A scalar triple product’s absolute value gives parallelepiped volume.

A line through $P$ in nonzero direction $\vv v$ is $\vv r(t)=P+t\vv v$ . A plane through $P$ with nonzero normal $\vv n$ is $\vv n\cdot(\vv r-P)=0$ . Point-to-plane distance for $Ax+By+Cz=D$ is $|Ax_0+By_0+Cz_0-D|/\sqrt{A^2+B^2+C^2}$ . Geometry and vector identities: [O3Vectors].

D. Why products measure geometry#

The dot product extracts the component of one vector along another; multiplying that projected length by the other’s length gives $\|\vv a\|\|\vv b\|\cos\theta$ . The projection formula scales the target direction by exactly this component. A plane’s normal dot every in-plane displacement is zero, explaining its equation.

The cross product’s magnitude is base times perpendicular height, the parallelogram area. Its direction provides an oriented normal. This is the same normal construction used later for tangent planes and surface flux; a sign error today becomes an orientation error in vector calculus.

E. Recognize the spatial object#

One equation in three coordinates usually describes a surface, not a curve. Two independent plane equations usually intersect in a line. A parametric vector equation in one parameter usually describes a curve. To recognize a surface, hold one coordinate fixed and examine the resulting traces. For example, $z=x^2+y^2$ has circular horizontal traces and parabolic vertical traces; $z=x^2-y^2$ has a saddle. A missing coordinate indicates a cylinder extending along that coordinate direction, not necessarily a circular cylinder.

F. Worked examples#

A plane through three points. Let $P=(1,0,0)$ , $Q=(0,1,0)$ , $R=(0,0,2)$ . In-plane vectors are $Q-P=\langle-1,1,0\rangle$ and $R-P=\langle-1,0,2\rangle$ . Their cross product is $\langle2,2,1\rangle$ , so the plane is $2(x-1)+2y+z=0$ , or $2x+2y+z=2$ . Substituting all three points checks the result.

A projection. For $\vv a=\langle3,1,0\rangle$ and $\vv b=\langle1,1,0\rangle$ , the projection is $(4/2)\vv b=\langle2,2,0\rangle$ . The leftover $\langle1,-1,0\rangle$ has dot product zero with $\vv b$ , confirming the decomposition into parallel and perpendicular parts.

A quadric trace. The surface $x^2+y^2-z^2=1$ has circles of radius $\sqrt{1+z^2}$ in horizontal planes, never empty. It is a one-sheet hyperboloid. In contrast, $z^2-x^2-y^2=1$ requires $|z|\ge1$ and has two separated sheets.

Computing bridge: Coordinates belong to a chosen basis#

A vector is not intrinsically tied to one list of coordinates. A game camera, a world frame, and an object’s local frame can assign different coordinate lists to the same geometric direction. A01 explains matrices as maps between such representations. When applying a change of frame, identify which coordinates are input, which are output, and whether you are moving an object or merely redescribing it. The order of matrix multiplication follows that choice and generally cannot be reversed.

From the engine: ZeroVectorAn echo is a mirror image

A gunshot in ZeroVector reaches the listener twice: directly, and after bouncing off the face of a building. The engine finds the bounce with this module’s plane geometry. The wall is a plane through a point $q$ with unit normal $n$ ; a point $p$ lies at signed distance $d = (p - q)\cdot n$ from it, and its mirror image is $p' = p - 2dn$ .

ZeroVector: first reflections of a soundC++

const f32 sourceSide = glm::dot(source - facade.point, normal);
const f32 earSide = glm::dot(ear - facade.point, normal);
if (!(sourceSide > 0.5f) || !(earSide > 0.5f)) continue;
const vec3 mirrored = source - 2.f * sourceSide * normal;
const f32 along = sourceSide / (sourceSide + earSide);
const vec3 point = mirrored + along * (ear - mirrored);
if (std::abs(glm::dot(point - facade.point, tangent)) > facade.halfWidth) continue;
if (point.y < facade.bottom || point.y > facade.top) continue;
const f32 path = glm::length(ear - mirrored);
const f32 extra = (path - directMeters) / kSpeedOfSound;

The straight line from the mirrored source to the ear crosses the wall at the fraction $d_s/(d_s + d_e)$ of the way along, where the two distances are the source’s and the ear’s, and its length equals the length of the bounced path. Put the wall on the plane $x = 0$ , the source 20 m out, and the ear 30 m out and 40 m further along. The image sits 20 m behind the wall, the bounced path is $\sqrt{50^2 + 40^2} \approx 64.03$ m against a direct 41.23 m, the bounce point is 16 m along the wall, and at 343 m/s the echo arrives 66.5 ms after the direct sound. The two guards in the middle reject a bounce point that falls off the edge of the wall.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M33.01FSolution

For $P=(1,-2,3)$ and $Q=(3,1,-1)$ , find $\overrightarrow{PQ}$ , its length, and a unit vector in that direction.

M33.02CSolution

Find the angle between $\vv a=\langle1,1,0\rangle$ and $\vv b=\langle1,0,1\rangle$ , and project $\vv a$ onto $\vv b$ .

M33.03CSolution

Find $\vv a\times\vv b$ for $\vv a=\langle1,2,0\rangle$ , $\vv b=\langle0,1,3\rangle$ , and the area of the triangle they span.

M33.04CSolution

Find the line through $(2,-1,0)$ parallel to $\langle1,3,-2\rangle$ . Determine where it intersects the plane $z=4$ .

M33.05TSolution

Find a plane through $(1,2,-1)$ with normal $\langle2,-1,3\rangle$ , and the distance from the origin to this plane.

M33.06TSolution

Find a parametric equation for the intersection of $x+y+z=2$ and $x-y=0$ .

M33.07CSolution

Identify each surface using traces: $x^2+y^2+z^2=9$ , $z=x^2+y^2$ , and $y=x^2$ in three-dimensional space.

M33.08SSolution

Decide whether the lines $\vv r_1(t)=\langle t,0,0\rangle$ and $\vv r_2(s)=\langle0,1,s\rangle$ are parallel, intersecting, or skew. Find their minimum distance.

Optional small hints

For M33.05, use the point-normal form before expanding. For M33.06, set $x=y=t$ . For M33.08, write the squared distance between two arbitrary points on the lines.

H. Complete solutions: Module 33#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M33.01Show solution

Subtract endpoint coordinates in the requested direction

The displacement is $Q-P=\boxed{\langle2,3,-4\rangle}$ . Its length is $\sqrt{4+9+16}=\boxed{\sqrt{29}}$ . Dividing by that nonzero length gives unit vector $\boxed{\langle2,3,-4\rangle/\sqrt{29}}$ . Its squared component sum is one, checking normalization.

Back to M33.01

M33.02Show solution

Dot product gives both angle and projection

The dot product is $1$ , and both lengths are $\sqrt2$ . Thus $\cos\theta=1/2$ , so $\theta=\boxed{\pi/3}$ . Projection is $(1/2)\langle1,0,1\rangle=\boxed{\langle1/2,0,1/2\rangle}$ . Subtracting it from $\vv a$ gives a vector perpendicular to $\vv b$ , as a dot-product check confirms.

Back to M33.02

M33.03Show solution

Cross product supplies an oriented normal and an area

The component formula gives $\vv a\times\vv b=\boxed{\langle6,-3,1\rangle}$ . Its dot products with $\vv a$ and $\vv b$ are zero. Its length is $\sqrt{36+9+1}=\sqrt{46}$ , so the triangle area is $\boxed{\sqrt{46}/2}$ square units. The cross product itself is a vector, not the scalar area.

Back to M33.03

M33.04Show solution

Use one shared parameter for all coordinates

The line is $\boxed{\vv r(t)=\langle2+t,-1+3t,-2t\rangle}$ . The condition $z=4$ gives $-2t=4$ , so $t=-2$ . The intersection point is $\boxed{(0,-7,4)}$ . Substituting a different parameter into each coordinate would no longer describe the same point on the line.

Back to M33.04

M33.05Show solution

Normal dot displacement gives the plane

The equation is $2(x-1)-(y-2)+3(z+1)=0$ , or $\boxed{2x-y+3z=-3}$ . The origin-to-plane distance is $|0-(-3)|/\sqrt{4+1+9}=\boxed{3/\sqrt{14}}$ . The numerator is an algebraic offset; dividing by normal length converts it to geometric distance.

Back to M33.05

M33.06Show solution

Two constraints leave one free parameter

The second plane gives $x=y=t$ . Substituting into the first gives $z=2-2t$ . Hence the intersection is $\boxed{\vv r(t)=\langle t,t,2-2t\rangle}$ , $t\in\R$ . Direct substitution satisfies both planes. Its direction $\langle1,1,-2\rangle$ is perpendicular to both plane normals.

Back to M33.06

M33.07Show solution

Missing coordinates and trace shapes classify the surfaces

The first is a sphere centered at the origin with radius $3$ . The second is an upward circular paraboloid: horizontal traces are circles for $z\ge0$ , and vertical traces are parabolas. The third is a parabolic cylinder, because every fixed $z$ slice has the same parabola $y=x^2$ and $z$ is unrestricted. Calling the third merely a planar parabola would ignore the free spatial coordinate.

Back to M33.07

M33.08Show solution

Nonparallel lines need not meet in space

Their directions $\langle1,0,0\rangle$ and $\langle0,0,1\rangle$ are not parallel. Every point on the first has $y=0$ and every point on the second has $y=1$ , so they never intersect: they are skew. Squared distance between their points is $t^2+1+s^2\ge1$ , with equality at $t=s=0$ . The minimum distance is therefore $\boxed1$ , realized along the segment from $(0,0,0)$ to $(0,1,0)$ , perpendicular to both directions.

Back to M33.08

I. Retention and readiness#

Retrieve M31.06 and M04.05. Explain whether a dot product, cross product, norm, or set of equations is the requested object. Advance when you can verify a proposed normal by dot products and a proposed point by substitution.