Dawson College · 201-SN4-RE

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Linear Algebra and Vector Geometry: one-to-one online tutoring and exam preparation.

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Course overview

Dawson 201-SN4-RE links linear systems and matrix calculations to vector geometry and basic mathematical arguments.

Start here

Try four systems and vector-geometry checks
Classify a parameterized system

201-SN4-RE topics to review

Linear systemsMatricesVectorsLines and planesGeometric distances

Common difficulties

Distinguishing a contradictory row from a redundant row when classifying a system’s solutions
Connecting row reduction to the geometry of intersecting planes and direction vectors

Preparing for this course

Practise solving simultaneous equations, keeping fractions exact and multiplying a matrix by a column vector. Review coordinates, vector components and dot products. Move between equations and an augmented matrix without changing variable order.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with Dawson College.

Check the official course catalogue

Try a course problem

Let $x,y,z$ be the real unknowns in $x+y+z=4$ , $2x-y+z=1$ , and $3x+2z=k$ . Here $k$ is a fixed real parameter specifying the right-hand side of the third equation. For which values of $k$ does the system have solutions? Use an augmented matrix to justify your answer, then express every solution as a point plus a multiple of a direction vector. Define the free parameter used in that expression.

Show the reasoning

1. Eliminate with the right-hand side included. Let $R_1,R_2,R_3$ denote the augmented matrix rows; each contains the coefficients of $x,y,z$ , followed by the right-hand side. Initially the rows are $[1,1,1\mid4]$ , $[2,-1,1\mid1]$ , and $[3,0,2\mid k]$ . The operations $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-3R_1$ give $[0,-3,-1\mid-7]$ and $[0,-3,-1\mid k-12]$ . Subtracting the new second row from the third gives $[0,0,0\mid k-5]$ . 2. Interpret the last row. It says $0=k-5$ . If $k\ne5$ , this is false and the system has no solution. If $k=5$ , it is the identity $0=0$ , imposing no extra restriction. The two remaining equations are $x+y+z=4$ and $3y+z=7$ . 3. Introduce a free parameter explicitly. Let $t=y-2$ , which may be any real number because neither remaining equation fixes $y$ . Then $y=2+t$ . The second equation gives $z=7-3(2+t)=1-3t$ ; the first then gives $x=4-(2+t)-(1-3t)=1+2t$ . Thus every solution for $k=5$ is $(x,y,z)=(1,2,1)+t(2,1,-3)$ , with $t\in\mathbb{R}$ . The vector $(1,2,1)$ locates a point on the solution line, while $(2,1,-3)$ gives its direction. The free parameter $t$ selects points on that line; it is different from the fixed parameter $k$ in the original system. 4. Check all three equations. Substitution gives $(1+2t)+(2+t)+(1-3t)=4$ , $2(1+2t)-(2+t)+(1-3t)=1$ , and $3(1+2t)+2(1-3t)=5$ . All dependence on $t$ cancels. A zero coefficient row is contradictory only when its right-hand side is nonzero.

Four checks: connect systems, vectors and geometry

Can you turn a geometric condition into equations, then explain what the answer means? These four original Nablio questions progress from intersecting lines to perpendicular distances and parameter cases. Each has a hint, a worked solution and a changed problem to retry.

Dawson’s official mathematics course list places linear systems, vectors, lines and planes, areas, volumes, distances and basic mathematical arguments within 201-SN4-RE. This is a selected practice set, not an official exam or a complete course review. Use your current section outline to choose the topics and assessment rules that apply.

Attempt each question before opening its solution. Record the first step you cannot justify, use the relevant review link, then close the answer and try the changed question. All coordinates and parameters are real. The three coordinate axes are perpendicular and use the same length unit; distances, areas and volumes use the corresponding units. A direction vector is nonzero. The symbols $\cdot$ , $\times$ and $\|\mathbf v\|$ denote the dot product, the right-handed cross product and a vector’s Euclidean length.

1. Decide whether two space lines meet

Two lines have vector equations

\begin{aligned}L:\ \mathbf r(t)&=(1,-1,2)+t(2,1,-1),\\M:\ \mathbf q(s)&=(1,2,-3)+s(1,-1,2),\end{aligned}

where $t,s\in\mathbb R$ are independent parameters. Do the lines intersect? Set up a system for $t$ and $s$ , check all three coordinates and explain whether the intersection is unique. A student says that different direction vectors guarantee an intersection. Is that reasoning sufficient in three dimensions?

Show a hint: compare the same coordinates

A common point must have the same x-, y- and z-coordinates on both lines. The parameters label points on different lines; they do not have to be equal. Solve two coordinate equations, then test the third.

Show the worked solution: an algebraic candidate needs a geometric check

Set up three equations for two parameters

Equating coordinates gives $1+2t=1+s$ , $-1+t=2-s$ and $2-t=-3+2s$ . Therefore,

\begin{aligned}2t-s&=0,\\t+s&=3,\\t+2s&=5.\end{aligned}

From the first equation, $s=2t$ . The second becomes $3t=3$ , so $t=1$ and $s=2$ . These are candidate parameter values, not yet a verified space intersection.

Check the remaining coordinate and both lines

The third equation holds: $1+2(2)=5$ . Substituting into the original vector equations gives

\begin{aligned}\mathbf r(1)&=(3,0,1),\\\mathbf q(2)&=(3,0,1).\end{aligned}

The intersection is the single point $(3,0,1)$ . Any common point must satisfy the first two equations, which force this one pair $(t,s)=(1,2)$ ; the third equation confirms that the candidate works. This establishes uniqueness, not just existence.

Distinguish nonparallel directions from intersecting lines

The vectors $(2,1,-1)$ and $(1,-1,2)$ are not scalar multiples: the x-components would require a factor of 2, but the y-components would then disagree. Thus the lines are not parallel. That alone is not enough for an intersection in three dimensions: nonparallel lines can be skew, meaning they do not meet. Also, merely being different vectors would not prove nonparallelism, since one could be a multiple of the other.

First error to watch for: setting $t=s$ , or treating a match in two coordinates as a match in space.

Retry: change only one coordinate of M

Replace M by $M':\mathbf q(s)=(1,2,-2)+s(1,-1,2)$ , leaving L unchanged. Decide whether the new pair intersects, is parallel or is skew. Explain why checking a third coordinate matters.

Check the changed lines

The x- and y-equations have not changed, so the only candidate remains $t=1,s=2$ . But now the z-coordinates are $2-1=1$ on L and $-2+2(2)=2$ on $M'$ . They disagree. No other pair can satisfy the first two equations, so there is no intersection. The direction vectors are unchanged and are not parallel: the lines are skew, not parallel. In the new system, the third equation is $t+2s=4$ , while the first two force $t+2s=5$ .

If the equation setup or solution test was difficult, review reversible operations and checking a system. For systems whose unknowns are coordinates rather than line parameters, return to the parameterized system already on this page.

2. Find the closest point on a line

Let $A=(1,-1,0)$ , $B=(3,0,2)$ and $P=(4,2,0)$ . Find the closest point H to P on the whole line through A and B, then find the distance from P to that line. Explain why your point is closest. This is a line problem, not a problem about the segment from A to B.

Show a hint: use a perpendicular remainder

Set $\mathbf d=B-A$ , $\mathbf u=P-A$ , and $H=A+t\mathbf d$ . The shortest connecting segment is perpendicular to the line, so $(\mathbf u-t\mathbf d)\cdot\mathbf d=0$ . Solve for the scalar t before computing the point.

Show the worked solution: project a displacement, not a position

Identify the two different vectors

The line direction and the displacement from A to P are

\mathbf d=(2,1,2),\qquad \mathbf u=(3,3,0).

The vector $\mathbf u$ , not the position vector $(4,2,0)$ , is what we resolve along this line from A. Since $\mathbf d\ne\mathbf0$ , its squared length is positive.

Find and check the perpendicular foot

Expanding the perpendicular condition gives $\mathbf u\cdot\mathbf d-t(\mathbf d\cdot\mathbf d)=0$ . Hence

\begin{aligned}t&=\frac{\mathbf u\cdot\mathbf d}{\mathbf d\cdot\mathbf d}\\&=\frac{3(2)+3(1)+0(2)}{2^2+1^2+2^2}=1.\end{aligned}

Thus $H=A+\mathbf d=(3,0,2)=B$ . The displacement from H to P is $P-H=(1,2,-2)$ , and its dot product with $\mathbf d$ is $2+2-4=0$ . The point lies on the line and the connecting segment is perpendicular to it. The distance is

\|P-H\|=\sqrt{1^2+2^2+(-2)^2}=3.

Explain why perpendicular means closest

For any other point Q on the line, $Q-H$ is parallel to $\mathbf d$ , while $P-H$ is perpendicular to it. Pythagoras therefore gives

\|P-Q\|^2=\|P-H\|^2+\|Q-H\|^2.

The second term is nonnegative and vanishes only when Q=H. Thus H is the unique closest point. The distance is 3 coordinate length units. The coefficient t=1 and the projection vector $t\mathbf d$ are not themselves this distance.

First error to watch for: dividing by $\|\mathbf d\|$ instead of $\|\mathbf d\|^2$ , or forgetting to add A after finding the projected displacement.

Retry: allow the foot to pass B

Keep A and B, but replace P by $R=(6,3,2)$ . Find the closest point on the whole line and the distance. Is the foot between A and B?

Check the new foot and distance

Now $R-A=(5,4,2)$ . Its dot product with $\mathbf d$ is $10+4+4=18$ , so $t=18/9=2$ . The foot is $H_R=A+2\mathbf d=(5,1,4)$ . The remainder $R-H_R=(1,2,-2)$ again has zero dot product with $\mathbf d$ and length 3.

A corresponds to t=0 and B to t=1; t=2 lies beyond B, outside the segment. It is still a valid closest point on the whole line because t may be any real number. A restriction to the segment would be a different question.

For a shorter explanation of the dot product and projection coefficient, review dot product versus projection. That is a reusable concept section in a MATH 204 guide, not a statement that the courses have identical requirements.

3. Build a plane and measure area and volume

Take $A=(1,0,-1)$ , $B=(3,1,-1)$ , $C=(1,2,1)$ and $D=(2,0,0)$ . Find a Cartesian equation for the plane through A, B and C, and the area of triangle ABC. Then find the volume of the parallelepiped whose three edges from A are $\overrightarrow{AB}$ , $\overrightarrow{AC}$ and $\overrightarrow{AD}$ . Explain why the three points determine a plane.

Show a hint: one cross product has two roles

Form $\mathbf u=B-A$ and $\mathbf v=C-A$ . The cross product is perpendicular to both directions; its length is the area of their parallelogram. A triangle uses half that area. The parallelepiped volume is $|(\mathbf u\times\mathbf v)\cdot(D-A)|$ .

Show the worked solution: keep direction and scale separate

Construct directions from a common point

\begin{aligned}\mathbf u&=(2,1,0),\\\mathbf v&=(0,2,2),\\\mathbf w=D-A&=(1,0,1).\end{aligned}

For components in x,y,z order,

\begin{aligned}(\mathbf u\times\mathbf v)_x&=u_yv_z-u_zv_y,\\(\mathbf u\times\mathbf v)_y&=u_zv_x-u_xv_z,\\(\mathbf u\times\mathbf v)_z&=u_xv_y-u_yv_x.\end{aligned}

Here the three components are $1(2)-0(2)=2$ , $0(0)-2(2)=-4$ and $2(2)-1(0)=4$ . Thus $\mathbf N=\mathbf u\times\mathbf v=(2,-4,4)$ . This is nonzero, so the two directions are not parallel and A, B, C are not collinear. They determine one plane.

Use a simpler normal for the equation

We may use $\mathbf n=\mathbf N/2=(1,-2,2)$ as a normal. A point $(x,y,z)$ lies in the plane exactly when its displacement from A is perpendicular to $\mathbf n$ :

\begin{aligned}(x-1)-2y+2(z+1)&=0,\\x-2y+2z&=-1.\end{aligned}

Check A, B and C: their left-hand sides are respectively $1-2=-1$ , $3-2-2=-1$ and $1-4+2=-1$ . Also $\mathbf n\cdot\mathbf u=2-2=0$ and $\mathbf n\cdot\mathbf v=-4+4=0$ .

Use the original cross-product magnitude for area and volume

The parallelogram area is $\|\mathbf N\|=\sqrt{4+16+16}=6$ . Therefore triangle ABC has area $6/2=3$ square coordinate units. The parallelepiped volume is

|\mathbf N\cdot\mathbf w|=|2(1)-4(0)+4(1)|=6

cubic coordinate units. This is a parallelepiped, not a tetrahedron. As an independent check, D is distance $|2-(-1)|/\|\mathbf n\|=3/3=1$ from the plane: base area 6 times height 1 gives volume 6.

First error to watch for: replacing the cross product by a simpler normal before computing area. Multiplying a normal by a nonzero scalar gives a normal to the same plane, but generally changes its length. Here $\|\mathbf n\|=3$ is not the parallelogram area.

Retry: stretch one edge

Keep A, B and D, but use $C'=(1,4,3)$ . Without recomputing every component, predict the plane, triangle area and parallelepiped volume. Justify the factors by comparing $C'-A$ with $C-A$ .

Check the stretched configuration

Since $C'-A=(0,4,4)=2\mathbf v$ , the new cross product is $\mathbf u\times(2\mathbf v)=2\mathbf N$ , whose length is 12. The new triangle area is 6 and the parallelepiped volume is $|(2\mathbf N)\cdot\mathbf w|=12$ . The plane remains $x-2y+2z=-1$ : its two independent directions span the same plane through A. Check $C'$ : $1-2(4)+2(3)=-1$ . Doubling an edge in the base doubles the area and volume here, not the height.

If the normal-vector meaning is unclear, use the plane-intersection guide to connect plane equations, normals and solution sets. The cross-product area and volume calculations are developed here.

4. Classify a line–plane intersection before dividing

Let $\Pi:x-2y+2z=-1$ and consider the family of lines

\ell_a:\ \mathbf r(t)=(2,0,0)+t(2,1,a),\qquad t\in\mathbb R.

Here a is a fixed real number that selects a line, while t selects a point on that line. For which a does the line meet the plane, and where? If there is no intersection, find the distance from the line to the plane. Explain why a line direction perpendicular to the plane’s normal does not by itself tell you whether the line lies in the plane.

Show a hint: keep the zero case

Substitute x=2+2t, y=t and z=at into the plane equation. Before dividing by a coefficient, decide whether it can be zero. When a line runs parallel to a plane, test its starting point to distinguish containment from no intersection.

Show the worked solution: one equation, two cases

Schematic of three line–plane positions: one crossing point for the original family when a is nonzero; no common point when a is zero; and a line contained in the plane when a is zero after the retry changes the starting point. — Schematic, not a coordinate plot: do not read angles or distances from it. Shaded patches show only part of each infinite plane. The first two panels describe the original family; the third uses the changed starting point in the retry. A direction parallel to a plane can give either no common point or a contained line.

Substitute and interpret the coefficient

The normal is $\mathbf n=(1,-2,2)$ , and the direction $\mathbf d_a=(2,1,a)$ is nonzero for every a. Substitution gives

\begin{aligned}(2+2t)-2t+2at&=-1,\\2at&=-3.\end{aligned}

The coefficient $2a=\mathbf n\cdot\mathbf d_a$ measures whether travelling along the line changes the left-hand side of the plane equation.

Case a≠0: one intersection

Division is now valid: $t=-3/(2a)$ . The intersection point has coordinates

\begin{aligned}x&=2-\frac3a,\\y&=-\frac3{2a},\\z&=-\frac32.\end{aligned}

The plane check is $2-3/a+3/a-3=-1$ . There is one parameter value and hence one intersection. For example, a=1 gives $(-1,-3/2,-3/2)$ .

Case a=0: no intersection and a positive distance

The equation becomes $0=-3$ , which is impossible. The direction is $(2,1,0)$ , perpendicular to the normal, but the starting point $P=(2,0,0)$ is not in the plane. The line is parallel to the plane and disjoint from it.

Find the perpendicular foot from P by writing $H=P-\lambda\mathbf n$ . Requiring H to lie on $\Pi$ gives $2-9\lambda=-1$ , so $\lambda=1/3$ . Therefore,

\begin{aligned}H&=(5/3,2/3,-2/3),\\\|P-H\|&=\tfrac13\|\mathbf n\|=1.\end{aligned}

Check the plane: $5/3-4/3-4/3=-1$ . The vector $P-H=\mathbf n/3$ is perpendicular to it. For any point $S=P+t(2,1,0)$ on the line, $\mathbf n\cdot S=2$ , so subtracting the same normal vector $\mathbf n/3$ reaches the plane; this displacement has length 1. Thus the distance from the entire line to the plane is 1 coordinate unit, not just a distance computed from a conveniently chosen point. A perpendicular segment is shortest by the same Pythagorean reasoning used in question 2.

First error to watch for: dividing by a before checking a=0, or calling a line contained merely because its direction is perpendicular to the normal.

Retry: change the starting point

Replace the starting point by $Q=(1,1,0)$ , keeping direction $(2,1,a)$ and plane $\Pi$ . Classify the intersections for a=0 and a≠0. State the line–plane distance in both cases.

Check containment against crossing

Now $\mathbf n\cdot Q=1-2=-1$ , so Q lies in the plane. Substitution gives $-1+2at=-1$ , or $2at=0$ . If a=0, every real t works: the whole line lies in the plane, so there are infinitely many common points. If a≠0, only t=0 works: the unique intersection is Q. The distance between the line and plane is zero in both cases because they have common points. Zero distance alone does not distinguish crossing from containment.

If you lost the fixed/free parameter distinction, revisit the original system example. If the geometric classification was difficult, compare the two-plane solution sets; two planes and a line–plane pair have different possible intersection shapes.

Use your first error to choose what to practise next

Wrong equations or a missed coordinate: redo question 1 with the parameters labelled separately.
Wrong object returned: in question 2, identify the scalar coefficient, projected vector, foot and distance before calculating.
Wrong scale or units: in question 3, keep the original cross product for area and volume even if you simplify the normal.
A missing exceptional case: in question 4, classify the parameter equation before dividing.

For tutoring, bring your first attempted setup and the step you cannot explain. You can also start without an attempt. Work through one example together, then use the changed problem to check whether you can choose and justify the method yourself.

How tutoring can help

Your tutor can follow your row operations and help you translate the final rows into a solution set. For line and plane problems, you can revisit your diagram together to distinguish a point, a direction vector and a normal vector before choosing a calculation.

Exam preparation for 201-SN4-RE

Check your section’s outline for the assessed topics and evaluation rules. Practise systems with one solution, no solution and infinitely many solutions, together with the assigned matrix and vector-geometry questions. For each system, connect the final matrix rows to the solution set. For a geometry problem, identify the points, directions or normals used and explain what the result represents.

We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Independent practice

Free practice and study guides for 201-SN4-RE

8 min

Row Operations: Why the Solutions Stay the Same

Practise reversible row operations and use the final rows to distinguish one, no or infinitely many solutions.

20 min

Plane Intersections: Row Reduction and Geometry

Connect an augmented matrix to the intersection of two planes, using free parameters to describe every shared point.

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201-SN4-RE · official course reference