Intro to Vectors

A vector is an ordered set of numbers that represents length (magnitude) and direction.

Notation: A vector is typically denoted either in bold, v, or with an arrow above it ~v.We usually use the same letter for the different components of a vector. So a vector in 3-space with3 components might be written ~v =< v1, v2, v3 > or ~v = (v1, v2, v3). The <> and () brackets areused interchangeably. I can also use square brackets, [], and write them as column matrices.

When considering a vector ~u = (u1, u2) in R2, we must understand that u1 represents the dis-placement in the x-direction and u2 represents the displacement in the y-direction - it does NOTrepresent a location; just a length and a direction.

Equality of Vectors: Two vectors are said to be equal if their components are equal; thatis, if they have the same direction and magnitude.

If we want for a vector to be located in a specific place, it is necessary to give the vector aninitial point, which indicates where the vector will start. In this case, the point where it ends willbe called the terminal point.

A vector between two points: If P1(x1, y1) is the initial point and P2 = (x2, y2) is the terminal

point for the vector ~u, then−−−→P1P2 = (x2 − x1, y2 − y1)

If a vector is placed so that its initial point is the origin, O, with a terminal point of P , thenit is called a position vector. Unless we are specifically given initial points or terminal points, wewill think of any vector in Rn is a position vector.

Vector Operations

Addition/Subtraction of Vectors: If ~u = (u1, u2, ...un) and ~v = (v1, v2, ..., vn) arevectors then ~u± ~v = (u1 ± v1, u2 ± v2, ..., un ± vn)This means to first do the length and direction of ~u and then do the length and direction of ~v(or vice versa).

Scalar Multiplication of Vectors: If ~v = (v1, v2, ...vn) is a vector and k is a scalar,then k~v = (kv1, kv2, ..., kvn)

Scalar multiplication of a vector does the following:

1) Stretches it (makes it longer) if k > 12) Shrinks it (makes it shorter) if 0 < k < 13) Changes its direction if k < 0

Definition: Two vectors are said to be parallel if they point in the same or opposite directions(if the angle between them is 0)

Example: The vector ~u = (3, 5,−2, 4) is parallel to the vector ~v = (−6,−10, 4,−8) because theypoint in opposite directions. This can be seen because ~w is a scalar multiple of ~v.

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Properties of Vectors

Let ~u,~v, and ~w be vectors in Rn, and let a and b be scalars. Then:

a) ~u + ~v = ~v + ~ub) (~u + ~v) + ~w = ~u + (~v + ~w)c) a(b~u) = (ab)~ud) (a + b)~u = a~u + b~ue) a(~u + ~v) = a~u + a~v

f) ~u +~0 = ~ug) 1(~u) = ~u

h) a(~0) = ~0, and 0(~u) = ~0

i) ~u + (−~u) = ~0

j) If a~u = ~0, then either a = 0 or ~u = 0

The Magnitude (or Norm) of a Vector

The magnitude of a vector ~v = (v1, v2, ..., vn) is given by ‖~v‖ =√

v21 + v22 + ... + v2n

Definition: A vector of norm 1 is called a unit vector.

Given a non-zero vector ~v in Rn, ~u =1

‖~v‖~v is a unit vector in the same direction as ~v.

Theorem: If ~v is any non-zero vector in Rn, then

a) ‖~v‖ ≥ 0b) ‖~v‖ = 0⇔ ~v = 0c) ‖k~v‖ = |k| ‖~v‖

The Dot ProductIf ~u = (u1, u2, ..., un) and ~v = (v1, v2, ..., vn) are vectors in Rn then the dot product of ~u and ~v isgiven by ~u · ~v = u1v1 + u2v2 + ... + unvn.

Example: Find the dot product of ~u = (1, 3, 2, 0) and ~v = (6,−3, 1, 2).

~u · ~v = (1)(6) + (3)(−3) + (2)(1) + (0)(2) = 6− 9 + 2 = −1

Theorem: Two non-zero vectors ~u and ~v are said to be orthogonal (or perpendicular) if~u · ~v = 0. We write ~u ⊥ ~v

Definition: A non-empty set of vectors in Rn is said to be an orthogonal set if ALLpairs of vectors in the set are orthogonal.

Example: Do the vectors ~u = (1, 2, 3), ~v = (−4, 5,−2), and ~w = (3, 6,−5) form an orthogonal set?

Check each pair separately:~u · ~v = (1)(−4) + (2)(5) + (3)(−2) = −4 + 10− 6 = 0, so ~u ⊥ ~v~u · ~w = (1)(3) + (2)(6) + (3)(−5) = 3 + 12− 15 = 0, so ~u ⊥ ~w

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~v · ~w = (−4)(3) + (5)(6) + (−2)(−5) = −12 + 30 + 10 = 28 6= 0, so ~v is not perpendicular to ~w.Because ~v and ~w are not orthogonal, these vectors do not form an orthogonal set.

The Cross ProductIf If ~u = (u1, u2, u3) and ~v = (v1, v2, v3) are vectors in R3 then the cross product of ~u and ~v is

given by ~u× ~v =

(∣∣∣∣u2 v2u3 v3

∣∣∣∣ ,− ∣∣∣∣u1 v1u3 v3

∣∣∣∣ , ∣∣∣∣u1 v1u2 v2

∣∣∣∣)

Example: Find the cross product of ~u = (1, 2, 3) and ~v = (−2, 1, 4)

~u× ~v =

(∣∣∣∣2 13 4

∣∣∣∣ ,− ∣∣∣∣1 −23 4

∣∣∣∣ , ∣∣∣∣1 −22 1

∣∣∣∣) = (5,−10, 5)

Theorem: If ~u,~v are two non-parallel non-zero vectors in R3, then ~w = ~u×~v is orthogonal toboth ~u and ~v.

Equations of Planes and Lines

Remember that the general equation of a plane in R3 is ax + by + cz + d = 0 and the generalequation for a line in R2 is ax + by + c = 0.

Point Normal Equations

A line in R2 containing a point P (x0, y0) with normal ~n = (a, b) can be expressed:a(x− x0) + b(y − y0) = 0.

A plane in R3 containing a point (x0, y0, z0) with normal ~n = (a, b, c) can be expressed:a(x− x0) + b(y − y0) + c(z − z0) = 0

Theorem:a) If a and b are constants that are not both 0, then an equation of the form ax + by + c = 0represents a line in R2 with normal ~n = (a, b).

b)If a, b, and c are constant that are not all 0, then an equation of the form ax+ by+ cz+d = 0represents a plane in R3 with normal ~n = (a, b, c).

Example: Find a vector orthogonal to the line 2x + 6y − 3 = 0.We know that the normal vector represents a vector orthogonal to the line, and we can see thatthe normal is ~n = (2, 6).

Example: Find a vector orthogonal to the plane 7x + 6y − z = 0.We know that the normal vector represents a vector orthogonal to the plane, and we can see thatthe normal is ~n = (7, 6,−1).

Theorem:a) If the normal vectors of two lines are multiples of each other, then those lines are either parallelor the same line.b) If the normal vectors of two lines/planes are orthogonal (with dot product=0), then thoselines/planes must also be orthogonal.

Vector Equations

A line in R3 can be written in vector form as:

xyz

=

x0y0z0

+ t

abc

where P0 = (x0, x0, z0) is a point on the line and ~v = (a, b, c) is a vector parallel to the line.

A plane in R3 can be written in vector form as:

xyz

=

x0y0z0

+ t

a1b1c1

+ s

a2b2c2

where P0 = (x0, y0, z0) is a point on the plane and ~v1 = (a1, b1, c1) and ~v2 = (a2, b2, c2) arevectors parallel to the plane.

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To find the vector equation of a line in R3 through two points, P and Q

Step 1: Find the vector−−→PQ. This vector will be parallel to the line.

Step 2: Set up the vector equation as

xyz

= P + t−−→PQ or

xyz

= Q + t−−→PQ

To find the general equation of a plane in R3 through three points, P , Q, and R

Step 1: Find two vectors on the plane,−−→PQ and

−→PR. (Note that you could also use

−−→QR

Step 2: Take the cross product of the two vectors to find a vector normal to the plane: ~n =−−→PQ×

−→PR

Step 3: Now that you have a normal, plug the normal along with a point into the point-normalequation (it doesn’t matter which point you use).

To find the vector equation of a plane in R3 through three points, P , Q, and R

Step 1: Find two vectors on the plane,−−→PQ and

−→PR. (Note that you could also use

−−→QR

Step 2: As these two vectors are on the plane, they must be parallel to the plane, so use them to

set up the vector equation

xyz

= P + t−−→PQ + s

−→PR.

Note that you could choose P , Q, or R as your point.

Intersection of PLANES

(A) (B) parallel planes (C) 3 planes intersecting (same normal) at a single point line of intersection P1 P1 P2 ● P2 n (D) 3 planes intersecting (E) 2 parallel planes (F) no pairs of parallel planes on a line 1 transversal inconsistent (Tent Case) inconsistent case P3 P1 P2 P1 P3 P2 Examples:

inconsistent system

(A) (B) or same plane (C)

4 2 87 2 4 2 4

3 10 10 303 2 2 4 2 10 2 4 2 8

7 4 2 28

( , , )

x y zx y z x y z x y z

x y zx y z x y z x y z

x y z

x y z

+ + = + + = + − = + − = + + = − − = + − = + − = + + == ( ) ( )

(D) (E)

9 19 10 1 54 4 3 3 3, , ( , , ) , ,

3 4 12 2 4

2 5 2 20 3

4 11 10 44

t t x y z

x y z x y z

x y z

x y z

− =

+ + = + − = + + = + + =

( )inconsistent system inconsistent system

(F)

4 2 5

10 3 13 6

2 4 2 10 5 22 4 20

( , , ) 14 ,4 6 ,

x y z

x y z x y z

x y z x y z

x y z t t t

− + + = + + = − + = + − = − + =

= −

Note: 2 of 3 planes are parallel Note: no two of the planes are parallel

Linear Combinations and Spans

Linear Combinations

Definition: If ~u1, ~u2, ..., ~uk are vectors in Rn and c1, c2, ..., cn are scalars, then the vector~v = c1 ~u1 + c2 ~u2 + ... + ck ~uk is called a linear combination of the vectors ~u1,,. . ., ~uk.

Example: Given vectors ~u1 = (3, 4, 6) and ~v2 = (−1, 2, 8), then ~v = 3 ~u1 − 2 ~u2 = (11, 8, 12)is a linear combination of ~u1 and ~u2.

Is ~v a linear combination of given vectors?To determine if a vector ~v is a linear combination of some vectors ~u1,,. . ., ~uk, you must setup a system of linear equations and solve in the normal manner.

Step 1: Set up the system of equations c1 ~u1 + c2 ~u2 + ... + c3 ~uk = ~vStep 2: Write the system of equations as an augmented matrix (with each vector as a column

vector).Step 3: Reduce the augmented matrix to RREFStep 4: If the system is consistent, read the solution off the matrix; if the matrix is inconsistent,

then ~v cannot be written as a linear combination of the given vectors.

Example: Is the vector ~v = (2, 6, 9, 1) a linear combination of ~u1 = (1, 2, 1, 3) and ~u2 =(−1, 0, 4, 1)?

Step 1: Are there a c1 and a c2 such that c1 ~u1 + c2 ~u2 = ~v?

Step 2:

1 −1 22 0 61 4 93 1 1

Step 3: →

1 −1 20 2 20 5 70 4 −5

→

1 −1 20 2 20 5 70 0 −9

Step 4: This matrix produces a contradiction, so we can see that the system is inconsistent. There-

fore now, ~v cannot be written as a linear combination of ~u1 and ~u2.

Example: Is the vector ~v = (2, 6, 7) a linear combination of ~u1 = (1, 2, 1) and ~u2 = (−1, 0, 4)?

Step 1: Are there a c1 and a c2 such that c1 ~u1 + c2 ~u2 = ~v?

Step 2:

1 −1 22 0 61 4 7

Step 3:

R2−2R1−−−−−→R3−R1−−−−→

1 −1 20 2 20 5 5

R2(12)

−−−−→R3−5R2−−−−−→

1 −1 20 1 10 0 0

R1+R2−−−−→1 0 3

0 1 10 0 0

Step 4: This matrix is consistent, and we can see that c1 = 3 and c2 = 1. So ~v = 3 ~u1 + 1 ~u2

That is, (2, 6, 7) = 3(1, 2, 1) + 1(−1, 0, 4).1

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Theorem: AX = B is consistent if and only if B is expressible as a linear combination of thecolumn vectors of A.

Spans

Definition: The span of a set of vectors ~u1, ~u2, ..., ~uk in Rn is the set of all vectors c1 ~u1 +c2 ~u2 + ... + ck ~uk that are linear combinations of the given vectors.

Example: Given ~v = (−1, 1, 20, 16), ~u1 = (1, 3, 6, 4), and ~u2 = 2, 4,−1, 2), we can see that ~vis in span{u1, u2} because it can be written as a linear combination of ~u1 and ~u2 in the followingway: ~v = 3 ~u1 − 2 ~u2.This can be shown by following the steps above to determine if ~v is a linear combination of ~u1 and ~u2.

Theorem: Because we can always just let all of the weights (constants) be 0, we can see that

every span contains the zero vector, ~0.

Describing the Span of Vectors - the span of vectors is almost always infinite:

In R2:

* span{(0, 0, 0)} = (0, 0, 0)

* span{~u} = the set of all scalar multiples of ~u = a line in R2

* span{~u,~v} =the set of all linear combinations of ~u and ~v= R2 if ~u and ~v are not collinear (thatis, they are not scalar multiples; they are linearly independent)

In R3:

* span{(0, 0, 0)} = (0, 0, 0)

* span{~u} =the set of all scalar multiples of ~u = a line in R3

* span{~u,~v} =the set of all linear combinations of ~u and ~v = a plane in R3 if ~u and ~v are notcollinear (that is, they are not scalar multiples; they are linearly independent)

* span{~u,~v, ~w} =the set of all linear combinations of ~u,~v, and ~w = R3 if ~u,~v, and ~w are notcoplanar (they are linearly independent).

Steps to determine the span of vectors in Rn

Step 1: Set up an augmented matrix with the vectors as column vectors and variables on theright side of the augmentation line.

Step 2: Reduce the matrix to RREFStep 3: If there is a pivot in every ROW, then the span of the given vectors is Rn

If there is a row of the form[0 0 ... 0 | x, y, z

], then the equation formed from

setting the x, y, z equal to 0 will describe the span.

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Theorem: Given vectors ~u1, ~u2, ..., ~uk in Rn, if k = n (that is, the number of vectors is the sameas the number of elements in the vector) and if the determinant of matrix formed from the columnvectors is not 0 (that is, they are linearly independent), then the vectors span Rn.

Example: Describe the span of the vectors ~u1 = (−1, 4, 3), ~u2 = (3,−2, 6), and ~u3 = (1, 0, 3).

That is, find an equation relating x, y, and z for general vector ~b = (x, y, z) in span{ ~u1, ~u2, ~u3}

Step 1:

−1 3 1 x4 −2 0 y3 6 3 z

Step 2:

R2+4R1−−−−−→R3+3R1−−−−−→

−1 3 1 x0 10 4 4x + y0 15 6 3x + z

(R1)(−1)−−−−−→(R2)(

110

)−−−−−→

1 −3 −1 −x0 1 2/5 4x+y

100 15 6 3x + z

R1+3R2−−−−−→

R3−15R2−−−−−−→1 0 1/5 2x+3y10

0 1 2/5 −26x+y10

0 0 0 −6x−3y+2z2

Step 3: There is clearly not a pivot in every row, so these three vectors are coplanar (and thus do

not span R3). We must therefore form an equation from the right side of the row of 0s bysetting it equal to 0.

−6x−3y+2z2 = 0→ −6x− 3y + 2z = 0

So span{ ~u1, ~u2, ~u3} is the plane −6x− 3y + 2z = 0.

Another way this question could be asked:

Example: Find the plane spanned by the vectors ~u = (1, 3, 3) and ~v = (−1, 2, 0).

Step 1:

1 −1 x3 2 y3 0 z

Step 2:

R2−3R1−−−−−→R3−3R1−−−−−→

1 −1 x0 5 y − 3x0 3 z − 3x

R2(

15)

−−−−→

1 −1 x

0 5 y−3x5

0 3 −3x + 3y − 8z

R1+R2−−−−→

R3−3R2−−−−−→

1 0 2x+y5

0 1 y−3x5

0 0 −6x−3y+5z5

Step 3: −6x−3y+5z

5 = 0→ −6x− 3y + 5z = 0, So the plane spanned by these vectors is−6x− 3y + 5z = 0.

Linear Independence and Linear Dependence

Definition: The set of k vectors { ~u1, ~u2, ..., ~uk} in Rn is said to be linearly dependent if one ormore of the vectors can be written as a linear combination of the other vectors.The set of vectors is called linearly independent if it is not linearly dependent.A set {~u} consisting of a single vector is called linearly dependent iff ~u is the zero vector.

In other words, the set of k vectors { ~u1, ~u2, ..., ~uk} in Rn is said to be linearly dependent ifthere exist scalars c1, c2, ...ck, with at least one of the scalars not 0, such that

c1 ~u1 + c2 ~u2 + ... + ck ~uk = 0

This equation is called a dependency equation.

They are linearly independent iff the only scalars that satisfy c1 ~u1 + c2 ~u2 + ... + ck ~uk = 0are ALL 0!!

Example: The set of vectors {(1, 2), (3, 0), (−1, 4)} is linearly dependent because−2(1, 2) + (3, 0) + (−1, 4) = 0

That is, we were able to form a dependency equation in which at least one the the constants wasn’t 0.

Tips for Determining Linear Dependency Let { ~u1, ~u2, ..., ~uk} be a set of k vectors inRn. Then:

1. If k > n, then the vectors are linearly dependent.2. If ~0 ∈ { ~u1, ~u2, ..., ~uk}, then the vectors are linearly dependent.3. If k = 2 and the vectors are proportional (scalar multiples), then they are linearly dependent.

If they are not scalar multiples, they are linearly independent.4. If k = n and det(U) = 0, then the vectors are linearly dependent. If det(U) 6= 0, then the

vectors are linearly independent.(U is the matrix formed with the vectors in the set as the columns)

Steps for Determining if Vectors are Linearly Independent (L.I.) or LinearlyDependent (L.D.)To determine if a set of vectors { ~u1, ~u2, ..., ~uk} in Rn are L.I. or L.D.,

Step 1: Determine if the Tips for Determining Dependency apply.Step 2: If they don’t, create the matrix U whose columns are ~u1, ~u2, ..., ~uk.Step 3: Solve the system UX = ~0 where X = (c1, c2, ..., ck). This can most easily done by row

reducing the augmented matrix with 0 on the right of the augmentation.Step 4: If the system has only the trivial solution c1 = 0, c2 = 0, ..., ck = 0, then the vectors

are linearly independent. (In this case, every column has a pivot).If the system has infinitely many (non-trivial) solutions, then the vectors are linearlydependent. (In this case, there will be at least one column with no pivot.)

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Theorem: Given vectors ~u1, ~u2, ..., ~uk in Rn, if k = n (that is, the number of vectors is thesame as the number of elements in the vector) and if the vectors are linearly independent (thedeterminant of matrix formed from the column vectors is not 0), then the vectors span Rn.

Important Note: If the vectors do not span Rn, we can easily determine if they span a line or aplane by looking at the number of linearly independent vectors. The number of linearly indepen-dent vectors will be the number of pivots in the row reduced form.

Example: Are the vectors ~u1 = (3, 4, 1), ~u2 = (−1, 3, 2), and u3 = (4, 1,−1) linearly dependentor independent? Do they span a plane, a line, or R3?

Step 1: In this case, k = n, so we can set up U and find the determinant.

|U | =

∣∣∣∣∣∣3 −1 44 3 11 2 −1

∣∣∣∣∣∣ = 3

∣∣∣∣3 12 −1

∣∣∣∣−4

∣∣∣∣−1 42 −1

∣∣∣∣+1

∣∣∣∣−1 43 1

∣∣∣∣ = (3)(−5)−4(−7)+(1)(−13) = 0

We can see that det(U) = 0, so we know that these vectors are linearly dependent.Because the three vectors are linearly dependent, they do not span R3. To determine theirspan, we must continue the steps.

Step 2: U =

3 −1 44 3 11 2 −1

Step 3:

R1−3R3−−−−−→R2−4R3−−−−−→

0 −7 70 −5 51 2 −1

R1↔R3−−−−−→1 2 −1

0 −7 70 −5 5

R2(1/−7)−−−−−−→1 2 −1

0 1 −10 −5 5

R1−2R2−−−−−→R3+5R2−−−−−→

1 0 −30 1 −10 0 0

Step 4: This system clearly has infinite solutions, so we can conclude the same thing as in step 1:

the vectors are linearly dependent.

We see, though, that there are 2 pivots in the reduced matrix. From this we conclude that thereare two linearly independent vectors (see above Important Note). We know that the span of twolinearly independent vectors in R3 is a plane (see notes on Span).

Bases and Basis Selection

Definition: Let S be a subspace of Rn. The set { ~u1, ~u2, ..., ~uk} of vectors in S is called abasis for S if both of the following are true:

1) the vectors ~u1, ~u2, ..., ~uk are linearly independent and2) S = span{ ~u1, ~u2, ..., ~uk}Each of the individual vectors is called a basis vector.

Basically, a basis of S is the smallest set of vectors needed to span S.

Theorem: BASES ARE NOT UNIQUE! However, all bases for S will contain the same num-ber of vectors.All bases of a line through the origin will contain 1 vector.All bases of a plane through the origin or R2 will contain 2 vectors.All bases of R3 will contain 3 vectors.All bases of Rn will contain n vectors.

Theorem: Let S be a subspace of Rn. Then every basis of S has the same number of vec-tors, and the number of vectors in a basis for S is called the dimension of S.

The above can be easily summarized in the following chart.

Subspace Dimension

{~0} 0line through O 1

plane through O 2Rn n

Theorem: If U =[~u1 ~u2 ... ~un

]and det(U) 6= 0, then { ~u1, ~u2, ..., ~un} is linearly independent

and spans Rn and is therefore a basis for Rn.Let S be a k-dimensional subspace (need k basis vectors to span all of S) of Rn. If a set of k vectorsin S either spans S or is linearly independent, then it is both (that is, one implies the other) andis therefore a basis for S.

Example: Which of the following sets of vectors form a basis for R3?

a) {(1, 3, 5), (−2, 6, 1)} does not form a basis for R3 because they do not span R3. In this case,they span a plane in R3 because they are linearly independent (not scalar multiples).

b) {(1, 3, 5), (−2, 6, 1), (0, 4, 2), (−3, 0, 1)} does not form a basis for R3 because the vectors are lin-early dependent (k > n : the number of vectors > the number of components)

c) {(1, 3, 5), (−2, 6, 1), (0, 4, 2)} has the correct number of vectors. To determine if the set formsa basis, we must just check to see if the vectors are linearly independent. We can do this byevaluating the determinant.

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2 ∣∣∣∣∣∣1 −2 03 6 25 1 2

∣∣∣∣∣∣→∣∣∣∣∣∣

1 −2 0−7 4 05 1 2

∣∣∣∣∣∣ = 2

∣∣∣∣ 1 −2−7 4

∣∣∣∣ = 2(4− 14) = −20 6= 0.

The determinant does not equal 0, so the vectors are linearly independent, and this set of vectorsis therefore a basis for R3.

d) {(1, 3, 5), (−2, 6, 1), (7,−3, 13)} has the correct number of vectors. To determine if the set formsa basis, we must just check to see if the vectors are linearly independent. We can do this byevaluating the determinant.∣∣∣∣∣∣1 −2 73 6 −35 1 13

∣∣∣∣∣∣→∣∣∣∣∣∣1 −2 70 12 −240 11 −22

∣∣∣∣∣∣ = 1

∣∣∣∣12 −2411 −22

∣∣∣∣ = ((12)(−22)− (−24)(11)) = 0.

The determinant equals 0, so the vectors are linearly dependent, and this set of vectors is not abasis for R3.

Example: Find a basis for S = {(x, y, z) ∈ R3|x + y − z = 0}Since we know that a basis is a set of linearly independent vectors that span the space, we needonly find the two (because it’s a plane) linearly independent vectors in the span of this space.

Find the parametric solutions of[1 1 −1|0

]:

x = −t + s

y = t

z = s

, so

S = span{(−1, 1, 0), (1, 0, 1)} and therefore {(−1, 1, 0), (1, 0, 1)} is a basis for S.