Calculate with three-dimensional vectors and test independence

Use the existing plane-vector foundations

Vector equality, endpoint order, scalar multiplication and the dot-product definition are already established

A vector describes magnitude and direction

A directed segment from A to B represents the vector AB. Translating the entire segment does not change its vector. Equal vectors have equal components, hence equal magnitude and direction when nonzero. A position vector OA depends on the chosen origin; the displacement AB=b−a does not. The zero vector has magnitude zero and no defined direction

Scalar multiplication and magnitude

For scalar λ, λ(x,y)=(λx,λy). For a nonzero vector, positive λ preserves its direction and negative λ reverses it. Length is multiplied by |λ|. If λ=0 the result has no direction. Algebraic rules such as u+v=v+u and λ(u+v)=λu+λv follow component by component

The scalar product is a number

For a=(a₁,a₂) and b=(b₁,b₂), a·b=a₁b₁+a₂b₂. Multiply matching components and add. The result is a scalar; a·a=|a|²≥0, with equality only for the zero vector. Commutativity, distributivity and compatibility with scalar multiplication follow by expanding components

Use three orthogonal unit directions

In an orthonormal Cartesian frame, i=(1,0,0), j=(0,1,0), k=(0,0,1). A vector a=(x,y,z) means xi+yj+zk. The coordinates are signed scalar components; xi,yj,zk are vector components. The origin changes position vectors, while displacements between fixed points stay unchanged

Subtract all three coordinates

For A=(1,-2,3), B=(-2,4,1), the displacement AB is (-3,6,-2). Its negative is BA. Points and position vectors are related, but a point is not itself a free directed displacement

For A=(1,-2,3) and B=(-2,4,1), find the third component of BA

Combine spatial vectors component by component

For a=(2,-1,3), b=(-1,4,0), 2a-3b=(7,-14,6). Every component receives its scalar multiplier. The zero vector has no direction; multiplying by a negative scalar reverses direction and multiplies magnitude by its absolute value

For a=(2,-1,3) and b=(-1,4,0), find the second component of 2a-3b

Apply Pythagoras twice

For a=(x,y,z), the squared length of its projection onto the x-y plane is x²+y². A second perpendicular triangle gives |a|²=x²+y²+z². Thus (2,-3,6) has magnitude 7 and unit vector (2/7,-3/7,6/7)

Find the magnitude of (2,-3,6)

Find the second component of the same-direction unit vector for (2,-3,6)

Retain exact radical components

For (1,2,2), the norm is 3. For (1,1,2), it is √6. Practice asks for a positive square-free radicand and rational coefficients; zero coefficients remain zero. The zero vector cannot be normalised

Dot products use the orthonormal basis

The basis vectors have squared magnitude 1 and pairwise dot products 0. Expanding (xi+yj+zk)·(ui+vj+wk) leaves xu+yv+zw. For a=(1,2,-1), b=(2,-1,2), the dot product is -2 and cosθ=-2/(3√6)=-√6/9.

Find the dot product of a=(1,2,-1) and b=(2,-1,2)

For a=(1,2,-1) and b=(2,-1,2), find the coefficient of √6 in the cosine of their included angle

What can be concluded from a·b=0 if a is the zero vector?

Solve a perpendicularity condition without dividing by a coordinate that may vanish

For a=(t,2,-1), b=(3,-2,4), a·b=3t-8. Setting it to zero gives t=8/3. Then both vectors are nonzero, so the directions are perpendicular. If the unknown has zero coefficient, the equation can instead have no solution or every value as a solution

For a=(t,2,-1) and b=(3,-2,4), find t so that the nonzero vectors are perpendicular

Direction cosines are signed coordinate components of a unit vector

For nonzero a=(x,y,z), its angles with the positive coordinate axes satisfy cosα=x/|a|, cosβ=y/|a|, cosγ=z/|a|. Their squared cosines sum to 1. The three angles need not sum to 180°. Rectangular scalar components along i,j,k are x,y,z.

Scalar resolute is signed

The scalar resolute of a along nonzero b is a·b/|b|. It is negative when a points partly against the chosen direction. For a=(2,-1,2), b=(0,3,0), it is -1. Its absolute value is the nonnegative length of the projection

For a=(2,-1,2), find its signed scalar resolute along b=(0,3,0)

Vector resolute is a complete vector

Multiply the signed scalar resolute by the unit direction b/|b|. For a=(2,1,-2), b=(0,3,0), the parallel vector is P=(0,1,0) and the perpendicular remainder is R=(2,0,-2). Their sum is a and R·b=0.

For a=(2,1,-2) and b=(0,3,0), in this view within their plane, what spatial vector does P represent?

Find the squared magnitude of R=(2,0,-2)

Reversing the direction changes the scalar but not the vector projection

Replacing b by -b changes a·b/|b| to its negative. In the vector formula both the dot product and b reverse, so their product stays the same. More generally every nonzero scalar multiple of b gives the same vector projection onto the line. The direction must remain nonzero

Linear independence is a condition on all coefficients

Vectors v1,...,vn are independent when c1v1+...+cnvn=0 implies c1=...=cn=0. They are dependent when at least one nonzero coefficient gives zero. A set containing the zero vector is dependent. Two nonzero vectors are dependent exactly when they are parallel

Three nonparallel pairs do not guarantee independence

u=(1,0,0), v=(0,1,0), w=(1,1,0) are pairwise nonparallel, but u+v-w=0. They span a plane, so the triple is dependent. Three vectors are independent exactly when they do not all lie in a plane through the origin. Four vectors in R³ must be dependent

Are (1,0,0), (0,1,0), (1,1,0) linearly independent?

Use component equations to prove uniqueness

For u=(1,1,0), v=(0,1,1), w=(1,0,1), solve c=αu+βv+γw for c=(5,1,2). The equations α+γ=5, α+β=1, β+γ=2 give α=2, β=-1, γ=3. The corresponding homogeneous equations have only the zero solution, so this is a basis and the representation is unique

For u=(1,1,0), v=(0,1,1), w=(1,0,1) and c=(5,1,2), find β in c=αu+βv+γw.

Find a parameter where independence changes

For u=(1,0,2), v=(0,1,-1), w=(2,-3,t), the first two coordinates force w=2u-3v if w lies in their span. The third component must therefore be t=7. At t=7 the triple is dependent; at every other t it is independent

For u=(1,0,2), v=(0,1,-1) and w=(2,-3,t), find the value of t for which the three vectors are dependent

Derive spatial magnitude and dot-product rules, ungraded

Derive the norm formula using perpendicular triangles and the component dot-product formula using the orthonormal basis. Deduce the direction-cosine identity and explain precisely why an angle or same-direction unit vector requires a nonzero vector

Model: state the geometric and algebraic hypotheses

The x-y projection has squared length x²+y²; the perpendicular z contribution adds z². In the basis expansion, i·i=j·j=k·k=1 and all cross-basis dot products are 0, leaving matching coordinate products. Dividing x²+y²+z²=|a|² by the positive |a|² gives cos²α+cos²β+cos²γ=1. If a=0 the required divisions are undefined and no direction is selected

Prove the projection decomposition and closest-point property, ungraded

For b≠0, derive the unique decomposition a=λb+r with r·b=0. Prove that λb is the closest point to a on the line spanned by b. Explain what changes if b is reversed and why b=0 is excluded

Model: use orthogonality, then complete the squared distance

Dot the decomposition with b to find the unique coefficient. The remainder is perpendicular to b, so the cross term vanishes in the squared distance to another point on the line. The positive final coefficient gives the unique minimum at t=λ. Reversing b changes λ but leaves the projection vector fixed

Justify dependence and basis claims, ungraded

Give full coefficient arguments for these statements: a set containing zero is dependent; two nonzero vectors are dependent exactly when parallel; three pairwise nonparallel vectors can be dependent; and coordinates relative to three independent vectors are unique. Explain why a common endpoint is needed when interpreting their span geometrically

Model: supply nontrivial relations and subtract representations

Give coefficient 1 to a zero vector and 0 to the rest. A scalar multiple produces a two-vector relation; conversely a nontrivial two-vector relation with nonzero vectors yields a nonzero multiplier. The triple i,j,i+j has relation i+j-(i+j)=0. If two basis representations agree, their difference is a homogeneous relation; independence forces every coefficient difference to vanish. Translate free vectors to a common tail when interpreting the line or plane of their linear span

Solve and classify a changing three-vector system, ungraded

For u=(1,0,2), v=(0,1,-1), w=(2,-3,t), solve αu+βv+γw=0 for t=7 and t≠7. Then describe which targets have unique coordinates in each case, using all three equations rather than a numerical determinant alone

Model: isolate the remaining pivot

The first two equations give α=-2γ and β=3γ. The third becomes (t-7)γ=0. For t≠7, every coefficient is zero and the three vectors form a basis of R³. For t=7, every γ gives the relation (-2γ,3γ,γ); targets in the span of u,v have infinitely many triples of coefficients, while targets outside that plane have none

Construct and verify a spatial resolute diagram, ungraded

For a=(2,1,-2), b=(0,3,0), draw the vectors from a common origin, the parallel vector P and perpendicular remainder R. Label all spatial coordinates. Explain the two-dimensional view used in this lesson and verify its lengths and right angle algebraically

Model: distinguish the viewing coordinates from world coordinates

P=(0,1,0) and R=(2,0,-2), with R·b=0. A unit direction along R is (1/√2,0,-1/√2), perpendicular to b/|b|=(0,1,0). In this orthonormal plane basis, a has coordinates (1,√8), b has coordinates (3,0), and P=(1,0). The figure uses those plane coordinates; it is not a world x-y projection or a full three-dimensional drawing. A learner-created spatial sketch remains ungraded

Assessment boundary

Structured practice checks the stated components, exact radical forms and bounded dependence conclusions. It does not certify independent geometric proof, a complete spatial sketch, arbitrary symbolic linear algebra or a validated physical model. Cross products, lines and planes follow after these foundations