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Concept 04 / 06Vector Algebra
Components, Direction Cosines and Linear Independence of Vectors
Resolving vectors along i, j, k, magnitude in components, direction cosines and direction ratios with l^2 + m^2 + n^2 = 1, component-wise operations, linear combinations, collinearity and coplanarity tests, and expressing a vector in terms of three non-coplanar vectors.
Figure 1
Components. With unit vectors i^,j^,k^ along the x, y and z axes, every vector can be written uniquely as
r=xi^+yj^+zk^,
where x,y,z are its components (the coordinates of its head when its tail is at the origin). Then ∣r∣=x2+y2+z2, and addition, subtraction and scalar multiplication act componentwise. Two vectors are equal iff their components are equal, and parallel iff their components are proportional.
Direction cosines. If r makes angles α,β,γ with the positive x, y and z axes, its direction cosines are l=cosα, m=cosβ, n=cosγ. Since x=∣r∣cosα and so on,
l=∣r∣x, m=∣r∣y, n=∣r∣z, and l2+m2+n2=1.
So r^=li^+mj^+nk^: the direction cosines are simply the components of the unit vector. Equivalently, cos2α+cos2β+cos2γ=1 and sin2α+sin2β+sin2γ=2. Direction ratios are any three numbers a,b,c proportional to l,m,n; to convert, divide by a2+b2+c2 (with a ± sign, since a line has two opposite directions). A vector cannot make 45∘ with all three axes, since 3×21=1; a vector equally inclined to all three axes has l=m=n=±31.
Linear combinations and independence. Vectors a1,…,an are linearly independent if x1a1+⋯+xnan=0 forces every xi=0; otherwise they are dependent.
Two vectors are dependent iff they are collinear.
Three vectors are dependent iff they are coplanar. In components, this means the determinant of their components is zero (equivalently, their scalar triple product is zero).
Any four vectors in space are dependent.
Basis. Any two non-collinear vectors a,b span their plane: every vector r in that plane is uniquely xa+yb. Any three non-coplanar vectors span space: every r is uniquely xa+yb+zc. The uniqueness is what allows 'comparing coefficients' in vector equations: if a,b,c are non-coplanar and xa+yb+zc=x′a+y′b+z′c, then x=x′, y=y′, z=z′. To find the coefficients, write everything in i^,j^,k^ components and solve three linear equations.
Key Takeaways:
- ∣xi^+yj^+zk^∣=x2+y2+z2.
- Direction cosines are the components of the unit vector; l2+m2+n2=1.
- sin2α+sin2β+sin2γ=2.
- Direction ratios a,b,c: l=a2+b2+c2±a.
- Two vectors dependent ⟺ collinear; three dependent ⟺ coplanar ⟺ component determinant =0.
- Three non-coplanar vectors form a basis; coefficients are unique.
- Pitfall: comparing coefficients with respect to vectors that are coplanar.
Brain-Teaser 1: A vector makes angles 60∘ and 45∘ with the x- and y-axes. Find the angle with the z-axis and the vector of magnitude 4 in that direction.
Solution: cos260∘+cos245∘+cos2γ=1 gives 41+21+cos2γ=1, so cosγ=±21 and γ=60∘ or 120∘. The vectors are 4(21i^+21j^±21k^)=2i^+22j^±2k^. The identity l2+m2+n2=1 determines the third angle only up to the sign of its cosine.
Brain-Teaser 2: For what λ are i^+j^+k^, 2i^+λj^+3k^ and i^+2j^+3k^ coplanar?
Solution: Set the component determinant to zero: 1211λ2133=1(3λ−6)−1(6−3)+1(4−λ)=3λ−6−3+4−λ=2λ−5. So λ=25. At that value one vector is a combination of the other two: here 2i^+25j^+3k^=23(i^+j^+k^)+21(i^+2j^+3k^), which can be checked component by component.
Brain-Teaser 3: Express r=4i^+5j^+6k^ as xa+yb+zc with a=i^+j^, b=j^+k^, c=k^+i^.
Solution: Components: x+z=4, x+y=5, y+z=6. Adding: 2(x+y+z)=15, so x+y+z=215. Then y=215−4=27, z=215−5=25, x=215−6=23. So r=23a+27b+25c. The coefficients are unique because a,b,c are non-coplanar (their determinant is 2=0).
