Use scalar products, angles and projections

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

For a=(2,−3), b=(4,1), find a·b.

Find a·a for a=(2,−3)

Connect the product to an included angle

For nonzero vectors, a·b=|a||b|cosθ, where 0≤θ≤π is the included angle. Both norms in the denominator must be nonzero. A positive product gives an angle less than 90°, zero gives 90°, and a negative product gives an angle greater than 90°. Endpoints 0° and 180° represent same and opposite directions. For other angles, evaluate inverse cosine in the requested degree or radian mode; retain the exact norm product until final rounding

For a=(1,0), b=(−1,1), find the coefficient c in cosθ=c√2.

For a=(1,0) and b=(−1,1), find the included angle in degrees

If a=0 and b≠0, what can you conclude from a·b=0?

Perpendicular and parallel tests are different

Two nonzero vectors are perpendicular exactly when a·b=0. They are parallel exactly when one is a nonzero scalar multiple of the other; in the plane this is equivalent to a₁b₂−a₂b₁=0. Same and opposite directions are both parallel. Never divide by a zero component when testing proportionality

For a=(t,2), b=(3,−6), find t so that a·b=0.

For a=(0,2) and b=(0,−3), which statement holds?

A scalar projection is signed

For b≠0, comp_b(a)=a·b/|b| is the signed component of a in the direction b. It is negative if the included angle is obtuse. The nonnegative length of the projection is its absolute value. Reversing b reverses this scalar sign

Let a=(−2,4), b=(3,0). Find a·b.

For a=(−2,4) and b=(3,0), find the signed scalar projection of a onto b.

For a=(−2,4) and b=(3,0), find the nonnegative length of the projection of a onto b.

A vector projection has two components

Multiply the signed scalar projection by the unit vector in direction b. Equivalently, divide a·b by |b|² and multiply by b. This projects onto the entire line spanned by b, not only the ray in its positive direction

For a=(−2,4), b=(3,0), find the x-component of proj_b(a)

Replace b by −b. What happens to the vector projection of a onto its line?

Decompose into perpendicular parts

Let p=proj_b(a) and r=a−p. Then p is a scalar multiple of b (possibly zero), r·b=0, and a=p+r. Check both components of the sum and the zero scalar product. Since p·r=0, |a|²=|p|²+|r|². The projection is valid for a=0, provided b≠0. A zero part has zero scalar product with b; no geometric angle is assigned to that zero vector

For a=(5,1) and b=(1,2), let p be the projection of a onto b. Find λ in p=λb.

For a=(5,1) and b=(1,2), let p be the projection of a onto b. Find the y-component of p.

For a=(5,1) and b=(1,2), let p be the projection of a onto b and r=a−p. Find the x-component of r.

Derive the angle law, ungraded

Derive a·b=|a||b|cosθ for nonzero a,b using the cosine rule and a component expansion of |a−b|². Explain the zero-vector exception and justify the perpendicularity test. This derivation is ungraded

Model: compare two expressions for the same length

The cosine rule gives |a−b|²=|a|²+|b|²−2|a||b|cosθ. Expanding components gives |a−b|²=|a|²+|b|²−2a·b. Equating and cancelling gives the formula. For nonzero vectors both norms are positive, so a·b=0 exactly when cosθ=0, hence θ=π/2. The zero vector has no direction, so the geometric angle statement cannot be extended to it

Prove and sketch the projection, ungraded

For b≠0, derive λ such that r=a−λb is perpendicular to b. Prove that λb is the unique closest point to a on the line through the origin in direction b. Sketch a=(−2,4), b=(3,0), including the foot on the opposite side of the origin. The derivation and sketch are ungraded

Model: impose orthogonality and compare squared distances

Set 0=(a−λb)·b=a·b−λ(b·b), so λ=(a·b)/(b·b). Since b·b>0 this is unique. Put p=λb, r=a−p. For any point tb on the line, a−tb=r+(λ−t)b; the cross term is zero, so |a−tb|²=|r|²+(λ−t)²|b|². This is minimised uniquely at t=λ. In the sketch p=(−2,0),r=(0,4); signed component −2,projection length 2 and shortest distance 4.