F3, F4 - Scalar products and angles

F3, F4 - Scalar products and angles

Can multiplying two vectors give a negative number? F3 introduces the scalar product and connects that number to an angle; F4 uses a zero scalar product to detect perpendicular directions. You will distinguish the angle between directed vectors from the smaller angle between unoriented lines.

Dot product in components and magnitude-angle form

Can multiplying two vectors give a negative number? Take a=(1,2,2)\mathbf a=(1,2,2) and b=(2,−1,−2)\mathbf b=(2,-1,-2). Multiply matching components and add: 1(2)+2(−1)+2(−2)=−41(2)+2(-1)+2(-2)=-4. This operation produces a scalar, a number with no direction, rather than another vector. It is called the scalar product or dot product, written a⋅b\mathbf a\cdot\mathbf b.

To see its connection with geometry, put the vectors tail to tail. The cosine rule gives

∣a−b∣2=∣a∣2+∣b∣2−2∣a∣∣b∣cos⁡θ|\mathbf a-\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2-2|\mathbf a||\mathbf b|\cos\theta

Expanding the squared components of a−b\mathbf a-\mathbf b gives the same expression with the final term −2(a1b1+a2b2+a3b3)-2(a_1b_1+a_2b_2+a_3b_3). Comparing them explains why the algebra measures the angle.

Scalar product

For a=(a1,a2,a3)\mathbf a=(a_1,a_2,a_3), b=(b1,b2,b3)\mathbf b=(b_1,b_2,b_3),

a⋅b=a1b1+a2b2+a3b3\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3

For nonzero a,b\mathbf a,\mathbf b, it also equals ∣a∣∣b∣cos⁡θ|\mathbf a||\mathbf b|\cos\theta, where 0≤θ≤π0\le\theta\le\pi. Thus

θ=cos⁡−1 ⁣(a⋅b∣a∣∣b∣)\theta=\cos^{-1}\!\left(\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}\right)

Not in the formula booklet: learn it.

The component formula is valid even for a zero vector. The angle formula is not: a zero vector has no direction and would give a zero denominator. Notice also that a⋅a=∣a∣2\mathbf a\cdot\mathbf a=|\mathbf a|^2. The dot product is symmetric, so reversing the order of its two factors does not change its value.

Worked example 1: a negative dot product

Question. For a=(1,2,2)\mathbf a=(1,2,2), b=(2,−1,−2)\mathbf b=(2,-1,-2), find a⋅b\mathbf a\cdot\mathbf b and the angle between the vectors, to one decimal place.

Plan. Calculate the component product and the two magnitudes, then use inverse cosine.

a⋅b=2−2−4=−4sum matching products∣a∣=1+4+4=3,∣b∣=4+1+4=3cos⁡θ=−49divide by the magnitude product\begin{aligned} \mathbf a\cdot\mathbf b&=2-2-4=-4 && \text{sum matching products}\\ |\mathbf a|&=\sqrt{1+4+4}=3,\quad |\mathbf b|=\sqrt{4+1+4}=3\\ \cos\theta&=-\frac49 && \text{divide by the magnitude product} \end{aligned}

Answer. a⋅b=−4\mathbf a\cdot\mathbf b=-4, θ=116.4∘\theta=116.4^\circ, to one decimal place.

Check. a−b=(−1,3,4)\mathbf a-\mathbf b=(-1,3,4), whose squared length is 2626. The cosine rule gives 9+9−2(3)(3)(−4/9)=269+9-2(3)(3)(-4/9)=26, independently confirming the cosine.

Now reverse only the second vector. Its magnitude stays the same, but the dot product changes sign.

Worked example 2: reversing one vector

Question. Find the dot product and angle for a=(1,2,2)\mathbf a=(1,2,2), c=(−2,1,2)=−b\mathbf c=(-2,1,2)=-\mathbf b.

Plan. Recalculate with the reversed direction and compare the angle.

a⋅c=−2+2+4=4∣c∣=3cos⁡ϕ=49\begin{aligned} \mathbf a\cdot\mathbf c&=-2+2+4=4\\ |\mathbf c|&=3\\ \cos\phi&=\frac49 \end{aligned}

Answer. The dot product is 44, and ϕ=63.6∘\phi=63.6^\circ, to one decimal place.

Check. a−c=(3,1,0)\mathbf a-\mathbf c=(3,1,0), with squared length 1010. The cosine rule gives 18−18(4/9)=1018-18(4/9)=10. The two angles are supplementary, as reversing one arrow predicts.

[DIAGRAM: asset_name: Reversing a vector changes the angle; asset_slug: aqa7367-vector-angle; caption: Reversing one vector changes an obtuse vector angle to its acute supplement; description: Two equal-scale diagrams in the plane spanned by the vectors. The blue vector a points right and has length three. In the left panel the clay vector b has length three and forms an angle of 116.4 degrees with a. In the right panel the clay vector minus b is reversed and forms 63.6 degrees with a. The dot products are labelled minus four and plus four. These are direction diagrams, not projections onto coordinate axes.]
Reversing one vector changes an obtuse vector angle to its acute supplement

The negative answer to the opening question has a geometric meaning: an obtuse vector angle has negative cosine. For nonzero vectors, a positive dot product means 0≤θ<90∘0\le\theta<90^\circ; a negative product means 90∘<θ≤180∘90^\circ<\theta\le180^\circ; and zero means a right angle. The endpoint cases are equally directed vectors at 0∘0^\circ and oppositely directed vectors at 180∘180^\circ. All these possibilities belong to the same formula.

Perpendicularity

For nonzero vectors, perpendicular directions make an angle of 90∘90^\circ, whose cosine is zero. Since both magnitudes are positive, the dot-product formula gives a quick test in either direction.

Perpendicular vectors

Two nonzero vectors a,b\mathbf a,\mathbf b are perpendicular exactly when a⋅b=0\mathbf a\cdot\mathbf b=0. Their matching component products cancel in the sum; the individual products need not be zero.

This turns a geometric condition into an ordinary algebraic equation, useful when a component is unknown.

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