CP 6.3, 6.4 - The scalar product and normal vectors

CP 6.3, 6.4 - The scalar product and normal vectors

The scalar product turns two vectors into a number that measures their alignment. You will calculate it, use zero scalar products to test perpendicularity (CP 6.4), and express a plane using a normal vector and the equation r.n=k\mathbf r.\mathbf n=k (CP 6.3).

A number that measures alignment

Can two vectors with no zero components be perpendicular? The coordinate axes make right angles look easy; the scalar product will detect them in any direction.

Try u=(1,2,2)T\mathbf u=(1,2,2)^{\mathrm T} and v=(2,−1,0)T\mathbf v=(2,-1,0)^{\mathrm T}. Multiplying corresponding entries and adding gives 2−2+0=02-2+0=0. That cancellation is the signal we will explain.

You already know

For u=(u1,u2,u3)T\mathbf u=(u_1,u_2,u_3)^{\mathrm T}, its magnitude is ∣u∣=u12+u22+u32|\mathbf u|=\sqrt{u_1^2+u_2^2+u_3^2}. A plane has two non-parallel direction vectors; a line has one non-zero direction vector.

The scalar product is also called the dot product; Pearson writes the dot on the line between the vectors.

Scalar product

For u=(u1,u2,u3)T\mathbf u=(u_1,u_2,u_3)^{\mathrm T} and v=(v1,v2,v3)T\mathbf v=(v_1,v_2,v_3)^{\mathrm T},

u.v=u1v1+u2v2+u3v3\mathbf u.\mathbf v=u_1v_1+u_2v_2+u_3v_3

The result is a scalar, not a vector. For example, (1,2,2)T.(2,−1,3)T=6(1,2,2)^{\mathrm T}.(2,-1,3)^{\mathrm T}=6.

Why should coordinate multiplication know about angles? In the triangle formed by u,v\mathbf u,\mathbf v, the cosine rule gives ∣u−v∣2=∣u∣2+∣v∣2−2∣u∣∣v∣cos⁡θ|\mathbf u-\mathbf v|^2=|\mathbf u|^2+|\mathbf v|^2-2|\mathbf u||\mathbf v|\cos\theta. Expanding the same squared length in coordinates gives the first two terms minus 2(u1v1+u2v2+u3v3)2(u_1v_1+u_2v_2+u_3v_3). Comparing the two gives the link.

Geometric meaning of the scalar product

For non-zero vectors and their angle 0≤θ≤π0\le\theta\le\pi,

u.v=∣u∣∣v∣cos⁡θ\mathbf u.\mathbf v=|\mathbf u||\mathbf v|\cos\theta

Thus a positive product means an acute vector angle, zero means a right angle, and a negative product means an obtuse vector angle. The coordinate rule also gives u.u=∣u∣2\mathbf u.\mathbf u=|\mathbf u|^2. Not in the formula booklet: learn it.

Calculate with signs intact: a negative component contributes its signed product.

Worked example 1: a positive scalar product

Question. Find u.v\mathbf u.\mathbf v for u=(1,2,2)T\mathbf u=(1,2,2)^{\mathrm T}, v=(2,−1,3)T\mathbf v=(2,-1,3)^{\mathrm T}, and state whether their angle is acute or obtuse.

Plan. Add the three coordinate products and use the sign.

u.v=1(2)+2(−1)+2(3)corresponding coordinates=2−2+6=6keep the negative sign\begin{aligned} \mathbf u.\mathbf v&=1(2)+2(-1)+2(3) && \text{corresponding coordinates} \\ &=2-2+6=6 && \text{keep the negative sign} \end{aligned}

Why this step? The geometric formula divides by positive magnitudes, so the product and cos⁡θ\cos\theta have the same sign.

Answer. The scalar product is 66; the vector angle is acute.

Check. ∣u∣=3|\mathbf u|=3, ∣v∣=14|\mathbf v|=\sqrt{14}, and 0<6/(314)<10<6/(3\sqrt{14})<1, consistent with an acute angle.

Worked example 2 changes only the last component of v\mathbf v, from 33 to −3-3.

Worked example 2: a negative scalar product

Question. Keep u\mathbf u, but take v=(2,−1,−3)T\mathbf v=(2,-1,-3)^{\mathrm T}. Find the scalar product and classify the vector angle.

Plan. Recalculate the affected coordinate product and interpret its sign.

u.v=1(2)+2(−1)+2(−3)signed coordinate products=2−2−6=−6collect\begin{aligned} \mathbf u.\mathbf v&=1(2)+2(-1)+2(-3) && \text{signed coordinate products} \\ &=2-2-6=-6 && \text{collect} \end{aligned}

Why this step? A magnitude is non-negative, but a scalar product can be negative because it includes directional information.

Answer. u.v=−6\mathbf u.\mathbf v=-6, so the vector angle is obtuse.

Check. The lengths are unchanged; the cosine has the opposite sign, so the new vector angle is supplementary to the previous one.

That last observation is useful: equal vector lengths do not imply equal scalar products. Turning a vector can change the sign without changing either length.

Watch out

Wrong: u.v=(u1v1,u2v2,u3v3)T\mathbf u.\mathbf v=(u_1v_1,u_2v_2,u_3v_3)^{\mathrm T}. Right: add those products to get one real number. The word “scalar” tells you the type of answer.

Try a new last coordinate.

Testing perpendicularity with a zero product

For non-zero vectors, the scalar product vanishes exactly when the vectors are perpendicular.

In u.v=∣u∣∣v∣cos⁡θ\mathbf u.\mathbf v=|\mathbf u||\mathbf v|\cos\theta, both magnitudes are positive, so the only way to obtain zero is cos⁡θ=0\cos\theta=0. Within 0≤θ≤π0\le\theta\le\pi, that means θ=π/2\theta=\pi/2.

Perpendicular-vector test

For u≠0\mathbf u\ne\mathbf0, v≠0\mathbf v\ne\mathbf0,

u⊥v⟺u.v=0\mathbf u\perp\mathbf v\quad\Longleftrightarrow\quad\mathbf u.\mathbf v=0

The zero vector has zero scalar product with every vector, but has no direction and no defined angle. Not in the formula booklet: learn it.

A parameter in a perpendicularity problem therefore enters a simple equation, rather than an inverse-cosine calculation.

Worked example 3: a component fixed by perpendicularity

Question. Find kk if (1,2,2)T(1,2,2)^{\mathrm T} is perpendicular to (3,k,−2)T(3,k,-2)^{\mathrm T}.

Plan. Set their scalar product to zero and solve the linear equation.

3+2k−4=0perpendicularity2k=1collect constantsk=12solve\begin{aligned} 3+2k-4&=0 && \text{perpendicularity} \\ 2k&=1 && \text{collect constants} \\ k&=\frac12 && \text{solve} \end{aligned}

Why this step? Zero is a condition on the complete scalar product, not on each coordinate product separately.

Answer. k=1/2k=1/2.

Check. 3+2(1/2)−4=03+2(1/2)-4=0, and neither vector is zero.

Worked example 4 changes only the final component of the second vector, from −2-2 to 22.

Worked example 4: the sign changes the required component

Question. Find kk if (1,2,2)T(1,2,2)^{\mathrm T} is perpendicular to (3,k,2)T(3,k,2)^{\mathrm T}.

Plan. Update the last coordinate product, keeping the zero-product condition.

3+2k+4=0new signed product2k=−7collectk=−72solve\begin{aligned} 3+2k+4&=0 && \text{new signed product} \\ 2k&=-7 && \text{collect} \\ k&=-\frac72 && \text{solve} \end{aligned}

Why this step? The positive last product now needs a negative middle product to cancel it.

Answer. k=−7/2k=-7/2.

Check. 3−7+4=03-7+4=0. Both vectors have all three components non-zero, yet are perpendicular: the opening puzzle is solved.

For lines, perpendicular directions can occur even when the lines are skew. A right angle between their directions does not by itself prove that the lines intersect.

Watch out

Wrong: “Both directions have non-zero components, so they cannot be perpendicular.” Right: test the sum of coordinate products; cancellation can occur across all three terms.

The next check varies one component again.

Normal vectors and the equation of a plane

A normal vector points perpendicular to a plane, giving a shorter alternative to its two spanning directions.

If AA and a general point RR are on the plane, then AR→=r−a\overrightarrow{AR}=\mathbf r-\mathbf a is perpendicular to a normal n\mathbf n. Hence (r−a).n=0(\mathbf r-\mathbf a).\mathbf n=0. Expanding the coordinate products gives r.n=a.n\mathbf r.\mathbf n=\mathbf a.\mathbf n.

Normal vector

A normal to a plane is a non-zero vector perpendicular to every direction in the plane. It is enough to check perpendicularity to two non-parallel spanning directions: every plane displacement is a linear combination of them. A vector parallel to the plane is not a normal.

The distributive rule n.(su+tv)=s(n.u)+t(n.v)\mathbf n.(s\mathbf u+t\mathbf v)=s(\mathbf n.\mathbf u)+t(\mathbf n.\mathbf v) follows directly from coordinate expansion, so two zero products make every such product zero.

[DIAGRAM: asset_name: A normal perpendicular to two plane directions; asset_slug: 9fm0-cp41-plane-normal; caption: Checking two independent directions establishes a normal to the whole plane; description: An exact spatial diagram shows the plane through A at (1,2,3), with spanning directions u at (1,0,1) and v at (0,1,2). A clay normal arrow is the vector n at (1,2,minus 1). A side panel gives n dot u equals 1 minus 1 equals 0, and n dot v equals 2 minus 2 equals 0]
Checking two independent directions establishes a normal to the whole plane

Normal form of a plane

For non-zero n=(n1,n2,n3)T\mathbf n=(n_1,n_2,n_3)^{\mathrm T}, the plane through AA has

r.n=a.n=k⟺n1x+n2y+n3z=k\mathbf r.\mathbf n=\mathbf a.\mathbf n=k \quad\Longleftrightarrow\quad n_1x+n_2y+n_3z=k

Thus the coefficients of a Cartesian plane equation form a normal vector. Its Cartesian normal form is supplied in the booklet; the scalar-product expression and calculation rule are not printed. Not in the formula booklet: learn it.

To find a normal from two directions, use two scalar-product equations. You do not need a vector product.

Worked example 5: constructing a normal from two directions

Question. Find a normal and a Cartesian equation of the plane through (1,2,3)(1,2,3), parallel to (1,0,1)T(1,0,1)^{\mathrm T} and (0,1,2)T(0,1,2)^{\mathrm T}.

Plan. Set a general normal's products with both directions to zero, choose a convenient scale, then use the given point.

n=(p,q,s)T,p+s=0,q+2s=0two perpendicularity testss=−1,p=1,q=2choose a non-zero scalek=(1,2,3)T.(1,2,−1)T=1+4−3=2use the point\begin{aligned} \mathbf n&=(p,q,s)^{\mathrm T},&p+s&=0,&q+2s&=0 && \text{two perpendicularity tests} \\ s&=-1,&p&=1,&q&=2 && \text{choose a non-zero scale} \\ k&=(1,2,3)^{\mathrm T}.(1,2,-1)^{\mathrm T}=1+4-3=2 && \text{use the point} \end{aligned}

Why this step? A normal's length does not matter; choosing s=−1s=-1 removes unnecessary fractions.

Answer. One normal is (1,2,−1)T(1,2,-1)^{\mathrm T}; the plane is r.(1,2,−1)T=2\mathbf r.(1,2,-1)^{\mathrm T}=2, or x+2y−z=2x+2y-z=2.

Check. The two direction products are 1−1=01-1=0 and 2−2=02-2=0, while the anchor satisfies the equation.

Worked example 6 changes only the second spanning direction's last component from 22 to 33.

Worked example 6: the normal changes when the plane tilts

Question. Find a normal and Cartesian equation through (1,2,3)(1,2,3), parallel to (1,0,1)T(1,0,1)^{\mathrm T} and (0,1,3)T(0,1,3)^{\mathrm T}.

Plan. Update the second zero-product equation and then recalculate the constant using the same anchor.

p+s=0,q+3s=0normal to both directionss=−1,p=1,q=3choose scalek=1+3(2)−3=4anchor scalar product\begin{aligned} p+s&=0,&q+3s&=0 && \text{normal to both directions} \\ s&=-1,&p&=1,&q&=3 && \text{choose scale} \\ k&=1+3(2)-3=4 && \text{anchor scalar product} \end{aligned}

Why this step? The Cartesian constant belongs to the chosen normal: when the normal changes, recompute that constant too.

Answer. n=(1,3,−1)T\mathbf n=(1,3,-1)^{\mathrm T}, with plane x+3y−z=4x+3y-z=4.

Check. Both direction products vanish, and 1+3(2)−3=41+3(2)-3=4.

This gives another way to recognise equivalent plane equations: multiplying both the normal and the constant by the same non-zero number leaves the plane unchanged.

Watch out

Wrong: use a position vector of a point in the plane as its normal. Right: a normal must be perpendicular to the spanning directions; only then use a point to find kk.

The final check gives a proposed normal, so show both zero products before writing the plane.

Quick Recap

  • Multiply corresponding components and add to calculate a scalar product.
  • u.v=∣u∣∣v∣cos⁡θ\mathbf u.\mathbf v=|\mathbf u||\mathbf v|\cos\theta; for non-zero vectors a zero product means perpendicularity.
  • A plane normal is perpendicular to two independent spanning directions.
  • The plane through a\mathbf a with normal n\mathbf n is r.n=a.n\mathbf r.\mathbf n=\mathbf a.\mathbf n.

Vectors with no zero components can be perpendicular: (1,2,2)T.(3,−7/2,2)T=3−7+4=0(1,2,2)^{\mathrm T}.(3,-7/2,2)^{\mathrm T}=3-7+4=0.