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The scalar product

Space geometry and vectors, grade 12 (4 units) Lesson 14 of 20Learning

The scalar product

The angle between two vectors

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The angle between two vectors is measured with both leaving the same point. It is between 0∘\displaystyle 0^{\circ} and 180∘\displaystyle 180^{\circ} (in this chapter no vector is the zero vector).

  • Collinear vectors in the same direction: 0∘\displaystyle 0^{\circ}; in opposite directions: 180∘\displaystyle 180^{\circ}.

αABC

Careful: in the triangle ABC\displaystyle ABC, the angle between AB→\displaystyle \overrightarrow{AB} and BC→\displaystyle \overrightarrow{BC} is not ∠B\displaystyle \angle B. BC→\displaystyle \overrightarrow{BC} leaves B\displaystyle B, and AB→\displaystyle \overrightarrow{AB} arrives at it; for both to leave B\displaystyle B, carry AB\displaystyle AB on past B\displaystyle B. The angle is α=180∘−∠B\displaystyle {\alpha}={180^{\circ}-\angle B}.

The cosine of a right and an obtuse angle

In grade 10 you learnt sin⁡(180∘−α)=sin⁡α\displaystyle {\sin(180^{\circ}-\alpha)}={\sin\alpha}. The cosine of an obtuse angle is defined thus:

cos⁡(180∘−α)\displaystyle \cos(180^{\circ}-\alpha) =\displaystyle = −cos⁡α\displaystyle -\cos\alpha
  • cos⁡90∘=0\displaystyle {\cos90^{\circ}}={0}, and cos⁡180∘=−1\displaystyle {\cos180^{\circ}}={-1}.
  • cos⁡120∘=−cos⁡60∘=−12\displaystyle {\cos120^{\circ}}={-\cos60^{\circ}}={-\frac{1}{2}}; cos⁡135∘=−22\displaystyle {\cos135^{\circ}}={-\frac{\sqrt{2}}{2}}; cos⁡150∘=−32\displaystyle {\cos150^{\circ}}={-\frac{\sqrt{3}}{2}}.
  • An acute angle – a positive cosine; an obtuse angle – a negative cosine. The calculator does this itself, and cos⁡−1\displaystyle \cos^{-1} of a negative number gives an obtuse angle.
The definition of the scalar product

The scalar product of u⃗\displaystyle \vec{u} and v⃗\displaystyle \vec{v}, with the angle α\displaystyle \alpha between them, is a number:

u⃗⋅v⃗\displaystyle \vec{u}\cdot\vec{v} =\displaystyle = ∣u⃗∣⋅∣v⃗∣⋅cos⁡α\displaystyle |\vec{u}|\cdot|\vec{v}|\cdot\cos\alpha
  • α\displaystyle \alpha acute – the product is positive; α=90∘\displaystyle {\alpha}={90^{\circ}} – the product is 0; α\displaystyle \alpha obtuse – the product is negative.
  • The angle from the product: cos⁡α=u⃗⋅v⃗∣u⃗∣⋅∣v⃗∣\displaystyle {\cos\alpha}={\frac{\vec{u}\cdot\vec{v}}{|\vec{u}|\cdot|\vec{v}|}}.
  • Example: ∣u⃗∣=4\displaystyle {|\vec{u}|}={4}, ∣v⃗∣=3\displaystyle {|\vec{v}|}={3}, and the angle between them 60∘\displaystyle 60^{\circ}: u⃗⋅v⃗=4⋅3⋅12=6\displaystyle {\vec{u}\cdot\vec{v}}={4\cdot3\cdot\frac{1}{2}}={6}.
The properties of the scalar product
  • A vector’s length: the angle between u⃗\displaystyle \vec{u} and itself is 0∘\displaystyle 0^{\circ}, so u⃗⋅u⃗=u⃗2=∣u⃗∣2\displaystyle {\vec{u}\cdot\vec{u}}={\vec{u}^2}={|\vec{u}|^2}.
∣u⃗∣\displaystyle |\vec{u}| =\displaystyle = u⃗2\displaystyle \sqrt{\vec{u}^2}
u⃗⋅v⃗\displaystyle \vec{u}\cdot\vec{v} =\displaystyle = v⃗⋅u⃗\displaystyle \vec{v}\cdot\vec{u}
u⃗⋅(tv⃗)\displaystyle \vec{u}\cdot(t\vec{v}) =\displaystyle = t(u⃗⋅v⃗)\displaystyle t(\vec{u}\cdot\vec{v})
u⃗⋅(v⃗+w⃗)\displaystyle \vec{u}\cdot(\vec{v}+\vec{w}) =\displaystyle = u⃗⋅v⃗+u⃗⋅w⃗\displaystyle \vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}

The distributive law holds (without proof), so the short multiplication formulas hold too – the short scalar products:

(u⃗+v⃗)2\displaystyle (\vec{u}+\vec{v})^2 =\displaystyle = u⃗2+2u⃗⋅v⃗+v⃗2\displaystyle \vec{u}^2+2\vec{u}\cdot\vec{v}+\vec{v}^2
(u⃗+v⃗)⋅(u⃗−v⃗)\displaystyle (\vec{u}+\vec{v})\cdot(\vec{u}-\vec{v}) =\displaystyle = u⃗2−v⃗2\displaystyle \vec{u}^2-\vec{v}^2
Example: the length of a sum

∣u⃗∣=4\displaystyle {|\vec{u}|}={4}, ∣v⃗∣=3\displaystyle {|\vec{v}|}={3}, and the angle between them is 60∘\displaystyle 60^{\circ}, so u⃗⋅v⃗=6\displaystyle {\vec{u}\cdot\vec{v}}={6}.
Find ∣u⃗+v⃗∣\displaystyle |\vec{u}+\vec{v}| and ∣u⃗−v⃗∣\displaystyle |\vec{u}-\vec{v}|.

∣u⃗+v⃗∣2=u⃗2+2u⃗⋅v⃗+v⃗2=16+12+9=37,∣u⃗+v⃗∣=37|\vec{u}+\vec{v}|^2=\vec{u}^2+2\vec{u}\cdot\vec{v}+\vec{v}^2=16+12+9=37,\qquad|\vec{u}+\vec{v}|=\sqrt{37}∣u⃗−v⃗∣2=16−12+9=13,∣u⃗−v⃗∣=13|\vec{u}-\vec{v}|^2=16-12+9=13,\qquad|\vec{u}-\vec{v}|=\sqrt{13}

In the card the calculation is written in numbers: 42+2⋅6+32\displaystyle \sqrt{4^{2}+2\cdot6+3^{2}}.

Exercise 1

∣u⃗∣=5\displaystyle {|\vec{u}|}={5}, ∣v⃗∣=2\displaystyle {|\vec{v}|}={2}. Find u⃗⋅v⃗\displaystyle \vec{u}\cdot\vec{v} exactly, when the angle between them is:

a
60∘\displaystyle 60^{\circ}

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b
    90∘\displaystyle 90^{\circ}

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

      c
      120∘\displaystyle 120^{\circ}

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        d
        135∘\displaystyle 135^{\circ}

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

          Exercise 2

          ∣u⃗∣=3\displaystyle {|\vec{u}|}={3}, ∣v⃗∣=5\displaystyle {|\vec{v}|}={5}, and u⃗⋅v⃗=−7.5\displaystyle {\vec{u}\cdot\vec{v}}={-7.5}.
          What is the angle between u⃗\displaystyle \vec{u} and v⃗\displaystyle \vec{v}?

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

            Exercise 3

            ∣u⃗∣=2\displaystyle {|\vec{u}|}={2}, ∣v⃗∣=3\displaystyle {|\vec{v}|}={3}, and the angle between them is 60∘\displaystyle 60^{\circ}.
            Find exactly:

            a
            ∣u⃗+v⃗∣\displaystyle |\vec{u}+\vec{v}|

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

              b
              ∣2u⃗−v⃗∣\displaystyle |2\vec{u}-\vec{v}|

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                c
                (u⃗+v⃗)⋅(u⃗−v⃗)\displaystyle (\vec{u}+\vec{v})\cdot(\vec{u}-\vec{v})

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                  Exercise 4

                  4ABC

                  ABC\displaystyle ABC is an equilateral triangle of side 4. Find:

                  a
                  AB→⋅AC→\displaystyle \overrightarrow{AB}\cdot\overrightarrow{AC}

                    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                    b
                    AB→⋅BC→\displaystyle \overrightarrow{AB}\cdot\overrightarrow{BC}

                      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power