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The scalar product

The scalar product

The definition

uvα

The scalar product of two vectors is a number:

u‾⋅v‾\displaystyle \underline{u}\cdot\underline{v} =\displaystyle = ∣u‾∣⋅∣v‾∣⋅cos⁡α\displaystyle |\underline{u}|\cdot|\underline{v}|\cdot\cos\alpha
  • α\displaystyle \alpha is the angle between the vectors when they leave the same point, 0∘≤α≤180∘\displaystyle 0^{\circ}\le\alpha\le180^{\circ}.
  • The dot on the left is an operation between vectors; on the right – a product of numbers.
  • For example: ∣u‾∣=3\displaystyle {|\underline{u}|}={3}, ∣v‾∣=2\displaystyle {|\underline{v}|}={2}, α=60∘\displaystyle {\alpha}={60^{\circ}}: u‾⋅v‾=3⋅2⋅12=3\displaystyle {\underline{u}\cdot\underline{v}}={3\cdot2\cdot\frac{1}{2}}={3}.
The sign and the angle

The lengths are positive, so the sign of u‾⋅v‾\displaystyle \underline{u}\cdot\underline{v} is the sign of cos⁡α\displaystyle \cos\alpha:

  • u‾⋅v‾>0\displaystyle \underline{u}\cdot\underline{v}>0 – the angle is acute;
  • u‾⋅v‾<0\displaystyle \underline{u}\cdot\underline{v}<0 – the angle is obtuse;
  • u‾⋅v‾=0\displaystyle {\underline{u}\cdot\underline{v}}={0} – the angle is right, or one of the vectors is 0‾\displaystyle \underline{0}.
u‾⊥v‾  ⟺  u‾⋅v‾\displaystyle \underline{u}\perp\underline{v}\iff\underline{u}\cdot\underline{v} =\displaystyle = 0\displaystyle 0

When the product is known, the angle comes from cos⁡α=u‾⋅v‾∣u‾∣⋅∣v‾∣\displaystyle {\cos\alpha}={\frac{\underline{u}\cdot\underline{v}}{|\underline{u}|\cdot|\underline{v}|}}.

The laws of the product
  • u‾⋅v‾=v‾⋅u‾\displaystyle {\underline{u}\cdot\underline{v}}={\underline{v}\cdot\underline{u}};
  • (tu‾)⋅v‾=t(u‾⋅v‾)\displaystyle {(t\underline{u})\cdot\underline{v}}={t(\underline{u}\cdot\underline{v})};
  • u‾⋅(v‾+w‾)=u‾⋅v‾+u‾⋅w‾\displaystyle {\underline{u}\cdot(\underline{v}+\underline{w})}={\underline{u}\cdot\underline{v}+\underline{u}\cdot\underline{w}};
  • u‾⋅u‾=∣u‾∣2\displaystyle {\underline{u}\cdot\underline{u}}={|\underline{u}|^2}, since the angle between a vector and itself is 0∘\displaystyle 0^{\circ}. We write u‾2\displaystyle \underline{u}^2.

So brackets are opened as in algebra, and the short multiplication formulas still hold:

(u‾+v‾)2\displaystyle (\underline{u}+\underline{v})^2 =\displaystyle = u‾2+2u‾⋅v‾+v‾2\displaystyle \underline{u}^2+2\underline{u}\cdot\underline{v}+\underline{v}^2

and (u‾+v‾)⋅(u‾−v‾)=u‾2−v‾2\displaystyle {(\underline{u}+\underline{v})\cdot(\underline{u}-\underline{v})}={\underline{u}^2-\underline{v}^2}. This is how a length is found: ∣w‾∣=w‾2\displaystyle {|\underline{w}|}={\sqrt{\underline{w}^2}}.

Example: a length and an angle

∣u‾∣=3\displaystyle {|\underline{u}|}={3}, ∣v‾∣=2\displaystyle {|\underline{v}|}={2}, and the angle between them is 60∘\displaystyle 60^{\circ}.
Find ∣u‾+v‾∣\displaystyle |\underline{u}+\underline{v}| and the angle between u‾\displaystyle \underline{u} and u‾+v‾\displaystyle \underline{u}+\underline{v}.

The solution below first finds u‾⋅v‾\displaystyle \underline{u}\cdot\underline{v}, and then opens brackets.

  • u‾⋅v‾=3⋅2⋅cos⁡60∘=3\displaystyle {\underline{u}\cdot\underline{v}}={3\cdot2\cdot\cos60^{\circ}}={3}.
  • (u‾+v‾)2=u‾2+2u‾⋅v‾+v‾2=9+6+4=19\displaystyle {(\underline{u}+\underline{v})^2}={\underline{u}^2+2\underline{u}\cdot\underline{v}+\underline{v}^2}={9+6+4}={19}, so ∣u‾+v‾∣=19\displaystyle {|\underline{u}+\underline{v}|}={\sqrt{19}}.
  • u‾⋅(u‾+v‾)=u‾2+u‾⋅v‾=9+3=12\displaystyle {\underline{u}\cdot(\underline{u}+\underline{v})}={\underline{u}^2+\underline{u}\cdot\underline{v}}={9+3}={12}.
  • cos⁡β=12319\displaystyle {\cos\beta}={\frac{12}{3\sqrt{19}}}, so β≈23.41∘\displaystyle \beta\approx23.41^{\circ}.
Exercise 1

∣u‾∣=4\displaystyle {|\underline{u}|}={4}, ∣v‾∣=5\displaystyle {|\underline{v}|}={5}, and the angle between them is 120∘\displaystyle 120^{\circ}.

a
u‾⋅v‾\displaystyle \underline{u}\cdot\underline{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
    ∣u‾+v‾∣\displaystyle |\underline{u}+\underline{v}|, exactly

      / – 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
      Open the brackets

        / – 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
        (2u‾−v‾)⋅(u‾+3v‾)\displaystyle (2\underline{u}-\underline{v})\cdot(\underline{u}+3\underline{v}) – put in the numbers

          / – 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

          ∣u‾∣=∣v‾∣=1\displaystyle {|\underline{u}|}={|\underline{v}|}={1}, and ∣u‾+v‾∣=3\displaystyle {|\underline{u}+\underline{v}|}={\sqrt{3}}.

          a
          u‾⋅v‾\displaystyle \underline{u}\cdot\underline{v} (hint: (u‾+v‾)2=3\displaystyle {(\underline{u}+\underline{v})^2}={3})

            / – 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
            The angle between u‾\displaystyle \underline{u} and v‾\displaystyle \underline{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

              ∣u‾∣=2\displaystyle {|\underline{u}|}={2}, ∣v‾∣=3\displaystyle {|\underline{v}|}={3}, and the angle between them is 60∘\displaystyle 60^{\circ}.
              Find t\displaystyle t for which u‾+tv‾\displaystyle \underline{u}+t\underline{v} is perpendicular to u‾−v‾\displaystyle \underline{u}-\underline{v}. (Hint: open (u‾+tv‾)⋅(u‾−v‾)\displaystyle (\underline{u}+t\underline{v})\cdot(\underline{u}-\underline{v}), put in the numbers, and set it equal to 0.)

                Exercise 4

                True or false? Explain.

                1. If u‾⋅v‾=u‾⋅w‾\displaystyle {\underline{u}\cdot\underline{v}}={\underline{u}\cdot\underline{w}} and u‾≠0‾\displaystyle \underline{u}\neq\underline{0}, then v‾=w‾\displaystyle {\underline{v}}={\underline{w}}.
                2. If ∣u‾+v‾∣=∣u‾−v‾∣\displaystyle {|\underline{u}+\underline{v}|}={|\underline{u}-\underline{v}|}, then u‾⊥v‾\displaystyle \underline{u}\perp\underline{v}.