Scalar product of vectors

If we have two vectors, \vec a and \vec b, we write the scalar product as:

\vec a \cdot \vec b

Because of the big dot operator which we use to mean ‘scalar product’, it is also called the dot product.

Scalar product rule for 2D

For vectors \vec a=\begin{pmatrix}a_1\\a_2\end{pmatrix} and b=\begin{pmatrix}b_1\\b_2\end{pmatrix}, then \vec a\cdot\vec b=a_1b_1 + a_2b_2.

Scalar product rule for 3D

For 3D, it’s exactly the same.

For vectors \vec a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} and b=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}, then \vec a\cdot\vec b=a_1b_1 + a_2b_2 + a_3b_3.

Geometric definition

\vec a \cdot \vec b = |\vec a| |\vec b| \cos \theta

Proof

• The origin is at point O.
• Point A is at (a_1, a_2).
• Point B is at (b_1, b_2).
• let \vec a=|OA|
• let \vec b=|OB|
• let \theta be the angle between \vec a and \vec b.
• | \vec a | = \sqrt{a_1^2 + a_2^2}
• | \vec a |^2 = a_1^2 + a_2^2
• | \vec b | = \sqrt{b_1^2 + b_2^2}
• | \vec b |^2 = b_1^2 + b_2^2
• The cosine rule is \cos C=\frac{a^2 + b^2 - c^2}{2ab}.
• \cos C=\frac{|\vec a|^2 + |\vec b|^2 - |\vec{AB}|^2}{2 |\vec a| |\vec b|}
• \cos C=\frac{(a_1^2 + a_2^2) + (b_1^2 + b_2^2) - ((b_1 - a_1)^2 + (b_2 - a_2)^2)}{2 \sqrt{a_1^2 + a_2^2} \sqrt{b_1^2 + b_2^2}}
• \cos C=\frac{(a_1^2 + a_2^2) + (b_1^2 + b_2^2) - (b_1^2 - 2a_1b_1 + a_1^2 + b_2^2 - 2a_2b_2 + a_2^2)}{2 \sqrt{a_1^2 + a_2^2} \sqrt{b_1^2 + b_2^2}}
• \cos C=\frac{2a_1b_1 + 2a_2b_2}{2 \sqrt{a_1^2 + a_2^2} \sqrt{b_1^2 + b_2^2}}
• \cos C=\frac{a_1b_1 + a_2b_2}{\sqrt{a_1^2 + a_2^2} \sqrt{b_1^2 + b_2^2}}
• a_1b_1+a_2b_2=a\cdot b
• \cos C=\frac{\vec a \cdot \vec b}{|\vec a| |\vec b|}

Applying the rules: finding the angle between two vectors

Find the angle between \vec a=\begin{pmatrix}3\\4\end{pmatrix} and \vec b=\begin{pmatrix}5\\-12\end{pmatrix}

• \cos\theta=\frac{\vec a \cdot \vec b}{|\vec a| |\vec b|}
• \vec a \cdot \vec b = 3 \times 5 + 4 \times -12
  • = 15 - 48=-33
• |\vec a| = \sqrt{3^2 + 4^2}
  • = \sqrt{9 + 16}
  • = \sqrt{25}=5
  • so |\vec a|=5
• |\vec b| = \sqrt{5^2 + (-12)^2}
  • = \sqrt{25 + 144}
  • = \sqrt{169}=13
  • so |\vec b|=13
• \cos\theta=\frac{-33}{5 \times 13}
  • =\frac{-33}{65}
• \theta=\cos^{-1}\left(\frac{-33}{65}\right)
  • \approx 120.5^\degree
• So the angle between the two vectors is 120.5^\circ - we’ve solved it!!

Find the angle between \vec a=\begin{pmatrix}3\\5\\6\end{pmatrix} and \vec b=\begin{pmatrix}4\\2\\1\end{pmatrix}

• \cos\theta=\frac{\vec a \cdot \vec b}{|\vec a| |\vec b|}
• \vec a \cdot \vec b = a_1b_1 + a_2b_2 + a_3b_3
• \vec a \cdot \vec b = 3\times4+5\times2+6\times1
  • =12+10+6=28
• |\vec a| = \sqrt{3^2 + 5^2 + 6^2}
  • = \sqrt{9 + 25 + 36}
  • = \sqrt{70}
• |\vec b| = \sqrt{4^2 + 2^2 + 1^2}
  • = \sqrt{16 + 4 + 1}
  • = \sqrt{21}
• \cos\theta=\frac{28}{\sqrt{70} \times \sqrt{21}}
  • =\frac{28}{\sqrt{1470}}
• \theta=\cos^{-1}\left(\frac{28}{\sqrt{1470}}\right)
  • \approx 51.2^\circ
• So the angle between the two vectors is 51.2^\circ :)

Find the angle between \vec a=\begin{pmatrix}3\\5\\6\end{pmatrix} and \vec b=\begin{pmatrix}4\\2\\1\end{pmatrix}

Checking if vectors are perpendicular

• \cos\theta=\frac{\vec a \cdot \vec b}{|\vec a| |\vec b|}
• If \vec a \perp \vec b, then \theta=90^\circ.
• \cos90=\frac{\vec a \cdot \vec b}{|\vec a| |\vec b|}
• 0=\frac{\vec a \cdot \vec b}{|\vec a| |\vec b|}
• \vec a \cdot \vec b=0
• So if the scalar product of two vectors is zero, then the vectors are perpendicular:

Key take-away: if \vec a \cdot \vec b=0, then \vec a \perp \vec b.

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