Scalar Product
Scalar Product
- The scalar product (dot product) \(\mathbf{a} \text{ }\mathbf{b}\) of two vectors is the sum of the products of the components.
-
\[\mathbf{a} \text{ } \mathbf{b}=\sum a_{i}b_{i}\]
Finding angles between Vectors
- We can find the unit vector of any vector by dividing the vector by it's modulus.
- And the unit vector of \(\mathbf{\hat{a}} \times\mathbf {\hat{b}}\) = \(\cos\theta\) of the angle formed between the two vectors.
- \(\frac{\mathbf{a}}{|a|}\times \frac{\mathbf{b}}{|b|}=\cos\theta\)
- \(\frac{1}{|a||b|}(\mathbf{a}\times \mathbf{b})=\cos\theta\)
- \(\mathbf{a}\times \mathbf{b}=|a||b|\cos\theta\)
Proof for Scalar Product (long)
- \(|\hat{b}-\hat{a}|^2=|\hat{a}|^2+|\hat{b}|^2-2|\hat{a}||\hat{b}|\cos\theta\)
- \(|\hat{b}-\hat{a}|^2=1+1-2\cos\theta\)
- \((\sqrt{ (b_{1}-a_{1}^2)+(b_{2}-a_{2})^2+(b_{3}-a_{3})^2 })^2=2-2\cos\theta\)
- \(b_{1}^2-2b_{1}a_{1}+a_{1}^2+b_{2}^2-2b_{2}a_{2}+a_{2}^2+b_{3}^2-2b_{3}a_{3}+a_{3}^2=2-2\cos\theta\)
- \(b_{1}^2+b_{2}^2+b_{3}^2+a_{1}^2+a_{2}^2+a_{3}^2-2(b_{1}a_{1}+b_{2}a_{2}+b_{3}a_{3})=2-2\cos\theta\)
- \(|b|^2+|a|^2-2\hat{a}\times \hat{b}=2-2\cos\theta\)
- \(2-2\hat{a}\times \hat{b}=2-2\cos\theta\)
- \(\hat{a}\times \hat{b}=\cos\theta\)
Find the acute angle between the vectors \(\mathbf{a}=\begin{pmatrix}5 \\ 3 \\ 1\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}1 \\ 0 \\ 5\end{pmatrix}\)
- Times the two vectors together and divide by the product of the moduli.
- \(\frac{(5\times1)+(3\times0)+(1\times5)}{|a|\times |b|}\)
- \(\frac{(5\times1)+(3\times0)+(1\times5)}{\sqrt{ 35 }\times \sqrt{ 26 }}\)
- \(=0.331496\dots\)
- \(\cos^{-1}(0.331496)=70.64^\circ\)
Find the angle between the vectors \(\mathbf{a}=\begin{pmatrix}2 \\ 4 \\ -1\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}0 \\ 1 \\ 8\end{pmatrix}\)
- \(\frac{(2\times0)+(4\times1)+(-1\times8)}{|a|\times |b|}\)
- \(\frac{(2\times0)+(4\times1)+(-1\times8)}{\sqrt{ 19 }\times \sqrt{ 65 }}=-0.108266\)
- \(\cos^{-1}(-0.108266)=96.215\)
Perpendicular vectors