Tag Archives: scalar product

April 7, 2025 · 9:09 am

Vector and Scalar Projection

The vector projection (vector resolution or vector component) of \mathbf{a} onto a non-zero vector \mathbf{b} is splitting \mathbf{a} into two vectors, one is parallel to \mathbf{b} (the vector projection) and one perpendicular to \mathbf{b}

In the above diagram \mathbf{a_1} is the vector projection of \mathbf{a} onto \mathbf{b} and \mathbf{a_2} is perpendicular to \mathbf{b}.

How do we find \mathbf{a_1} and \mathbf{a_2}?

Using right trigonometry,

cos(\theta)=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}

Remember the scalar product (dot product) of vectors is

(1)   \begin{equation*}\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|cos(\theta)\end{equation*}

Hence cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}

and, |\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\times|\mathbf{a}|

|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}

This is the scalar projection of \mathbf{a} onto \mathbf{b}

To find the vector projection we need to multiply by \mathbf{\hat{b}}, that is find a vector with the same magnitude as \mathbf{a_1} in the direction of \mathbf{b}.

The vector projection is

\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \mathbf{\hat{b}}

\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \frac{\mathbf{b}}{|\mathbf{b}|}

\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{(|\mathbf{b}|)^2}\times \mathbf{b}

Now for \mathbf{a_2}, we know \mathbf{a}=\mathbf{a_1}+\mathbf{a_2}

Hence, \mathbf{a_2}=\mathbf{a}-\mathbf{a_1}

Example

From Cambridge Year 11 Specialist Mathematics (Chapter 3)

(a) \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}

=\frac{\begin{pmatrix}4\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\end{pmatrix}}{2}\times(\mathbf{i}-\mathbf{j})

=\frac{3}{2}{(\mathbf{i}-\mathbf{j})

(b)4\mathbf{i}+\mathbf{j}-\frac{3}{2}{(\mathbf{i}-\mathbf{j})=\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}

(c)

The shortest distance (green vector) is the vector component of \mathbf{a} perpendicular to \mathbf{b}, i.e. \frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}

|\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}|=\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}

Leave a Comment

Filed under Right Trigonometry, Trigonometry, Vector Projection, Vectors, Year 11 Specialist Mathematics

Tagged as , , ,