AI HL · AHL 4
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Components of a vector: along and perpendicular
AHL 4.19
Drag the tips of a and b. A perpendicular from the tip of a to the line of b makes a right-angled triangle. Its two short sides are the two components.
In the plane a × b has only a k part, so its length is the size of a₁b₂ − a₂b₁. Click a tip, then use the arrow keys to move it one unit.
Drag a tip to change a vector.
Find the component of \(\mathbf{a}=4\mathbf{i} - 3\mathbf{j}\) in the direction of \(\mathbf{b}=2\mathbf{i} + 5\mathbf{j}\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a}\cdot\mathbf{b}=4\times2+(-3)\times5=-7\)
Direction means dot product.
\(|\mathbf{b}|=\sqrt{2^2+5^2}=\sqrt{29}\)
Divide by the length, not the length squared.
\(\dfrac{-7}{\sqrt{29}}=-1.30\)
The answer is negative. The vectors are more than \(90^\circ\) apart, so \(\mathbf{a}\) points partly against \(\mathbf{b}\).
A tractor pulls a log with a force \(\mathbf{F}=\begin{pmatrix}6\\-1\\3\end{pmatrix}\) kN. The direction along the track is \(\mathbf{d}=\begin{pmatrix}5\\2\\-2\end{pmatrix}\). Find the component of the force in the direction of the track.
Have a go first. Then open the working one line at a time.
\(\mathbf{F}\cdot\mathbf{d}=6\times5+(-1)\times2+3\times(-2)=22\)
Direction means dot.
\(|\mathbf{d}|=\sqrt{25+4+4}=\sqrt{33}\)
The direction vector may be any length. Dividing by its length takes care of that.
Component \(=\dfrac{\mathbf{F}\cdot\mathbf{d}}{|\mathbf{d}|}=\dfrac{22}{\sqrt{33}}=3.83\)
Units: kN.
Find the component of \(\mathbf{a}=\begin{pmatrix}2\\-3\\5\end{pmatrix}\) perpendicular to \(\mathbf{b}=\begin{pmatrix}1\\4\\-2\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}(-3)\times(-2)-5\times4\\ 5\times1-2\times(-2)\\ 2\times4-(-3)\times1\end{pmatrix}=\begin{pmatrix}-14\\9\\11\end{pmatrix}\)
Direction: dot. Perpendicular: cross.
\(|\mathbf{a}\times\mathbf{b}|=\sqrt{398}\)
Component \(=\dfrac{|\mathbf{a}\times\mathbf{b}|}{|\mathbf{b}|}=\dfrac{\sqrt{398}}{\sqrt{21}}=4.35\)
The perpendicular component is never negative.
Take \(\mathbf{a}=5\mathbf{i}+12\mathbf{j}\) and \(\mathbf{b}=3\mathbf{i}\). Find the component of \(\mathbf{a}\) perpendicular to \(\mathbf{b}\), then check that the two components and \(|\mathbf{a}|\) fit Pythagoras.
Have a go first. Then open the working one line at a time.
\(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}=\dfrac{15}{3}=5\)
b is along the x-axis, so this is the x-part of a.
\(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}0\\0\\-36\end{pmatrix}\), so \(\dfrac{36}{3}=12\)
This is the y-part of a. The length of b cancels.
\(5^2+12^2=169=13^2=|\mathbf{a}|^2\)
The two components are the two shorter sides of a right-angled triangle whose hypotenuse is \(|\mathbf{a}|\). Use this to check any answer.