Question

find the angle between ( 4i - j ) and ( j ).
the angle between ( 4i - j ) and ( j ) is ( square ^ { circ } ).
(round to the nearest tenth as needed.)

Explanation:

Step1: Identify the vectors

Let \(\vec{u} = 4\mathbf{i} - \mathbf{j}\) and \(\vec{v}=\mathbf{j}\) (which can be written as \(0\mathbf{i}+ 1\mathbf{j}\)).

Step2: Use the dot - product formula for the angle between two vectors

The formula for the angle \(\theta\) between two vectors \(\vec{u}\) and \(\vec{v}\) is \(\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\vert\vec{u}\vert\vert\vec{v}\vert}\)

First, calculate the dot product \(\vec{u}\cdot\vec{v}\):
If \(\vec{u}=a_1\mathbf{i}+b_1\mathbf{j}\) and \(\vec{v}=a_2\mathbf{i}+b_2\mathbf{j}\), then \(\vec{u}\cdot\vec{v}=a_1a_2 + b_1b_2\)
For \(\vec{u} = 4\mathbf{i}-\mathbf{j}\) (\(a_1 = 4\), \(b_1=- 1\)) and \(\vec{v}=0\mathbf{i}+1\mathbf{j}\) (\(a_2 = 0\), \(b_2 = 1\))
\(\vec{u}\cdot\vec{v}=(4\times0)+(-1\times1)=- 1\)

Next, calculate the magnitude of \(\vec{u}\), \(\vert\vec{u}\vert=\sqrt{a_1^{2}+b_1^{2}}\)
\(\vert\vec{u}\vert=\sqrt{4^{2}+(-1)^{2}}=\sqrt{16 + 1}=\sqrt{17}\approx4.123\)

Then, calculate the magnitude of \(\vec{v}\), \(\vert\vec{v}\vert=\sqrt{a_2^{2}+b_2^{2}}\)
\(\vert\vec{v}\vert=\sqrt{0^{2}+1^{2}} = 1\)

Now, substitute these values into the cosine formula:
\(\cos\theta=\frac{-1}{\sqrt{17}\times1}=\frac{-1}{\sqrt{17}}\approx\frac{-1}{4.123}\approx - 0.2425\)

Step3: Find the angle \(\theta\)

To find \(\theta\), we take the arccosine of \(- 0.2425\)
\(\theta=\arccos(-0.2425)\)
Using a calculator, \(\theta\approx103.9^{\circ}\) (since \(\arccos(-0.2425)\) is in the second quadrant, and we calculate it as \(180^{\circ}-\arccos(0.2425)\), \(\arccos(0.2425)\approx76.1^{\circ}\), so \(180 - 76.1=103.9^{\circ}\))

Answer:

\(103.9\)