Question

find the angle between 2i - 5j and j. the angle between 2i - 5j and j is □°. (round to the nearest tenth as needed.)

Explanation:

Step1: Recall the formula for the angle between two vectors

The formula for the angle \(\theta\) between two vectors \(\vec{u} = a_1\vec{i}+b_1\vec{j}\) and \(\vec{v}=a_2\vec{i}+b_2\vec{j}\) is \(\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\vert\vec{u}\vert\vert\vec{v}\vert}\), where \(\vec{u}\cdot\vec{v}=a_1a_2 + b_1b_2\), \(\vert\vec{u}\vert=\sqrt{a_1^{2}+b_1^{2}}\) and \(\vert\vec{v}\vert=\sqrt{a_2^{2}+b_2^{2}}\).

Let \(\vec{u} = 2\vec{i}- 5\vec{j}\) and \(\vec{v}=\vec{i}+0\vec{j}\) (since \(\vec{i}\) can be written as \(1\vec{i}+0\vec{j}\)).

Step2: Calculate the dot product \(\vec{u}\cdot\vec{v}\)

\(\vec{u}\cdot\vec{v}=(2)(1)+(- 5)(0)=2 + 0=2\)

Step3: Calculate the magnitude of \(\vec{u}\)

\(\vert\vec{u}\vert=\sqrt{2^{2}+(-5)^{2}}=\sqrt{4 + 25}=\sqrt{29}\approx5.385\)

Step4: Calculate the magnitude of \(\vec{v}\)

\(\vert\vec{v}\vert=\sqrt{1^{2}+0^{2}}=\sqrt{1}=1\)

Step5: Calculate \(\cos\theta\)

\(\cos\theta=\frac{2}{\sqrt{29}\times1}=\frac{2}{\sqrt{29}}\approx\frac{2}{5.385}\approx0.371\)

Step6: Find \(\theta\)

\(\theta=\arccos(0.371)\approx68.2^{\circ}\) (using a calculator to find the arccosine of \(0.371\) and rounding to the nearest tenth)

Answer:

\(68.2^{\circ}\)