Question

Find the angle between the given vectors to the nearest tenth of a degree.

Answer 1

$\bf \textit{angle between two vectors }\\ \quad \\ cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\\\ \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\\\\ -----------------------------\\\\ \theta=cos^{-1}\left[ \cfrac{\ \textless \ 8,4\ \textgreater \ \quad \cdot \quad \ \textless \ 9,-9\ \textgreater \ }{(\sqrt{8^2+4^2})(\sqrt{9^2+(-9)^2})} \right]$


$\bf \theta=cos^{-1}\left[ \cfrac{(8\cdot 9)+(4\cdot -9)}{(\sqrt{64+16})(\sqrt{81+81})} \right]\implies \theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right] \\\\\\ \theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right]\implies \theta=cos^{-1}\left[ \cfrac{36}{\sqrt{12960}}\right] \\\\\\ \theta=cos^{-1}\left[ \cfrac{36}{36\sqrt{10}}\right]\implies \theta\approx 71.565^o$

Answer 2

Answer:

$\theta=71.6^{\circ}$

Step-by-step explanation:

We are given that two vectors

u=<8,4> and v=<9,-9>

The given vectors can be write as

$\vec{u}=8\hat{i}+4\hat{j}$

$\vec{v}=9\hat{i}-9\hat{j}$

We have to find the angle between two given vectors

The formula to find out the angle between two vectors is given below

$cos\theta=\frac{\vec{u}\cdot\vec{v}}{\mid\vec{u}\mid\cdot \mid\vec{v}\mid}$

By applying this formula we have to find the angle between two vectors

$\mid{\vec{u}\mid=\sqrt{8^2+4^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt5$

$\mid\vec{v}\mid=\sqrt{9^2+(-9)^2}=\sqrt{81+81}=9\sqrt2$

$\vec{u}\cdot\vec{v}=(8\hat{i}+4\hat{j})\cdot(9\hat{i}-9\hat{j})=72-36=36$

Using  $\hat{i}\cdot\hat{i}=1,\hat{j}\cdot\hat{j}=1,\hat{k}\cdot\hat{k}=1$

Substituting  the values then we get

$cos\theta=\frac{36}{4\sqrt5\cdot9\sqrt2}$

$cos\theta=\frac{1}{\sqrt10}=\frac{1}{3.16}$

$cos\theta=0.3164$

$\theta=cos^{-1}(0.3164)=71.55^{\circ}$

$\theta=71.6^{\circ}$

Answer:$\theta=71.6^{\circ}$