Question

part a given two vectors $vec{a}=4.30hat{i}+7.20hat{j}$ and $vec{b}=4.90hat{i}-2.10hat{j}$, find the scalar product of the two vectors $vec{a}$ and $vec{b}$. $vec{a}cdotvec{b}=5.95$ previous answers correct part b find the angle between these two vectors. express your answer in degrees. $\theta =$ you have already submitted this answer. enter a new answer. no credit lost. try again.

Explanation:

Step1: Calculate magnitudes of vectors

For $\vec{A}=4.30\hat{i} + 7.20\hat{j}$, $|\vec{A}|=\sqrt{4.30^{2}+7.20^{2}}=\sqrt{18.49 + 51.84}=\sqrt{70.33}\approx8.386$.
For $\vec{B}=4.90\hat{i}-2.10\hat{j}$, $|\vec{B}|=\sqrt{4.90^{2}+(- 2.10)^{2}}=\sqrt{24.01+4.41}=\sqrt{28.42}\approx5.331$.

Step2: Use dot - product formula for angle

We know $\vec{A}\cdot\vec{B}=|\vec{A}||\vec{B}|\cos\theta$. Given $\vec{A}\cdot\vec{B} = 5.95$ (from part A). Then $\cos\theta=\frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}$.
Substitute the values: $\cos\theta=\frac{5.95}{8.386\times5.331}=\frac{5.95}{44.71}\approx0.133$.

Step3: Find the angle

$\theta=\arccos(0.133)\approx82.4^{\circ}$.

Answer:

$82.4$