Question

Given two vectors A--> = 4.20 i^+ 7.20 j^ and B--> = 5.70 i^− 2.40 j^ , find the scalar product of the two vectors A--> and B--> .
Find the angle between these two vectors.

Answer 1

Answer:

$\vec{A}\times \vec{B}=-51.12\hat{k}$

$\theta=83.2^{\circ}$

Explanation:

We are given that

$\vec{A}=4.2\hat{i}+7.2\hat{j}$

$\vec{B}=5.70\hat{i}-2.40\hat{j}$

We have to find the scalar product and the  angle between these two vectors

$\vec{A}\times \vec{B}=\begin{vmatrix}i&j&k\\4.2&7.2&0\\5.7&-2.4&0\end{vmatrix}$

$\vec{A}\times \vec{B}=\hat{k}(-10.08-41.04)=-51.12\hatk}$$\hat{k}$

Angle between two vectors is given by

$sin\theta=\frac{\mid a\times b\mid}{\mid a\mid \mi b\mid}$

Where $\theta$ in degrees

$\mid{\vec{A}}\mid=\sqrt{(4.2)^2+(7.2)^2}=8.3$

Using formula$\mid a\mid=\sqrt{x^2+y^2}$

Where x= Coefficient of unit vector i

y=Coefficient of unit vector j

$\mid{\vec{B}}\mid=\sqrt{5.7)^2+(-2.4)^2}=6.2$

$\mid{\vec{A}\times \vec{B}}\mid=\sqrt{(-51.12)^2}=51.12$

Using the formula

$sin\theta=\frac{51.12}{8.3\times 6.2}=0.993$

$\theta=sin^{-1}(0.993)=83.2$degrees

Hence, the angle between given two vectors=$83.2^{\circ}$