Answer :
Answer:
[tex]\vec{A}\times \vec{B}=-51.12\hat{k}[/tex]
[tex]\theta=83.2^{\circ}[/tex]
Explanation:
We are given that
[tex]\vec{A}=4.2\hat{i}+7.2\hat{j}[/tex]
[tex]\vec{B}=5.70\hat{i}-2.40\hat{j}[/tex]
We have to find the scalar product and the angle between these two vectors
[tex]\vec{A}\times \vec{B}=\begin{vmatrix}i&j&k\\4.2&7.2&0\\5.7&-2.4&0\end{vmatrix}[/tex]
[tex]\vec{A}\times \vec{B}=\hat{k}(-10.08-41.04)=-51.12\hatk}[/tex][tex]\hat{k}[/tex]
Angle between two vectors is given by
[tex]sin\theta=\frac{\mid a\times b\mid}{\mid a\mid \mi b\mid}[/tex]
Where [tex]\theta[/tex] in degrees
[tex]\mid{\vec{A}}\mid=\sqrt{(4.2)^2+(7.2)^2}=8.3[/tex]
Using formula[tex]\mid a\mid=\sqrt{x^2+y^2}[/tex]
Where x= Coefficient of unit vector i
y=Coefficient of unit vector j
[tex]\mid{\vec{B}}\mid=\sqrt{5.7)^2+(-2.4)^2}=6.2[/tex]
[tex]\mid{\vec{A}\times \vec{B}}\mid=\sqrt{(-51.12)^2}=51.12[/tex]
Using the formula
[tex]sin\theta=\frac{51.12}{8.3\times 6.2}=0.993[/tex]
[tex]\theta=sin^{-1}(0.993)=83.2[/tex]degrees
Hence, the angle between given two vectors=[tex]83.2^{\circ}[/tex]