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

`\vec{A}* \vec{B}=-51.12\hat{k}`

`\theta=83.2^(\circ)`

Step-by-step 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}* \vec{B}=\begin{vmatrix}i&j&k\\4.2&7.2&0\\5.7&-2.4&0\end{vmatrix}`

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

Angle between two vectors is given by

`sin\theta=(\mid a* b\mid)/(\mid a\mid \mi b\mid)`

Where
`\theta` in degrees

`\mid{\vec{A}}\mid=√((4.2)^2+(7.2)^2)=8.3`

Using formula
`\mid a\mid=√(x^2+y^2)`

Where x= Coefficient of unit vector i

y=Coefficient of unit vector j

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

`\mid{\vec{A}* \vec{B}}\mid=√((-51.12)^2)=51.12`

Using the formula

`sin\theta=(51.12)/(8.3* 6.2)=0.993`

`\theta=sin^(-1)(0.993)=83.2`degrees

Hence, the angle between given two vectors=
`83.2^(\circ)`