The angle between the \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is \(\boxed{\bf 175^{\circ}}\).
Further explanation:
Concept used:
The dot product of \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is calculated as follows:
\(\boxed{\overrightarrow{A}\cdot \overrightarrow{B}=|A||B|\ \text{cos}\theta}\) …… (1)
Here, \(|A|\) and \(|B|\) are the magnitude of \(\overrightarrow{A}\) and \(\overrightarrow{B}\) respectively.
Calculation:
The vector \(\overrightarrow{A}\) can be written as follows:
\(\boxed{\overrightarrow{A}=A_{x}\hat{i}+A_{y}\hat{j}}\)....(1)
The vector \(\overrightarrow{B}\) can be written as follows:
\(\boxed{\overrightarrow{B}=B_{x}\hat{i}+B_{y}\hat{j}}\)....(2)
The value of \(A_{x}\) is \(-3.20\),\(A_{y}\) is \(4.40\), \(B_{x}\) is \(2.70\) and \(B_{y}\) is \(-3.10\).
Now, substitute the value of \(A_{x}\) and \(A_{y}\) in equation (1).
\(A=-3.20\hat{i}+4.40\hat{j}\)
Substitute the value of \(B_{x}\) and \(B_{y}\) in equation (2).
\(B=2.70\hat{i}-3.10\hat{j}\)
The magnitude of vector \(\overrightarrow{A}\) can be calculated as follows:
\(\boxed{|A|=\sqrt{A^{2}_{x}+A^{2}_{y}}}\) …… (3)
Substitute the value of \(A_{x}\) and \(A_{y}\) equation (3) to obtain the magnitude of vector \(\overrightarrow{A}\) as follows:
\(\begin{aligned}|A|&=\sqrt{(-3.20)^{2}+(4.40)^{2}}\\&=\sqrt{10.24+19.36}\\&=\sqrt{29.6}\\&=5.44\end{aligned}\)
Similarly the magnitude of vector \(\overrightarrow{B}\) can be calculated as follows:
\(\boxed{|B|=\sqrt{B^{2}_{x}+B^{2}_{y}}}\) ……(4)
Substitute the value of \(B_{x}\) and \(B_{y}\) equation (4) to obtain the magnitude of vector \(\overrightarrow{B}\) as follows:
\(\begin{aligned}|B|&=\sqrt{(2.70)^{2}+(-3.10)^{2}}\\&=\sqrt{7.29+9.61}\\&=\sqrt{16.9}\\&=4.11\end{aligned}\)
Dot product of vector \(\overrightarrow{A}\) and vector \(\overrightarrow{B}\) can be calculated as follows:
\(\begin{aligned}\overrightarrow{A}\cdot \overrightarrow{B}&=\left(-3.20\hat{i}+4.40\hat{j}\right)\cdot \left(-3.20\hat{i}+4.40\hat{j}\right)\\&=(-3.20\cdot 2.70)-(4.40\cdot 3.10)\\&=-8.64-13.64\\&=-22.28\end{aligned}\)
Substitute \(\overrightarrow{A}\cdot \overrightarrow{B}=-22.28\), \(|A|=5.44\) and \(|B|=4.11\) in the equation (1) to obtain the angle between \(\overrightarrow{A}\) and \(\overrightarrow{B}\) as follows:
\(\begin{aligned}-22.28&=(5.44)(4.11)\ \text{cos}\theta\\-22.28&=22.3584\ \text{cos}\theta\\ \text{cos}\theta&=-\dfrac{22.28}{22.3584}\\ \text{cos}\theta&=-0.996\end{aligned}\)
Further simplify the above equation as follows:
\(\begin{aligned}\text{cos}\ \theta&=-0.996\\ \theta&=\text{cos}^{-1}(-0.996)\\ \theta&=174.87\\ \theta &\approx175\end{aligned}\)
Therefore, the angle between the \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is \(\boxed{\bf 175^{\circ}}\).
Learn more:
1. Learn more about triangles https://brainly.com/question/7437053
2. Learn more about slope of a line https://brainly.com/question/1473992
Answer details:
Grade: Senior school
Subject: Mathematics
Chapter: Vector
Keywords: Vector, dot product, angle, magnitude, AB=AB cos(theta), cross product, degree, radian, Ax1=-3.20, Ay1=4.40, Bx1=2.70, By1=-3.10, pair of vectors, angle between A and B.
