The product point of the two vectors is given by:
[tex] A.B = (-3.20i + 4.40j). (2.70i - 3.10j) [/tex]
[tex] A.B = (-3.20 * 2.70) + (4.40 * (- 3.10)) [/tex]
[tex] A.B = -8.64 -13.64 [/tex]
[tex] A.B = -22.28 [/tex]
Then, the module of both vectors is:
For vector A:
[tex] A = \sqrt{(- 3.20) ^ 2 + (4.40) ^ 2} [/tex]
[tex] A = 5.44 [/tex]
For vector B:
[tex] B = \sqrt{(2.70) ^ 2 + (- 3.10) ^ 2} [/tex]
[tex] B = 4.11 [/tex]
Finally the angle between the vectors is given by:
[tex] \alpha =Cos^{-1}(\frac{A.B}{|A||B|}) [/tex]
Substituting values:
[tex] \alpha =Cos^{-1}(\frac{-22.28}{|5.44||4.11|}) [/tex]
[tex] \alpha = 175 [/tex]
Answer:
The angle between the vectors is:
[tex] \alpha = 175 [/tex]
The angle between the [tex]\overrightarrow{A}[/tex] and [tex]\overrightarrow{B}[/tex] is [tex]\boxed{\bf 175^{\circ}}[/tex].
Further explanation:
Concept used:
The dot product of [tex]\overrightarrow{A}[/tex] and [tex]\overrightarrow{B}[/tex] is calculated as follows:
[tex]\boxed{\overrightarrow{A}\cdot \overrightarrow{B}=|A||B|\ \text{cos}\theta}[/tex] …… (1)
Here, [tex]|A|[/tex] and [tex]|B|[/tex] are the magnitude of [tex]\overrightarrow{A}[/tex] and [tex]\overrightarrow{B}[/tex] respectively.
Calculation:
The vector [tex]\overrightarrow{A}[/tex] can be written as follows:
[tex]\boxed{\overrightarrow{A}=A_{x}\hat{i}+A_{y}\hat{j}}[/tex]....(1)
The vector [tex]\overrightarrow{B}[/tex] can be written as follows:
[tex]\boxed{\overrightarrow{B}=B_{x}\hat{i}+B_{y}\hat{j}}[/tex]....(2)
The value of [tex]A_{x}[/tex] is [tex]-3.20[/tex],[tex]A_{y}[/tex] is [tex]4.40[/tex], [tex]B_{x}[/tex] is [tex]2.70[/tex] and [tex]B_{y}[/tex] is [tex]-3.10[/tex].
Now, substitute the value of [tex]A_{x}[/tex] and [tex]A_{y}[/tex] in equation (1).
[tex]A=-3.20\hat{i}+4.40\hat{j}[/tex]
Substitute the value of [tex]B_{x}[/tex] and [tex]B_{y}[/tex] in equation (2).
[tex]B=2.70\hat{i}-3.10\hat{j}[/tex]
The magnitude of vector [tex]\overrightarrow{A}[/tex] can be calculated as follows:
[tex]\boxed{|A|=\sqrt{A^{2}_{x}+A^{2}_{y}}}[/tex] …… (3)
Substitute the value of [tex]A_{x}[/tex] and [tex]A_{y}[/tex] equation (3) to obtain the magnitude of vector [tex]\overrightarrow{A}[/tex] as follows:
[tex]\begin{aligned}|A|&=\sqrt{(-3.20)^{2}+(4.40)^{2}}\\&=\sqrt{10.24+19.36}\\&=\sqrt{29.6}\\&=5.44\end{aligned}[/tex]
Similarly the magnitude of vector [tex]\overrightarrow{B}[/tex] can be calculated as follows:
[tex]\boxed{|B|=\sqrt{B^{2}_{x}+B^{2}_{y}}}[/tex] ……(4)
Substitute the value of [tex]B_{x}[/tex] and [tex]B_{y}[/tex] equation (4) to obtain the magnitude of vector [tex]\overrightarrow{B}[/tex] as follows:
[tex]\begin{aligned}|B|&=\sqrt{(2.70)^{2}+(-3.10)^{2}}\\&=\sqrt{7.29+9.61}\\&=\sqrt{16.9}\\&=4.11\end{aligned}[/tex]
Dot product of vector [tex]\overrightarrow{A}[/tex] and vector [tex]\overrightarrow{B}[/tex] can be calculated as follows:
[tex]\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}[/tex]
Substitute [tex]\overrightarrow{A}\cdot \overrightarrow{B}=-22.28[/tex], [tex]|A|=5.44[/tex] and [tex]|B|=4.11[/tex] in the equation (1) to obtain the angle between [tex]\overrightarrow{A}[/tex] and [tex]\overrightarrow{B}[/tex] as follows:
[tex]\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}[/tex]
Further simplify the above equation as follows:
[tex]\begin{aligned}\text{cos}\ \theta&=-0.996\\ \theta&=\text{cos}^{-1}(-0.996)\\ \theta&=174.87\\ \theta &\approx175\end{aligned}[/tex]
Therefore, the angle between the [tex]\overrightarrow{A}[/tex] and [tex]\overrightarrow{B}[/tex] is [tex]\boxed{\bf 175^{\circ}}[/tex].
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.
