a) The Dot Product is equal to 6.
b) The Angle between the two Vectors is approximately 82.875°.
c) The Scalar Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is approximately 0.894.
d) The Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is [tex]\vec v_{1, \vec v_{2}} = (-2.683, 5.367)[/tex].
Given two Vectors of the form [tex](x,y)[/tex], the Dot Product between two Vectors is defined by:
[tex]\vec v_{1} \,\bullet\,\vec v_{2} = \|\vec v_{1}\|\cdot \|\vec v_{2}\|\cdot \cos \theta = x_{1}\cdot x_{2} + y_{1}\cdot y_{2}[/tex] (1)
Where:
[tex]\|\vec v_{1}\|[/tex], [tex]\|\vec v_{2}\|[/tex] - Norm of the vectors.
[tex]\theta[/tex] - Angle between the two vectors, in sexagesimal degrees.
The Norm is defined by Pythagorean Theorem.
[tex]r = \sqrt{x^{2}+y^{2}}[/tex] (2)
The Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is:
[tex]\vec v_{1, \vec v_{2}} = \frac{\vec v_{1}\,\bullet \,\vec v_{2}}{\|\vec v_{2}\|^{2}}\cdot \vec v_{2}[/tex] (3)
Where the Scalar Projection is:
[tex]\|\vec v_{1,\vec v_{2}}\| = \|\vec v_{1}\|\cdot \cos \theta[/tex]
a) If we know that [tex]\vec v_{1} = (-6, 4)[/tex] and [tex]\vec v_{2} = (-3, 6)[/tex], then the Dot Product is:
[tex]\vec v_{1}\,\bullet\,\vec v_{2} = (-6)\cdot (-3) + (4)\cdot (6)[/tex]
[tex]\vec v_{1} \,\bullet \,\vec v_{2} = 6[/tex]
The Dot Product is equal to 6.
b) The angle between [tex]\vec v_{1}[/tex] and [tex]\vec v_{2}[/tex] is calculated from (1):
[tex]\theta = \cos^{-1} \frac{\vec v_{1}\,\bullet\,\vec v_{2}}{\|\vec v_{1}\|\cdot\|\vec v_{1}\|}[/tex]
If we know that [tex]\vec v_{1} = (-6, 4)[/tex] and [tex]\vec v_{2} = (-3, 6)[/tex], then the Angle between the two Vectors is:
[tex]\|v_{1}\| = \sqrt{(-6)^{2}+4^{2}}[/tex]
[tex]\|\vec v_{1}\| \approx 7.211[/tex]
[tex]\|\vec v_{2}\| = \sqrt{(-3)^{2}+6^{2}}[/tex]
[tex]\|\vec v_{2}\| \approx 6.708[/tex]
[tex]\theta = \cos^{-1} \frac{6}{(7.211)\cdot (6.708)}[/tex]
[tex]\theta \approx 82.875^{\circ}[/tex]
The Angle between the two Vectors is approximately 82.875°.
c) If we know that [tex]\|\vec v_{1}\| \approx 7.211[/tex] and [tex]\theta \approx 82.875^{\circ}[/tex], then the Scalar Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is:
[tex]\|v_{1,\vec v_{2}}\| = 7.211\cdot \cos 82.875^{\circ}[/tex]
[tex]\|\vec v_{1, \vec v_{2}}\| \approx 0.894[/tex]
The Scalar Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is approximately 0.894.
d) If we know that [tex]\vec v_{1} \,\bullet \,\vec v_{2} = 6[/tex], [tex]\|\vec v_{2}\| \approx 6.708[/tex] and [tex]\vec v_{2} = (-3, 6)[/tex], then the Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is:
[tex]\vec v_{1, \vec v_{2}} = \frac{6}{6.708}\cdot (-3, 6)[/tex]
[tex]\vec v_{1, \vec v_{2}} = (-2.683, 5.367)[/tex]
The Projection of [tex]\vec v_{1}[/tex] onto [tex]\vec v_{2}[/tex] is [tex]\vec v_{1, \vec v_{2}} = (-2.683, 5.367)[/tex].
Please see this question related to Dot Product: https://brainly.com/question/9795347
