Mathematics, 14.07.2019 17:30 johnsnow69

Let v1=(-6,4) and v2=(-3,6) compute the following. a. v1*v2 b. the angle between |v1| and |v2|. c. the scaler projection of v1 onto v2 d. the projection of v1 onto v2

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Answer from: kelis2012

a) The Dot Product is equal to 6.

b) The Angle between the two Vectors is approximately 82.875°.

c) The Scalar Projection of $\vec v_{1}$ onto $\vec v_{2}$ is approximately 0.894.

d) The Projection of $\vec v_{1}$ onto $\vec v_{2}$ is $\vec v_{1, \vec v_{2}} = (-2.683, 5.367)$ .

Given two Vectors of the form (x,y) , the Dot Product between two Vectors is defined by:

$\vec v_{1} \,\bullet\,\vec v_{2} = \|\vec v_{1}\|\cdot \|\vec v_{2}\|\cdot \cos \theta = x_{1}\cdot x_{2} + y_{1}\cdot y_{2}$ (1)

Where:

$\|\vec v_{1}\|$ , $\|\vec v_{2}\|$ - Norm of the vectors.

$\theta$ - Angle between the two vectors, in sexagesimal degrees.

The Norm is defined by Pythagorean Theorem.

$r = \sqrt{x^{2}+y^{2}}$ (2)

The Projection of $\vec v_{1}$ onto $\vec v_{2}$ is:

$\vec v_{1, \vec v_{2}} = \frac{\vec v_{1}\,\bullet \,\vec v_{2}}{\|\vec v_{2}\|^{2}}\cdot \vec v_{2}$ (3)

Where the Scalar Projection is:

$\|\vec v_{1,\vec v_{2}}\| = \|\vec v_{1}\|\cdot \cos \theta$

a) If we know that $\vec v_{1} = (-6, 4)$ and $\vec v_{2} = (-3, 6)$ , then the Dot Product is:

$\vec v_{1}\,\bullet\,\vec v_{2} = (-6)\cdot (-3) + (4)\cdot (6)$

$\vec v_{1} \,\bullet \,\vec v_{2} = 6$

The Dot Product is equal to 6.

b) The angle between $\vec v_{1}$ and $\vec v_{2}$ is calculated from (1):

$\theta = \cos^{-1} \frac{\vec v_{1}\,\bullet\,\vec v_{2}}{\|\vec v_{1}\|\cdot\|\vec v_{1}\|}$

If we know that $\vec v_{1} = (-6, 4)$ and $\vec v_{2} = (-3, 6)$ , then the Angle between the two Vectors is:

$\|v_{1}\| = \sqrt{(-6)^{2}+4^{2}}$

$\|\vec v_{1}\| \approx 7.211$

$\|\vec v_{2}\| = \sqrt{(-3)^{2}+6^{2}}$

$\|\vec v_{2}\| \approx 6.708$

$\theta = \cos^{-1} \frac{6}{(7.211)\cdot (6.708)}$

$\theta \approx 82.875^{\circ}$

The Angle between the two Vectors is approximately 82.875°.

c) If we know that $\|\vec v_{1}\| \approx 7.211$ and $\theta \approx 82.875^{\circ}$ , then the Scalar Projection of $\vec v_{1}$ onto $\vec v_{2}$ is:

$\|v_{1,\vec v_{2}}\| = 7.211\cdot \cos 82.875^{\circ}$

$\|\vec v_{1, \vec v_{2}}\| \approx 0.894$

The Scalar Projection of $\vec v_{1}$ onto $\vec v_{2}$ is approximately 0.894.

d) If we know that $\vec v_{1} \,\bullet \,\vec v_{2} = 6$ , $\|\vec v_{2}\| \approx 6.708$ and $\vec v_{2} = (-3, 6)$ , then the Projection of $\vec v_{1}$ onto $\vec v_{2}$ is:

$\vec v_{1, \vec v_{2}} = \frac{6}{6.708}\cdot (-3, 6)$

$\vec v_{1, \vec v_{2}} = (-2.683, 5.367)$

The Projection of $\vec v_{1}$ onto $\vec v_{2}$ is $\vec v_{1, \vec v_{2}} = (-2.683, 5.367)$ .

Please see this question related to Dot Product: link

Let v1=(-6,4) and v2=(-3,6) compute the following. a. v1*v2 b.the angle between |v1| and |v2|. c.the

Answer from: sylvia47

a. We assume you want the dot product. That is the sum of the products of the corresponding components.

... V1•V2 = (-6)(-3) + (4)(6) = 18 + 24 = 42

b. V1•V2 = |V1|·|V2|·cos(α)

... α = arccos(V1•V2/(|V1|·|V2|)

... α = arccos(42/√(((-6)²+4²)((-3)²+6²)))

... α = arccos(42/√2340)

... α ≈ 29.74°

c. The scalar projection of V1 onto V2 is the dot product of V1 with the unit vector in the V2 direction.

... V1•V2/|V2| = 42/√45 = 42/(3√5) = (14/5)√5

d. The projection of V1 onto V2 is the result of part c multiplied by the unit vector in the direction of V2.

... projection of V1 onto V2 = (14/5)√5·(-3, 6)/(3√5) = (14/5)(-1, 2) = (-14/5, 28/5)

Answer from: Quest

idk

step-by-step explanation:

use that smart brain

Answer from: Quest

easy its decreasing

step-by-step explanation:

because its top point is at the to of the graph on the left side (atleast from the veiw i see it) now if that top point wast in the lower part of the graph it would be increasing hope this have a nice day

Is this function increasing, decreasing, or constant?

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