Question

Find the angle between each pair of vectors below. (1 point each) a. a = 2x + 3y a = 4x + 2y b. a = x + y a = x – y c. a = 2x – 3y a = 5x + y d. a = x – 6y b = 2x – 2y Question 7 (2 points) Two cars are driving down separate streets in the same city. The velocity of car A is given by the vector a = (2x + 3y) m/s. The velocity of car B is given by the vector b = (5x + 2y) m/s. What is the angle between the paths of the two cars? (2 points) Question 8 (2 points) For three-dimensional vectors, the dot product is given by this equation: a·b = axbx + ayby + azbz a. Based on this definition, what is the dot product of the vectors shown below? (1 point) a = 2x + 2y – z b = –3x – y + 3z b. What is the angle between the two vectors above? (1 point) Question 9 (2 points) The velocity vectors of two airplanes are given below. a = (250x + 350y – 50z) m/s b = (100x – 500y + 50z) m/s What is the angle between the paths of the two airplanes? (2 points)

Answer 1

1a) The two vectors are:

a = 2x + 3y, b = 4x + 2y

The angle of each vector with respect to the horizontal is:

`\theta_a = arctan((3)/(2))=56.3^(\circ)`

`\theta_b = arctan((2)/(4))=26.6^(\circ)`

So, the angle between the two vectors is
`\theta=\theta_a - \theta_b = 56.3-26.6=26.7^(\circ)`

1b) The two vectors are:

a = x + y, b = x - y

The angle of each vector with respect to the horizontal is:

`\theta_a = arctan((1)/(1))=45.0^(\circ)`

`\theta_b = arctan((2)/(4))=-45.0^(\circ)`

So, the angle between the two vectors is
`\theta=\theta_a - \theta_b = 45.0-(-45.0)=90.0^(\circ)`

1c) The two vectors are:

a = 2x - 3y, b = 5x + y

The angle of each vector with respect to the horizontal is:

`\theta_a = arctan((-3)/(2))=-56.3^(\circ)`

`\theta_b = arctan((1)/(5))=11.3^(\circ)`

So, the angle between the two vectors is
`\theta=\theta_a - \theta_b = 11.3-(-56.3)=67.6^(\circ)`

1d) The two vectors are:

a = x - 6y, b = 2x - 2y

The angle of each vector with respect to the horizontal is:

`\theta_a = arctan((-6)/(1))=-80.5^(\circ)`

`\theta_b = arctan((-2)/(2))=-45.0^(\circ)`

So, the angle between the two vectors is
`\theta=\theta_a - \theta_b = -45.0-(-80.5)=35.5^(\circ)`

2) The two vectors are:

a = 2x + 3y, b = 5x + 2y

The angle of each vector with respect to the horizontal is:

`\theta_a = arctan((3)/(2))=56.3^(\circ)`

`\theta_b = arctan((2)/(5))=21.8^(\circ)`

So, the angle between the two vectors is
`\theta=\theta_a - \theta_b = 56.3-21.8=34.5^(\circ)`

3a) The two vectors are

a = 2x + 2y – z, b = –3x – y + 3z

The dot vector is:

`a\cdot b = 2 \cdot (-3) + 2 \cdot (-1) + (-1) \cdot 3 =-6 -2 -3=-11`

3b) the angle between two 3d vectors is given by

`cos \theta = (a \cdot b)/(|a| |b|)`

Let's calculate the magnitude of each vector:

`|a| = √(2^2 + 2^2 + (-1)^2)=√(9)=3`

`|b|=√((-3)^2+(-1)^2 +3^2)=√(19)=4.36`

So, if we apply the formula,

`\cos \theta = (-11)/(3 \cdot 4.36)=-0.84`

and the angle is

`\theta = arccos (-0.84)=147.1 ^(\circ)`

4) The two vectors are:

a = (250x + 350y – 50z), b = (100x – 500y + 50z)

the angle between two 3d vectors is given by

`cos \theta = (a \cdot b)/(|a| |b|)`

The dot vector here is

`a \cdot b = 250 \cdot 100 + 350 \cdot (-500) + (-50) \cdot 50 =-152500`

The magnitude of each vector is:

`|a| = √(250^2 + 350^2 + (-50)^2)=√(187500)=433.0`

`|b|=√((100)^2+(500)^2 +50^2)=√(262500)=512.3`

So, if we apply the formula,

`\cos \theta = (-152500)/(433.0 \cdot 512.3)=-0.687`

and the angle is

`\theta = arccos (-0.687)=133.4 ^(\circ)`