Question

1. Given vectors a =< −1, 2 > and b =< 3, 4 >, find 3a + 2b

a. <7, 16>
b. <-3, 8>
c. <3, 14>
d. <8, 21>

2. Find the magnitude of vector u = < 12, -4 >

a. 12.7
b. 8
c. 160
d. 11.3

3. Find the direction angle for v = < 5, - 8 >

a. 58°
b. 122°
c. 238°
d. 302°

4. Find the magnitude of the horizontal and vertical components of vector v, given |v| = 30 and θ = 25°

x-component =

y-component=

Answer 1

Answer:

1) c. <3, 14>

2) a. 12.7

3) a. 302°

4)

x-component = 27

y-component = 13

Step-by-step explanation:

1)

To find $\mathbf{a} + 2\mathbf{b}$, where $\mathbf{a} = <-1, 2>$ and $\mathbf{b} = <3, 4>$, we need to perform scalar multiplication and vector addition.

Given:

$\mathbf{a} = <-1, 2>$

$\mathbf{b} = <3, 4>$

We compute:

$3\mathbf{a} + 2\mathbf{b} = 3<-1, 2> + 2<3, 4>$

$= <-3, 6> + <6, 8>$

$= <-3 + 6, 6 + 8>$

$= <3, 14>$

Therefore, $3\mathbf{a} + 2\mathbf{b} = <3, 14>$.

$\dotfill$

2)

To find the magnitude of vector $\mathbf{u} = <12, -4>$, we use the formula for the magnitude (or length) of a vector in two dimensions:

$|\mathbf{u}| = \sqrt{u_x^2 + u_y^2}$

Given \( \mathbf{u} = <12, -4> [/tex], we substitute the components:

$|\mathbf{u}| = \sqrt{12^2 + (-4)^2}$

$|\mathbf{u}| = \sqrt{144 + 16}$

$|\mathbf{u}| = \sqrt{160}$

$|\mathbf{u}| = 4\sqrt{10}$

$] |\mathbf{u}| = 12.649110640673$

$] |\mathbf{u}| = 12.7$

Therefore, the magnitude of vector $\mathbf{u} = <12, -4>$ is $12.7$.

$\dotfill$

3)

The direction angle $\theta$ of a vector $\mathbf{v} = <a, b>$ is given by the formula:

$\theta = \arctan\left(\dfrac{b}{a}\right)$

Given $\mathbf{v} = <5, -8>$, we substitute the components:

$\theta = \arctan\left(\dfrac{-8}{5}\right)$

Now, calculate the arctangent using a calculator:

$\theta \approx −57.99461679191^\circ$

$\theta \approx -58°$

Note that the negative sign indicates that the angle is measured counterclockwise from the positive x-axis.

Therefore, the direction angle for $\mathbf{v} = <5, -8>$ is approximately $360 - 58 = 302^\circ$.

$\dotfill$

4)

The horizontal component ($v_x$) and vertical component ($v_y$) of a vector $\mathbf{v}$ can be calculated using the magnitude ($|\mathbf{v}|$) and the direction angle ($\theta$) using the following formulas:

$v_x = |\mathbf{v}| \cdot \cos(\theta)$

$v_y = |\mathbf{v}| \cdot \sin(\theta)$

Given $|\mathbf{v}| = 30$ and $\theta = 25^\circ$, we can substitute these values into the formulas:

$v_x = 30 \cdot \cos(25^\circ)$

$v_y = 30 \cdot \sin(25^\circ)$

Now, calculate these values:

$v_x \approx 30 \cdot 0.9063077870366 \\= 27.189233611099 \\= 27$

$v_y \approx 30 \cdot 0.4226182617406 \\= 12.678547852220\\ = 13$

Therefore, the horizontal component ($v_x$) is approximately $27.19$ and the vertical component ($v_y$) is approximately $12.68$.

Answer 2

Answer:

1)  < 3, 14 >

2) √(160) ≈ 12.6

3) 302°

4) x-component = 27.2, y-component = 12.7

Step-by-step explanation:

Question 1

Given vectors:

$\mathbf{a} = \langle -1, 2 \rangle$

$\mathbf{b} = \langle 3, 4 \rangle$

To find 3a + 2b, multiply each component of the vectors by their respective scalars:

$3\mathbf{a} + 2\mathbf{b} = 3 \cdot \langle -1, 2 \rangle + 2 \cdot \langle 3, 4 \rangle\\\\3\mathbf{a} + 2\mathbf{b} = \langle -3, 6 \rangle + \langle 6, 8 \rangle$

Now, add the corresponding components:

$3\mathbf{a} + 2\mathbf{b}= \langle -3 + 6, 6 + 8 \rangle\\\\3\mathbf{a} + 2\mathbf{b}= \langle 3, 14 \rangle$

Therefore:

$\Large\boxed{\boxed{3\mathbf{a} + 2\mathbf{b}= \langle 3, 14 \rangle}}$

$\dotfill$

Question 2

Given vector:

$\mathbf{u} = \langle 12, -4 \rangle$

The magnitude (or length) of a vector u = < x, y > in a two-dimensional space is given by the formula:

$|\mathbf{u}| = \sqrt{x^2 + y^2}$

For the vector u = < 12, -4 >, the magnitude is calculated as follows:

$|\mathbf{u}| = \sqrt{12^2 + (-4)^2}\\\\ |\mathbf{u}| = \sqrt{144 + 16}\\\\ |\mathbf{u}| = \sqrt{160}$

So, the magnitude of the vector u = < 12, -4 > is:

$\Large\boxed{\boxed{|\mathbf{u}| = \sqrt{160}\approx 12.6}}$

$\dotfill$

Question 3

A vector v = < x, y > makes an angle of tan⁻¹(y/x) with the horizontal.

In the case of vector v = < 5, -8 >, x = 5 and y = -8, so:

$\tan^{-1}\left(\dfrac{-8}{5}\right)=-57.99461679...^{\circ}=-58^{\circ}$

The direction is measured anticlockwise from the positive x-axis, so:

$\textsf{Direction}=360^{\circ}+(-58^{\circ})\\\\\textsf{Direction}=360^{\circ}-58^{\circ}\\\\\textsf{Direction}=302^{\circ}$

So, the direction of the vector v = < 5, -8 > is:

$\Large\boxed{\boxed{\textsf{Direction}=302^{\circ}}}$

$\dotfill$

Question 4

Given vector v with magnitude |v| = 30 and direction θ = 25°.

The vector forms a right-angled triangle, with the magnitude as the length of the hypotenuse.

Using trigonometry, we can find x and y:

$\cos 25^{\circ}=\dfrac{x}{30} \implies x=30\cos 25^{\circ}=27.2\\\\\\\sin 25^{\circ}=\dfrac{y}{30} \implies y=30\sin 25^{\circ}=12.7$

So, the horizontal and vertical components are:

$\Large\boxed{\boxed{\textsf{ x -component}=27.2}}$

$\Large\boxed{\boxed{\textsf{ y -component}=12.7}}$