Algebraic Vectors and Parametric Equations - Pre-Calculus

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Question

Express the following vector in component form:

Answer

When separating a vector into its component form, we are essentially creating a right triangle with the vector being the hypotenuse.

Therefore, we can find each component using the cos (for the x component) and sin (for the y component) functions:

$v_{x} = 27cos(55^{\circ}) = 15.5$

$v_y = 27sin(55^{\circ}) = 22.1$

We can now represent these two components together using the denotations i (for the x component) and j (for the y component).

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Question

Find , then find its magnitude. and are both vectors.

Answer

In vector addition, you simply add each component of the vectors to each other.

x component: .

y component: .

z component: .

The new vector is

.

To find the magnitude we use the formula,

$|a+b|=\sqrt{0^2+3^2+4^2}$

$|a+b|=\sqrt{9+16}$

$|a+b|=\sqrt{25}=5$

Thus its magnitude is 5.

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Question

Find the component form of the vector with

initial point

and

terminal point .

Answer

To find the vector in component form given the initial and terminal points, simply subtract the initial point from the terminal point.

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Question

Find the component form of the vector with

initial point

and

terminal point

Answer

To find the vector in component form given the initial and terminal points, simply subtract the initial point from the terminal point.

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Question

A bird flies 15 mph up at an angle of 45 degrees to the horizontal. What is the bird's velocity in component form?

Answer

Write the formula to find both the x and y-components of a vector.

$x=Vcos\theta$

$y=Vsin\theta$

Substitute the value of velocity and theta into the equations.

$x=15cos(45) = \frac{15\sqrt2}{2}$

$y=15sin(45) = \frac{15\sqrt2}{2}$

The vector is: $\left \langle \frac{15\sqrt2}{2},\frac{15\sqrt2}{2} \right \rangle$

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Question

Write this vector in component form:

Answer

In order to find the horizontal component, set up an equation involving cosine with 7 as the hypotenuse, since the side in the implied triangle that represents the horizontal component is adjacent to the 22-degree angle:

$\small cos(22)=\frac{x}{7}$ First, find the cosine of 22, then multiply by 7

$\small 6.49 \approx x$

To find the vertical component, set up an equation involving sine, since the side in the implied triangle that represents the vertical component is opposite the 22-degree angle:

$\small sin(22)=\frac{y}{7}$ First, find the sine of 22, then multiply by 7

$\small 2.62 \approx x$

We are almost done, but we need to make a small adjustment. The picture indicates that the vector points up and to the left, so the horizontal component, 6.49, should be negative:

$\small \small <-6.49, 2.62>$

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Question

Rewrite the vector $\vec{v}= 3 \hat{i} + 5 \hat{j}$ from Cartesian coordinates to polar coordinates $\vec{v} = \left \langle r \angle \theta \right \rangle$ .

Answer

To convert to polar form, we need to find the magnitude of the vector, , and the angle it forms with the positive -axis going counterclockwise, or $\theta$ . This is shown in the figure below.

We find the angle using trigonometric identities:

$\tan{\theta}= \frac{5}{3}$

Using a calculator,

$\theta = \arctan{\frac{5}{3}} \approx 59.04^\circ$

To find the magnitude of a vector, we add up the squares of each component and take the square root:

$r= \left | \vec{v}\right | = \sqrt{3^2+5^2}=\sqrt{34}$ .

So, our vector written in polar form is

$\vec{v} = \left \langle \sqrt{34} \angle 59.04^\circ \right \rangle$

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Question

Express the vector $\left \langle 1,-1\right \rangle$ in polar form.

Answer

To convert a point or a vector to its polar form, use the following equations to determine the magnitude and the direction.

$M=\sqrt{x^2+y^2}$

$\theta=tan^-^1 (\frac{y}{x})$

Substitute the vector $\left \langle 1,-1\right \rangle$ to the equations to find the magnitude and the direction.

$M=\sqrt{(1)^2+(-1)^2}= \sqrt2$

$\theta=tan^-^1 (\frac{-1}{1}) = tan^-^1 (-1) = -\frac{\pi}{4}$

The polar form is: $(M,\theta)= (\sqrt2, -\frac{\pi}{4})$

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Question

Express $\left \langle 3,6\right \rangle$ in polar form in degrees.

Answer

The polar form of the vector is: $(R,\theta)$

Find .

$R=\sqrt{ x^2+ y^2}=\sqrt{ 3^2+ 6^2}=\sqrt{9+36}=\sqrt{45}$

Find the angle.

$\theta= tan^-^1\left(\frac{y}{x}\right)= tan^-^1\left(\frac{6}{3}\right)=63.435$

$(R,\theta)=(\sqrt{45},63.435)$

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Question

Express the vector in polar form.

Answer

We know that converting into polar form requires using the formulas : $x=r\cos\theta$ and $y=r\sin\theta$ .

Solving for r will give us the equation:

$r=\frac{x}{\cos\theta}=\frac{y}{sin\theta}$

We can then solve this equation for theta thusly:

$\frac{y}{x}=\frac{sin\theta}{\cos\theta}\implies\tan^{-1}\left(\frac{x}{y}\right)=\theta$

We substitute the values of x and y found in the vector equation to get the angle measure:

$\theta=\tan^{-1}\left(\frac{15}{3}\right)\approx78.69^o$

Since we have already solved for the radius in terms of x and y and the angle, we substitute the proper values into the equation to get the radius.

$r=\frac{x}{\cos\theta}=\frac{3}{\cos78.69^o}\approx15.30$

Therefore, the vector expressed in polar form is:

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Question

Write the following vector in polar form: $\left \langle -3,-2\right \rangle$

Answer

To find the polar form of $\left \langle -3,-2\right \rangle$ , two formula will be needed since the polar form of a vector is defined as $(r,\theta)$ .

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

$tan(\theta) = \frac{y}{x}$

$r=\sqrt{x^2+y^2}=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}$

$\theta = tan^-^1(\frac{-2}{-3}) =33.69$

However, the direction of $\left \langle -3,-2\right \rangle$ is not in the first quadrant, but lies in the third quadrant. It is mandatory to add 180 degrees so that the angle corresponds to the correct quadrant.

$\theta = 33.69+180 = 213.69$

Therefore, the answer is:

$(r,\theta)= (\sqrt{13},213.69 )$

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Question

Write this vector in component form:

Answer

To figure out the horizontal component, set up an equation involving cosine, since that side of the implied triangle is adjacent to the 48-degree angle:

$\small cos(48)=\frac{x}{11}$ To solve for x, first find the cosine of 48, then multiply by 11:

$\small 7.36 \approx x$

To figure out the vertical component, set up an equation involving sine, since that side of the implied triangle is opposite the 48-degree angle:

$\small sin(48)=\frac{y}{11}$ to solve for y, just like x, first find the sine of 48, then multiply by 11:

$\small 8.17 \approx y$

Putting this in component form results in the vector $\small <7.36, 8.17>$

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Question

Write the vector $\small <2, -7>$ in polar form $\small (r,\theta)$ .

Answer

First, it could be helpful to draw the vector so that we can get a sense of what it looks like. The component form $\small <2, -7>$ means from the start to the end, it moves forward 2 and down 7:

We can now use the Pythagorean Theorem to solve for the magnitude:

$\small 2^2 + 7^2 = C^2$ note that if you had used -7 that would be perfect as well, since that would give you the exact same answer.

$\small 4 + 49 = C^2$

$\small 53 = C^2$ take the square root of both sides

$\small \sqrt{53}= C$ The magnitude is $\small \sqrt{53}$ .

Now to find the angle we should use trigonometric ratios. We can consider the angle being formed by the vector and the component 2, then we can place it in the right quadrant later on. We know that the tangent of that angle is $\small \frac{7}{2}$ :

$\small tan(\theta)= \frac{7}{2}$ now we can take $\small tan^{-1}$ of both sides to determine theta:

$\small \theta \approx 74.05^o$

We can see that the angle for this particular vector is pointing down and to the right, so the angle we want is in the 4th quadrant. This angle would be $\small 360^o - 74.05^o = 285.95^o$

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Question

Write the vector $\small <4,5>$ in polar form, $\small (r,\theta)$ .

Answer

It will be helpful to first draw the vector so we can see what quadrant the angle is in:

Since the vector is pointing up and to the right, it is in the first quadrant. To determine the angle, set up a trig equation with tangent, since the component 5 is opposite and the component 4 is adjacent to the angle we are looking for:

$\small tan(\theta)=\frac{5}{4}$ to solve for theta, take the inverse tangent of both sides:

$\small \theta \approx 51.34$

Now we have the direction, and we can solve for the magnitude using Pythagorean Theorem:

$\small 4^2 + 5^2 = C^2$

$\small 16 + 25 = C^2$

$\small 41 = C^2$ take the square root of both sides

$\small \sqrt{41}=C$

The vector in polar form is $\small (\sqrt{41}, 51.34^o)$

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Question

Find the directional vector of $\underset{BA}{\rightarrow}$ if points A and B are and , respectively.

Answer

To find vector $\underset{BA}{\rightarrow}$ , the point A is the terminal point and point B is the starting point.

The directional vector can be determined by subtracting the start from the terminal point.

$\left \langle \underset{BA}\rightarrow \right \rangle=(2,3)-(-3,2)=\left \langle 5,1\right \rangle$

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Question

Find the vector through the points

and .

Answer

The correct vector is given by the subtraction of the two points: .

Since the subtraction here is component-wise, it is given by the formula: .

This results in the vector .

The vector is also correct as it is a scalar multiple of the vector marked as correct, it is found by reversing the order of the subtraction of the two points.

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Question

Find $\overrightarrow{PQ}$ if and .

Answer

To find the direction vector going from to , subtract the x and y-coordinates of from .

$\overrightarrow{PQ}=<-5-4, 2-1>= <-9, 1>$

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Question

Find $\overrightarrow{MN}$ if and .

Answer

To find the direction vector from to , subtract the x- and y-coordinates of from .

$\overrightarrow{MN}=<2-(-2), 2-1>=<4, 1>$

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Question

Find the vector that has the initial point and the terminal point .

Answer

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{AB}=<0-2, 4-(-2)>=<-2, 6>$

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Question

Find the direction vector that has an initial point at and a terminal point of .

Answer

To find the directional vector, subtract the coordinates of the initial point from the coordinates of the terminal point.

$\overrightarrow{CD}=<0-1, 1-5>=<-1, -4>$

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