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

We find the angle using trigonometric identities:

\(\displaystyle \tan{\theta}= \frac{5}{3}\)

Using a calculator,

\(\displaystyle \theta = \arctan{\frac{5}{3}} \approx 59.04^\circ\)

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

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

So, our vector written in polar form is

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

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

\(\displaystyle M=\sqrt{x^2+y^2}\)

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

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

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

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

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

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

Find \(\displaystyle R\).

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

Find the angle.

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

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

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

Solving for r will give us the equation:

\(\displaystyle r=\frac{x}{\cos\theta}=\frac{y}{sin\theta}\)

We can then solve this equation for theta thusly:

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

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

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

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.

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

Therefore, the vector expressed in polar form is:

\(\displaystyle v=(15.30,78,69^o)\)

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

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

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

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

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

However, the direction of \(\displaystyle \left \langle -3,-2\right \rangle\) 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.

 \(\displaystyle \theta = 33.69+180 = 213.69\)

Therefore, the answer is:

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

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:

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

\(\displaystyle \small 7.36 \approx x\)

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:

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

\(\displaystyle \small 8.17 \approx y\)

Putting this in component form results in the vector \(\displaystyle \small < 7.36, 8.17>\)

First, it could be helpful to draw the vector so that we can get a sense of what it looks like. The component form \(\displaystyle \small < 2, -7>\) 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:

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

\(\displaystyle \small 4 + 49 = C^2\)

\(\displaystyle \small 53 = C^2\) take the square root of both sides

\(\displaystyle \small \sqrt{53}= C\) The magnitude is \(\displaystyle \small \sqrt{53}\).

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 \(\displaystyle \small \frac{7}{2}\):

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

\(\displaystyle \small \theta \approx 74.05^o\)

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 \(\displaystyle \small 360^o - 74.05^o = 285.95^o\)

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:

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

\(\displaystyle \small \theta \approx 51.34\)

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

\(\displaystyle \small 4^2 + 5^2 = C^2\)

\(\displaystyle \small 16 + 25 = C^2\)

\(\displaystyle \small 41 = C^2\) take the square root of both sides

\(\displaystyle \small \sqrt{41}=C\)

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