Geometric Vectors - Pre-Calculus

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Question

Express a vector with magnitude 2.24 directed 63.4° CCW from the x-axis in unit vector form.

Answer

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

$\text{magnitude}=2.24,\ \theta=63.4^o$

$x = 2.24 \cos(63.4^o) = 1.00$

$y = 2.24 \sin(63.4^o) = 2.00$

The resultant vector is: $\text{1.00i + 2.00j}$ .

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Question

What is the magnitude and angle for the following vector, measured CCW from the x-axis?

Answer

The magnitude of the vector is found using the distance formula:

$\sqrt{x^2+y^2}=\sqrt{(-2)^2+(3)^2}=\sqrt{13}=3.61$

To calculate the angle we must first find the inverse tangent of $\frac{y}{x}$ :

$\arctan(\frac{y}{x})=\arctan(\frac{3}{-2})=-56.3^o$

This angle value is the principal arctan, but it is in the fourth quadrant while our vector is in the second. We must add the angle 180° to this value to arrive at our final answer.

$\theta=180^o+(-56.3^o)=124^o$

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Question

Vector $\vec{A}$ has a magnitude of 3.61 and a direction 124° CCW from the x-axis. Express $2\vec{A}$ in unit vector form.

Answer

For vector $2\vec{A}$ , the magnitude is doubled, but the direction remains the same.

For our calculation, we use a magnitude of:

$m=2\times3.61=7.22$

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

$x = m \cos \theta = 7.22 \cos (124^o) = -4.00$

$y = m \sin \theta = 7.22 \sin (124^o) = 6.00$

The resultant vector is: .

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Question

$\vec{A}=-i-2j$

What are the magnitude and angle, CCW from the x-axis, of $2\vec{A}$ ?

Answer

When multiplying a vector by a constant (called scalar multiplication), we multiply each component by the constant.

$2\vec{A}=2(-1-2j)=-2i-4j$

The magnitude of this new vector is found with these new components:

$\sqrt{x^2+y^2}=\sqrt{(-2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47$

To calculate the angle we must first find the inverse tangent of $\frac{y}{x}$ :

$\arctan(\frac{y}{x})=\arctan(\frac{-4}{-2})=\arctan(2)=63.4^o$

This is the principal arctan, but it is in the first quadrant while our vector is in the third. We to add the angle 180° to this value to arrive at our final answer.

$\theta=180^o+63.4^o=243^o$

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Question

Vector $\vec{A}$ has a magnitude of 2.24 and is at an angle of 63.4° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 3.16 at an angle of 342° CCW from the x-axis.

Find $\vec{A}+\vec{B}$ by using the nose-to-tail graphical method.

Answer

First, construct the two vectors using ruler and protractor:

Place the tail of $\vec{B}$ at the nose of $\vec{A}$ :

Construct the resultant $\vec{R}$ from the tail of $\vec{A}$ to the nose of $\vec{B}$ :

With our ruler and protractor, we find that $\vec{R}$ is 4.12 at an angle of 14.0° CCW from the x-axis.

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Question

$\vec{A}=-2.00i+3.00j$

$\vec{B}=-1.00i-2.00j$

Find the magnitude and angle CCW from the x-axis of $\vec{A}-\vec{B}$ using the nose-to-tail graphical method.

Answer

Construct $\vec{A}$ and $\vec{B}$ from their x- and y-components:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Form $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$ by placing the tail of $-\vec{B}$ at the nose of $\vec{A}$ :

Construct and measure the resultant, $\vec{R}$ , from the tail of $\vec{A}$ to the nose of $-\vec{B}$ using a ruler and protractor:

$m_{\vec{R}}=5.10,\ \theta_{\vec{R}}=101^o$

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Question

Vector $\vec{A}$ has a magnitude of 2.24 and is at an angle of 63.4° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 3.61 and is at an angle of 124° CCW from the x-axis.

Find $\vec{A}-2\vec{B}$ by using the nose-to-tail graphical method.

Answer

First, construct the two vectors using ruler and protractor:

$2\vec{B}$ is twice the length of $\vec{B}$ , but in the same direction:

Since we are subtracting, reverse the direction of $2\vec{B}$ :

Form $\vec{A}-2\vec{B}=\vec{A}+(-2\vec{B})$ by placing the tail of $-2\vec{B}$ at the nose of $\vec{A}$ :

Construct and measure the resultant $\vec{R}$ from the tail of $\vec{A}$ to the nose of $-2\vec{B}$ with a ruler and protractor.

$m_{\vec{R}}=6.40,\ \theta_{\vec{R}}=321^o$

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Question

Vector $\vec{A}$ has a magnitude of 2.24 and is at an angle of 63.4° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 3.16 at an anlge of 342° CCW from the x-axis.

Find $\vec{A}+\vec{B}$ by using the parallelogram graphical method.

Answer

First, construct the two vectors using ruler and protractor:

Place the tails of both vectors at the same point:

Construct a parallelogram:

Construct and measure the resultant using ruler and protractor:

$m_{\vec{R}}=4.12,\ \theta_{\vec{R}}=14.0^o$

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Question

$\vec{A}=-1.00i-2.00j$

$\vec{B}=3.00i-1.00j$

Find $\vec{A}-\vec{B}$ using the parallelogram graphical method.

Answer

Construct $\vec{A}$ and $\vec{B}$ from their x- and y-components:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Place the tails of $\vec{A}$ and $-\vec{B}$ at the same point:

Construct a parallelogram:

Construct and measure the resultant using a ruler and protractor.

$m_{\vec{R}}=4.12,\ \theta_{\vec{R}}=194.^o$

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Question

Vector $\vec{A}$ has a magnitude of 3.61 and is at an angle of 124° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 2.24 at an anlge of 63.4° CCW from the x-axis.

Find $2\vec{A}-\vec{B}$ using the parallelogram graphical method.

Answer

First, construct the two vectors using ruler and protractor:

$2\vec{A}$ is twice the length of $\vec{A}$ , but in the same direction:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Place the tails of $2\vec{A}$ and $-\vec{B}$ at the same point:

Construct a parallelogram:

Construct and measure the resultant using ruler and protractor:

$m_{\vec{R}}=6.40,\ \theta_{\vec{R}}=141^o$

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Question

$\vec{A}=1.00i+2.00j$

$\vec{B}=3.00i-1.00j$

Find $\vec{A}+\vec{B}$ .

Answer

Finding the resultant requires us to add like components:

$\vec{R}=\vec{A}+\vec{B}$

$\vec{R}=(1.00i+2.00j)+(3.00i-1.00j)$

$\vec{R}=(1.00+3.00)i+(2.00-1.00)j$

$\vec{R}=4.00i+1.00j$

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Question

$\vec{A}=3.00i-1.00j$

$\vec{B}=-2.00i+3.00j$

Find $2\vec{A}+\vec{B}$ .

Answer

Finding the resultant requires us to add like components:

$\vec{R}=2\vec{A}+\vec{B}$

$\vec{R}=2(3.00i-1.00j)+(-2.00i+3.00j)$

$\vec{R}=(6.00i-2.00j)+(-2.00i+3.00j)$

$\vec{R}=(6.00-2.00)i+(-2.00+3.00)j$

$\vec{R}=4.00i+1.00j$

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Question

Evaluate

Answer

When adding two vectors, they need to be expanded into their components. Luckily, the problem statement gives us the vectors already in their component form. From here, we just need to remember that we can only add like components. So for this problem we get:

Now we can combine those values to write out the complete vector:

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Question

Find the norm of the vector $\vec{v}=[2,4,3]$ .

Answer

We find the norm of a vector by finding the sum of each element squared and then taking the square root.

$\left | \vec{v} \right |= \sqrt{2^2+4^2+3^2}=\sqrt{29}$ .

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Question

Find the product of the vector $\vec{r}= 11\hat{i}+7\hat{j}$ and the scalar .

Answer

When multiplying a vector by a scalar we multiply each component of the vector by the scalar and the result is a vector:

$a \vec{r}= 4 \cdot (11 \hat{i} + 7 \hat{j}) = 44 \hat{i}+28 \hat{j}$

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Question

Find the norm of the vector $\vec{r}=7\hat{i} +4 \hat{j}+5 \hat{k}$ .

Answer

We find the norm of a vector by finding the sum of each component squared and then taking the square root of that sum.

$\left | \vec{r}\right | = \sqrt{ 7^2+4^2+5^2 } = \sqrt{ 90 } = 3 \sqrt{10}$

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Question

This question refers to the previous question.

Simplify.

$4\cdot \left<3,6,9\right>$

Answer

In order to simplify this problem we need to multiply the scalar factor to each component of the vector.

In our case the scalar factor is

Thus,

$4\cdot \left<3,6,9\right>\rightarrow \left<4\cdot 3, 4\cdot 6, 4\cdot 9\right>$

$\rightarrow \left<12, 24, 36\right>$

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Question

Find the norm of the vector:

Answer

The norm of a vector is also known as the length of the vector. The norm is given by the formula:

$\sqrt{\sum_{i=1}^nv_i^2}=\left | v\right |$ .

Here, we have

$\left |v \right |=\sqrt{3^2+(-12)^2+5.6^2+2^2+(-9)^2}=\sqrt{269.36}$ ,

the correct answer.

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Question

Find the vector given by the product: $\frac{4}{7}<-3,5,14>$

Answer

Given a scalar k and a vector v, the vector given by their products is defined component-wise:

.

Here, our product is:

$\frac{4}{7}<-3,5,14>=<\frac{4}{7}(-3),\frac{4}{7}(5),\frac{4}{7}(14)>\ \ =<\frac{-12}{7},\frac{20}{7},8>$

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Question

Find the norm of vector $\left \langle -2,2\right \rangle$ .

Answer

Write the formula to find the norm, or the length the vector.

$\left | v\right |=\sqrt{a_{1}^2+b_{2}^2}$

Substitute the known values of the vector and solve.

$\left | v\right |=\sqrt{(-2)^2+2^2}= \sqrt{4+4}=\sqrt8=2\sqrt2$

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