Vectors - Trigonometry

Card 0 of 32

Question

Determine the magnitude of vector A.

Answer

We can use the pythagorean theorem to solve this problem. Using $\vec{A}$ as our hypotenuse, we can drop a vertical vector perpendicular to the x-axis. We will call this $\vec{A_y}$ and it is 4 units in length. We can also extend a vector from the origin that connects to $\vec{A_y}$ . We will call this $\vec{A_x}$ and it is 3 units in length.

Using the pythagorean theorem:

$|\vec{A}|^2 = \vec{A_x}^2 + \vec{A_y}^2$

$|\vec{A}| = \sqrt{\vec{A_x}^2 + \vec{A_y}^2}$

$|\vec{A}| = \sqrt{3^2 + 4^2}$

$|\vec{A}| = \sqrt{9 + 16}$

$|\vec{A}| = \sqrt{25}$

$|\vec{A}| = 5$

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Question

Which of the following is the correct term for the sum of two vectors?

Answer

When summing two vectors, you have both an x and y component and you sum these separately leaving you with a coordinate as your answer. This coordinate is called a resultant.

Compare your answer with the correct one above

Question

Determine the resultant of $\vec{A} = (2,5)$ and $\vec{B} = (1,9)$ .

Answer

When determining the resultant of two vectors, you are finding the sum of two vectors. To do this you must add the x component and the y component separately.

$\vec{A}+\vec{B} = (2,5) + (1,9)$

$\vec{A}+\vec{B} = (2 + 1, 5 + 9)$

$\vec{A} + \vec{B} = (3, 14)$

Compare your answer with the correct one above

Question

Consider the following graphs where $\vec{A}$ begins at the origin and ends at and $\vec{B} = (4, 1)$ . Which of the following depicts the correct resultant of these two vectors.

Answer

To find the resultant we must sum the two vectors:

$\vec{A} + \vec{B} = (-3, 5) + (4, 1)$

$\vec{A} + \vec{B} = (-3 + 4, 5 + 1)$

$\vec{A} + \vec{B} = (1, 6)$

Now we must graph the resultant

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Question

How many degrees above the x-axis is $\vec{A}$ ?

Answer

First, we must understand what we are solving for. We are solving for the angle that is formed by $\vec{A}$ and the x-axis. To do this, we can extend a vector from the origin which stops directly under the end of $\vec{A}$ . We will call this new vector $\vec{A_x}$ and it will be 7 units long. We will also extend a vector upwards that is perpendicular to the x-axis. We will call this $\vec{A_y}$ and it will be 3 units long.

Now we can use the relationship that $tan(\theta) = \frac{opposite}{adjacent}$ where $\vec{A_x}$ is the adjacent side and $\vec{A_y}$ is the opposite side.

$tan(\theta) = \frac{\vec{A_y}}{\vec{A_x}}$

$tan(\theta) = \frac{3}{7}$

$\theta = tan_{-1}(\frac{3}{7})$

$\theta = 23.2$

And so $\vec{A}$ is 23.2 degrees above the x-axis.

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Question

Find the difference of the two vectors, $\vec{A}$ which ends at and $\vec{B}$ ending at .

Answer

When finding the difference of two vectors, you must subtract the x and y components separately.

$\vec{A} - \vec{B} = (-2, 4) - (7, -6)$

$\vec{A} - \vec{B} = (-2-7, 4 - (-6))$

$\vec{A} - \vec{B} = (-9, 10)$

Compare your answer with the correct one above

Question

Which of the following is the correct depiction of the difference of vectors A and B?

Answer

To find the difference of two vectors we must consider the x and y components separately.

$\vec{A} - \vec{B} = (5, 8) - (-3, 10)$

$\vec{A} - \vec{B} = (5 - (-3), 8 - 10)$

$\vec{A} = \vec{B} = (8, -2)$

And then we must correctly graph this vector

Compare your answer with the correct one above

Question

True or False: The magnitude of a vector is the length of the vector.

Answer

When finding the magnitude of the vector, you use either the Pythagorean Theorem by forming a right triangle with the vector in question or you can use the distance formula. This is much more clear considering the distance vector that the magnitude of the vector is in fact the length of the vector.

Compare your answer with the correct one above

Question

Determine the magnitude of vector A.

Answer

We can use the pythagorean theorem to solve this problem. Using $\vec{A}$ as our hypotenuse, we can drop a vertical vector perpendicular to the x-axis. We will call this $\vec{A_y}$ and it is 4 units in length. We can also extend a vector from the origin that connects to $\vec{A_y}$ . We will call this $\vec{A_x}$ and it is 3 units in length.

Using the pythagorean theorem:

$|\vec{A}|^2 = \vec{A_x}^2 + \vec{A_y}^2$

$|\vec{A}| = \sqrt{\vec{A_x}^2 + \vec{A_y}^2}$

$|\vec{A}| = \sqrt{3^2 + 4^2}$

$|\vec{A}| = \sqrt{9 + 16}$

$|\vec{A}| = \sqrt{25}$

$|\vec{A}| = 5$

Compare your answer with the correct one above

Question

Which of the following is the correct term for the sum of two vectors?

Answer

When summing two vectors, you have both an x and y component and you sum these separately leaving you with a coordinate as your answer. This coordinate is called a resultant.

Compare your answer with the correct one above

Question

Determine the resultant of $\vec{A} = (2,5)$ and $\vec{B} = (1,9)$ .

Answer

When determining the resultant of two vectors, you are finding the sum of two vectors. To do this you must add the x component and the y component separately.

$\vec{A}+\vec{B} = (2,5) + (1,9)$

$\vec{A}+\vec{B} = (2 + 1, 5 + 9)$

$\vec{A} + \vec{B} = (3, 14)$

Compare your answer with the correct one above

Question

Consider the following graphs where $\vec{A}$ begins at the origin and ends at and $\vec{B} = (4, 1)$ . Which of the following depicts the correct resultant of these two vectors.

Answer

To find the resultant we must sum the two vectors:

$\vec{A} + \vec{B} = (-3, 5) + (4, 1)$

$\vec{A} + \vec{B} = (-3 + 4, 5 + 1)$

$\vec{A} + \vec{B} = (1, 6)$

Now we must graph the resultant

Compare your answer with the correct one above

Question

How many degrees above the x-axis is $\vec{A}$ ?

Answer

First, we must understand what we are solving for. We are solving for the angle that is formed by $\vec{A}$ and the x-axis. To do this, we can extend a vector from the origin which stops directly under the end of $\vec{A}$ . We will call this new vector $\vec{A_x}$ and it will be 7 units long. We will also extend a vector upwards that is perpendicular to the x-axis. We will call this $\vec{A_y}$ and it will be 3 units long.

Now we can use the relationship that $tan(\theta) = \frac{opposite}{adjacent}$ where $\vec{A_x}$ is the adjacent side and $\vec{A_y}$ is the opposite side.

$tan(\theta) = \frac{\vec{A_y}}{\vec{A_x}}$

$tan(\theta) = \frac{3}{7}$

$\theta = tan_{-1}(\frac{3}{7})$

$\theta = 23.2$

And so $\vec{A}$ is 23.2 degrees above the x-axis.

Compare your answer with the correct one above

Question

Find the difference of the two vectors, $\vec{A}$ which ends at and $\vec{B}$ ending at .

Answer

When finding the difference of two vectors, you must subtract the x and y components separately.

$\vec{A} - \vec{B} = (-2, 4) - (7, -6)$

$\vec{A} - \vec{B} = (-2-7, 4 - (-6))$

$\vec{A} - \vec{B} = (-9, 10)$

Compare your answer with the correct one above

Question

Which of the following is the correct depiction of the difference of vectors A and B?

Answer

To find the difference of two vectors we must consider the x and y components separately.

$\vec{A} - \vec{B} = (5, 8) - (-3, 10)$

$\vec{A} - \vec{B} = (5 - (-3), 8 - 10)$

$\vec{A} = \vec{B} = (8, -2)$

And then we must correctly graph this vector

Compare your answer with the correct one above

Question

True or False: The magnitude of a vector is the length of the vector.

Answer

When finding the magnitude of the vector, you use either the Pythagorean Theorem by forming a right triangle with the vector in question or you can use the distance formula. This is much more clear considering the distance vector that the magnitude of the vector is in fact the length of the vector.

Compare your answer with the correct one above

Question

Determine the magnitude of vector A.

Answer

We can use the pythagorean theorem to solve this problem. Using $\vec{A}$ as our hypotenuse, we can drop a vertical vector perpendicular to the x-axis. We will call this $\vec{A_y}$ and it is 4 units in length. We can also extend a vector from the origin that connects to $\vec{A_y}$ . We will call this $\vec{A_x}$ and it is 3 units in length.

Using the pythagorean theorem:

$|\vec{A}|^2 = \vec{A_x}^2 + \vec{A_y}^2$

$|\vec{A}| = \sqrt{\vec{A_x}^2 + \vec{A_y}^2}$

$|\vec{A}| = \sqrt{3^2 + 4^2}$

$|\vec{A}| = \sqrt{9 + 16}$

$|\vec{A}| = \sqrt{25}$

$|\vec{A}| = 5$

Compare your answer with the correct one above

Question

Which of the following is the correct term for the sum of two vectors?

Answer

When summing two vectors, you have both an x and y component and you sum these separately leaving you with a coordinate as your answer. This coordinate is called a resultant.

Compare your answer with the correct one above

Question

Determine the resultant of $\vec{A} = (2,5)$ and $\vec{B} = (1,9)$ .

Answer

When determining the resultant of two vectors, you are finding the sum of two vectors. To do this you must add the x component and the y component separately.

$\vec{A}+\vec{B} = (2,5) + (1,9)$

$\vec{A}+\vec{B} = (2 + 1, 5 + 9)$

$\vec{A} + \vec{B} = (3, 14)$

Compare your answer with the correct one above

Question

Consider the following graphs where $\vec{A}$ begins at the origin and ends at and $\vec{B} = (4, 1)$ . Which of the following depicts the correct resultant of these two vectors.

Answer

To find the resultant we must sum the two vectors:

$\vec{A} + \vec{B} = (-3, 5) + (4, 1)$

$\vec{A} + \vec{B} = (-3 + 4, 5 + 1)$

$\vec{A} + \vec{B} = (1, 6)$

Now we must graph the resultant

Compare your answer with the correct one above

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