# Find values so that the vector has the specified angle

• Calpalned
In summary, the question asks to find the values of x where the angle between two given vectors is 45 degrees. The equations used are the dot product, magnitude of vectors, and cosine of angle. The attempt at a solution involved setting the dot product and magnitudes equal to the cosine of 45 degrees, which simplified to a quadratic equation. The correctness of the final equation is uncertain.
Calpalned

## Homework Statement

Find the values of x such that the angle between the vectors <2, 1, -1>, and <1, x, 0> is 45 degrees.

## Homework Equations

<2, 1, -1> dot <1, x, 0> = 2 + x
|first vector| = (6)0.5
|2nd vector| = (1 + x2)1/2
Angle = cos-1(a dot b)/(|a||b|)
Cosine of 45 is the root of two divided by two.

## The Attempt at a Solution

I set (2 + x) divided by 6^0.5 and (1 + x2)^0.5 equal to 1 divided by the root of 2.
I eventually get 2√2 + x√2 = √6√1 + x2
Am I on the right track?

Calpalned said:

## Homework Statement

Find the values of x such that the angle between the vectors <2, 1, -1>, and <1, x, 0> is 45 degrees.

## Homework Equations

<2, 1, -1> dot <1, x, 0> = 2 + x
|first vector| = (6)0.5
|2nd vector| = (1 + x2)1/2
Angle = cos-1(a dot b)/(|a||b|)
Cosine of 45 is the root of two divided by two.

## The Attempt at a Solution

I set (2 + x) divided by 6^0.5 and (1 + x2)^0.5 equal to 1 divided by the root of 2.
Or equivalently, ##\frac{2 + x}{\sqrt{6}\sqrt{1 + x^2}} = \frac{\sqrt{2}}{2}##
Calpalned said:
I eventually get 2√2 + x√2 = √6√1 + x2
I'm not sure about the above, although it might be right. I get a quadratic with all integer coefficients.
Can you show what you did going from your equation above to this one?
Calpalned said:
Am I on the right track?

## 1. How do you find the values for a vector to have a specific angle?

To find the values for a vector to have a specific angle, you first need to determine the initial and terminal points of the vector. Then, you can use the formula θ = tan-1(y/x) to calculate the angle, where x and y are the horizontal and vertical components of the vector, respectively. You can then manipulate the values of x and y to achieve the desired angle.

## 2. Can the values for a vector have multiple angles?

Yes, it is possible for a vector to have multiple angles. This can happen when the vector is not in a standard position, meaning that its initial point is not at the origin and its terminal point is not on the positive x-axis. In this case, the vector can have both a positive and negative angle, with the angle being measured counterclockwise from the positive x-axis.

## 3. What is the difference between the angle of a vector and the direction of a vector?

The angle of a vector refers to the orientation of the vector in relation to the positive x-axis, measured in degrees or radians. On the other hand, the direction of a vector refers to the compass direction in which the vector is pointing, such as north, south, east, or west.

## 4. How does the magnitude of a vector affect its angle?

The magnitude of a vector does not directly affect its angle. However, if the magnitude of a vector is changed, the values of its x and y components will also change, which can result in a different angle. For example, if a vector with an angle of 45 degrees has a magnitude of 5, changing its magnitude to 10 will result in a different angle.

## 5. Are there any special cases when finding the values for a vector to have a specific angle?

Yes, there are a few special cases to consider when finding the values for a vector to have a specific angle. These include when the angle is 0 degrees or 180 degrees, when the vector is in a standard position, and when the vector is in a quadrantal position (where the terminal point lies on one of the axes). In these cases, the values for the vector can be easily determined without using the tangent function.

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