Question

A vector points into the third quadrant, and its x and y components are both negative. If its magnitude is equal to two times the magnitude of its x component, what is the angle between the vector and the positive x axis?

Answer 1

The angle between the vector and the positive x axis is approximately 60°.

Step-by-step explanation:

The vector points into the third quadrant, which means its x and y components are both negative. Let's assume that the magnitude of the vector is represented by |A| and the magnitude of the x component is represented by |Ax|. The given condition states that |A| = 2 * |Ax|. Using this information, we can find the angle between the vector and the positive x axis.

First, we need to find the values of |A| and |Ax|. Since |A| = 2 * |Ax|, we can substitute this into the Pythagorean theorem:

|A|² = |Ax|² + |Ay|²

Now let's substitute the given information into the equation:

(2 * |Ax|)² = |Ax|² + (|Ay|)²

Expanding and simplifying the equation:

4 * |Ax|² = |Ax|² + (|Ay|)²

3 * |Ax|² = (|Ay|)²

Now, we can take the square root of both sides:

√(3 * |Ax|²) = √((|Ay|)²)

√3 * |Ax| = |Ay|

Since both |Ax| and |Ay| are negative, we can ignore the negative sign. Therefore, |Ax| = -1 and |Ay| = -√3.

Now, we have all the information we need to find the angle between the vector and the positive x axis. We can use the formula: tan(A) = |Ay|/|Ax| = (-√3)/(-1) = √3

The angle A is the inverse tangent of √3: A = atan(√3) ≈ 60°

So, the angle between the vector and the positive x axis is approximately 60°.

Answer 2

`240^(\circ)`

Step-by-step explanation:

The vector lies in the 3rd quadrant, which means that both its x and y components are negative.

Let's analyze the magnitude now. We know that the magnitude of the vector (let's call it v) is 2 times the magnitude of the x-component:

`|v| = 2 |v_x|`

we know that the magnitude of the x-component is given by

`|v_x | = |v| cos \theta`

where here
`\theta` is measured as angle below the negative x-direction, since we are in the 3rd quadrant. Substituting into the previous equation,

`|v| = 2 |v| cos \theta\\cos \theta = (1)/(2)\\\theta = 60^(\circ)`

But this is the angle between the vector and the negative x axis: therefore, the angle between the vector and the positive x axis is

`\theta=60^(\circ) + 180^(\circ)=240^(\circ)`