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Find the magnitude and direction angle of the vector \(\boldsymbol{v}\). $$\mathbf{v}=\langle 5,5 \sqrt{3}\rangle$$

Short Answer

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Question: Find the magnitude and direction angle of the vector \(\boldsymbol{v}=\langle 5,5\sqrt{3} \rangle\). Answer: The magnitude of the vector \(\boldsymbol{v}\) is 10, and its direction angle is \(60^\circ\).

Step by step solution

01

Calculate the magnitude of the vector

To find the magnitude of a vector, we use the formula: $$\|v\|=\sqrt{(v_x^2+v_y^2)}$$ where \(v_x\) and \(v_y\) are the x and y components of the vector, respectively. For our vector \(\boldsymbol{v}=\langle 5,5\sqrt{3} \rangle\), \(v_x = 5\), and \(v_y=5\sqrt{3}\).
02

Plug the components into the formula and simplify

Now, we plug the components into the formula and simplify the result: $$\|v\|=\sqrt{(5^2+(5\sqrt{3})^2)}$$
03

Evaluate the square root

Calculate the square root to find the magnitude of the vector: $$\|v\|=\sqrt{(25+75)}=\sqrt{100}=10$$ So the magnitude of the vector \(\boldsymbol{v}\) is 10.
04

Calculate the direction angle

To find the direction angle \(\theta\) between the vector and the x-axis, we use the formula: $$\theta = \mathrm{arctan} (\frac{v_y}{v_x})$$ For our vector \(\boldsymbol{v}=\langle 5,5\sqrt{3} \rangle\), \(v_x = 5\), and \(v_y=5\sqrt{3}\).
05

Plug the components into the formula and simplify

Now, we plug the components into the formula and simplify the result: $$\theta=\mathrm{arctan}(\frac{5\sqrt{3}}{5})$$
06

Evaluate the arctan function

Evaluate the arctan function to find the direction angle: $$\theta=\mathrm{arctan}(\sqrt{3})$$ The arctan of \(\sqrt{3}\) corresponds to an angle of \(60^\circ\). So the direction angle of the vector \(\boldsymbol{v}\) is \(60^\circ\). The magnitude of the vector \(\boldsymbol{v}\) is 10, and its direction angle is \(60^\circ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnitude of a Vector
The magnitude of a vector is a measure of its length. For a vector represented in two dimensions as \(\vec{v} = \langle x, y \rangle\), its magnitude, denoted as \(||\vec{v}||\), can be determined using the Pythagorean theorem. This formula involves squaring the components of the vector and taking the square root of their sum, \(||\vec{v}|| = \sqrt{x^2 + y^2}\).

Let's consider a practical example with the vector \(\boldsymbol{v} = \langle 5, 5\sqrt{3} \rangle\). We square its components: \(5^2\) for the x-component and \(\left(5\sqrt{3}\right)^2\) for the y-component, getting 25 and 75 respectively. Adding these, we obtain 100. Taking the square root gives us the magnitude of 10. This process is not only fundamental in mathematics but also essential in physical sciences for assessing vector quantities like force, velocity, and displacement.
Direction Angle of a Vector
The direction angle of a vector in a coordinate plane is the angle it makes with the positive x-axis. To find this angle, \(\theta\), we typically use the arctangent function (\(\mathrm{arctan}\)) of the ratio of the vector's y-component to its x-component. The formula is \(\theta = \mathrm{arctan} (\frac{y}{x})\).

In the case of the vector \(\boldsymbol{v}\), we compute \(\theta = \mathrm{arctan}(\frac{5\sqrt{3}}{5})\). By simplifying the ratio, which is \(\sqrt{3}\), and referring to the arctangent values, we find that \(\theta = 60^\circ\). Understanding the direction angle is especially useful for vector addition, physics problems involving forces, or any situation where the direction of a vector is necessary for computation.
Trigonometric Functions
Trigonometric functions are a cornerstone of mathematics, deeply ingrained in the analysis of triangles and circles. They relate the angles of triangles to their side lengths. The primary trigonometric functions are sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)), with the latter being the ratio of the opposite side to the adjacent side in a right-angled triangle. The arctangent (\(\mathrm{arctan}\)) function we used earlier is simply the inverse of the tangent function, helping us find an angle when given the tangent's value.

For example, in our vector problem, we found the direction angle by calculating \(\mathrm{arctan}(\sqrt{3})\) which is the angle whose tangent is \(\sqrt{3}\). The angle corresponding to that value is \(60^\circ\). Knowing how to use these functions is vital in various disciplines, including engineering, navigation, and physics, where angles and lengths must be calculated precisely.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry that states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem can be expressed as \(a^2 + b^2 = c^2\), where \(c\) represents the hypotenuse, while \(a\) and \(b\) represent the other sides.

In the context of vectors, this theorem allows us to compute the magnitude of a vector from its components. As done with the vector \(\boldsymbol{v}\), the x and y components act as sides \(a\) and \(b\), while the magnitude is the hypotenuse \(c\). Applying the Pythagorean theorem, we calculate the vector's length, providing an essential tool for problems in algebra, trigonometry, and physics where distance and magnitude are key elements.

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Most popular questions from this chapter

Find the component form of the vector \(v\) whose magnitude and direction angle \(\theta\) are given. $$\|\mathbf{v}\|=10, \theta=225^{\circ}$$

Find a unit vector that has the same direction as \(v\). $$-7 \mathbf{i}+8 \mathbf{j}$$

The sum of two distinct complex numbers, \(a+b i\) and \(c+d i,\) can be found geometrically by means of the socalled parallelogram rule: Plot the points \(a+b i\) and \(c+d i\) in the complex plane, and form the parallelogram, three of whose vertices are \(0, a+b i,\) and \(c+d i,\) as in the figure. Then the fourth vertex of the parallelogram is the point whose coordinate is the sum $$(a+b i)+(c+d i)=(a+c)+(b+d) i$$ (GRAPH CAN'T COPY). Complete the following proof of the parallelogram rule when \(a \neq 0\) and \(c \neq 0\) (a) Find the slope of the line \(K\) from 0 to \(a+b i .[\text { Hint: } K\) contains the points \((0,0) \text { and }(a, b) .]\) (b) Find the slope of the line \(N\) from 0 to \(c+d i\) (c) Find the equation of the line \(L\) through \(a+b i\) and parallel to line \(N\) of part (b). [Hint: The point \((a, b)\) is on \(L\) find the slope of \(L\) by using part (b) and facts about the slope of parallel lines.] (d) Find the equation of the line \(M\) through \(c+d i\) and parallel to line \(K\) of part (a). (e) Label the lines \(K, L, M,\) and \(N\) in the figure. (f) Show by using substitution that the point \((a+c, b+d)\) satisfies both the equation of line \(L\) and the equation of line \(M .\) Therefore, \((a+c, b+d)\) lies on both \(L\) and \(M\) since the only point on both \(L\) and \(M\) is the fourth vertex of the parallelogram (see the figure), this vertex must be \((a+c, b+d) .\) Hence, this vertex has coordinate $$(a+c)+(b+d) i=(a+b i)+(c+d i)$$

Show that (1,2),(3,4),(5,2) are the vertices of a right triangle by considering the sides of the triangle as vectors.

Find \(u+v, v-u,\) and \(2 u-3 v\). $$\mathbf{u}=\sqrt{2} \mathbf{j}, \mathbf{v}=\sqrt{3} \mathbf{i}$$

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