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Chapter 7: Problem 74

Find the magnitude and direction angle of the vector v. $$\mathbf{v}=8\left(\cos 135^{\circ} \mathbf{i}+\sin 135^{\circ} \mathbf{j}\right)$$

Short Answer

Expert verified
The magnitude of the vector \(\mathbf{v}=8(\cos 135^\circ\mathbf{i} + \sin 135^\circ\mathbf{j})\) is 8 and its direction angle is 135 degrees.

Step by step solution

01

Calculate the magnitude of vector v

The magnitude of the vector, also known as its length or norm, is denoted by \(||\mathbf{v}||\) and is calculated by the formula \(\sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the components of the vector. Here, \(\mathbf{v}\) has components \(x=8\cos135^\circ\) and \(y=8\sin135^\circ\). Since \(\cos135^\circ = -\frac{1}{\sqrt{2}}\) and \(\sin135^\circ = \frac{1}{\sqrt{2}}\), the magnitude is: \(\sqrt{(8*(-1/\sqrt{2}))^2 + (8*1/\sqrt{2})^2}=8\).
02

Find the direction angle of vector v

The direction angle of a vector in the plane (usually denoted by \(\theta\)) can be found using the equation \(\tan\theta=\frac{y}{x}\). For the components \(x=8\cos135^\circ\) and \(y=8\sin135^\circ\), plug these into the equation and solve for \(\theta\). Since \(\cos135^\circ = -\frac{1}{\sqrt{2}}\) and \(\sin135^\circ = \frac{1}{\sqrt{2}}\), we find \(\tan\theta=\frac{1}{-1} = -1\). Looking up this value in the inverse tangent function, we find that \(\theta=135^\circ\) (in quadrant II where the tangent function is negative).
03

Summarize the results

The magnitude of the vector \(\mathbf{v}=8(\cos 135^\circ\mathbf{i} + \sin 135^\circ\mathbf{j})\) is 8, and its direction angle is 135 degrees. It's important to note that the direction angle is measured counterclockwise from the positive x-axis so it falls in the second quadrant.

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