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Chapter 3: Problem 18

Find the magnitude of the vector \(34 \hat{\imath}+13 \hat{\jmath} \mathrm{m}\) and determine its angle to the \(x\) -axis.

Short Answer

Expert verified
The magnitude of the vector is approximately 36.4 m and the angle with respect to the x-axis is approximately 0.37 radians or 21.04 degrees.

Step by step solution

01

Calculate the magnitude of the vector

The magnitude (or length) of a vector in 2D is found using the Pythagorean theorem: \(|V| = \sqrt{(\Delta x)^2 + (\Delta y)^2}\), where \(\Delta x\) and \(\Delta y\) are the components of the vector. Here, \(\Delta x = 34\) and \(\Delta y = 13\). Thus, the magnitude of the vector is \(\sqrt{(34)^2 + (13)^2}\).
02

Simplify the magnitude

To simplify, calculate (34)^2 + (13)^2 = 1156 + 169 = 1325. The square root of 1325 is approximately 36.4.
03

Calculate the angle

The angle \(\theta\) with the x-axis can be obtained using the inverse tangent function (or arctan). \(\theta = \arctan(\Delta y / \Delta x)\) where \(\Delta y = 13\) and \(\Delta x = 34\). Thus, \(\theta = \arctan(13 / 34)\).
04

Simplify the angle

Simplify \(\arctan(13 / 34)\) to get the angle in radians or degrees, depending on what is requested. Using a calculator, \(\arctan(13 / 34)\) equals approximately 0.37 radians or 21.04 degrees.

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