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Chapter 1: Problem 81

Find the magnitude and direction of each of the following vectors, which are given in terms of their \(x\) - and \(y\) -components: \(\vec{A}=(23.0,59.0),\) and \(\vec{B}=(90.0,-150.0)\)

Short Answer

Expert verified
Answer: The magnitude of \(\vec{A}\) is \(\sqrt{5186}\) and its direction is approximately \(68.56^\circ\). The magnitude of \(\vec{B}\) is \(\sqrt{31500}\) and its direction is approximately \(-59.04^\circ\).

Step by step solution

01

Find the magnitude of \(\vec{A}\)

To find the magnitude of \(\vec{A}=(23.0, 59.0)\), we use the formula, $$ |\vec{A}| = \sqrt{A_x^2 + A_y^2} $$ Where \(A_x = 23.0\) and \(A_y = 59.0\). Insert the values into the formula: $$ |\vec{A}| = \sqrt{(23.0)^2 + (59.0)^2} $$ Calculate the magnitude: $$ |\vec{A}| = \sqrt{5186} $$ The magnitude of \(\vec{A}\) is \(\sqrt{5186}\).
02

Find the direction of \(\vec{A}\)

To find the direction of \(\vec{A}\), we use the arctangent function. $$ \theta_A = \arctan \left( \frac{A_y}{A_x} \right) $$ Insert the values of components \(A_x\) and \(A_y\): $$ \theta_A = \arctan \left( \frac{59.0}{23.0} \right) $$ Calculate the direction: $$ \theta_A \approx 68.56^\circ $$ The direction of vector \(\vec{A}\) is approximately \(68.56^\circ\). Now, we will find the magnitude and direction of the vector \(\vec{B}\).
03

Find the magnitude of \(\vec{B}\)

To find the magnitude of \(\vec{B}=(90.0, -150.0)\), we again use the formula, $$ |\vec{B}| = \sqrt{B_x^2 + B_y^2} $$ Where \(B_x = 90.0\) and \(B_y = -150.0\). Insert the values into the formula: $$ |\vec{B}| = \sqrt{(90.0)^2 + (-150.0)^2} $$ Calculate the magnitude: $$ |\vec{B}| = \sqrt{31500} $$ The magnitude of \(\vec{B}\) is \(\sqrt{31500}\).
04

Find the direction of \(\vec{B}\)

To find the direction of \(\vec{B}\), we again use the arctangent function. $$ \theta_B = \arctan \left( \frac{B_y}{B_x} \right) $$ Insert the values of components \(B_x\) and \(B_y\): $$ \theta_B = \arctan \left( \frac{-150.0}{90.0} \right) $$ Calculate the direction: $$ \theta_B \approx -59.04^\circ $$ The direction of vector \(\vec{B}\) is approximately \(-59.04^\circ\). In conclusion, the magnitude and direction of vector \(\vec{A}\) are \(\sqrt{5186}\) and \(68.56^\circ\), while the magnitude and direction of vector \(\vec{B}\) are \(\sqrt{31500}\) and \(-59.04^\circ\).

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Most popular questions from this chapter

Find the magnitude and direction of \(-\vec{A}+\vec{B},\) where \(\vec{A}=(23.0,59.0), \vec{B}=(90.0,-150.0)\)

If \(\vec{A}\) and \(\vec{B}\) are vectors and \(\vec{B}=-\vec{A},\) which of the following is true? a) The magnitude of \(\vec{B}\) is equal to the negative of the magnitude of \(\vec{A}\). b) \(\vec{A}\) and \(\vec{B}\) are perpendicular. c) The direction angle of \(\vec{B}\) is equal to the direction angle of \(\vec{A}\) plus \(180^{\circ}\) d) \(\vec{A}+\vec{B}=2 \vec{A}\).

A position vector has a length of \(40.0 \mathrm{~m}\) and is at an angle of \(57.0^{\circ}\) above the \(x\) -axis. Find the vector's components.

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