Question 3.7: Translating Unit-Vector Information to Magnitude and Directi......
Translating Unit-Vector Information to Magnitude and Direction
Find the magnitude and direction of the vector \overrightarrow{ D }=(-3.5 \hat{\imath}+4.0 \hat{\jmath}) from Example 3.6.
INTERPRET and ANTICIPATE
Our work in Example 3.6 helps us anticipate the answer here. According to our sketch (Fig. 3.32), \vec{D} vector points into quadrant II, so we expect that 90^{\circ}<\theta<180^{\circ}
SOLVE
The scalar components are the coefficients in front of the unit vectors.
Find the magnitude from Equation 3.12.
A=\sqrt{A_x^2+A_y^2} (3.12)
\begin{aligned}& D=\sqrt{D_x^2+D_y^2} \\& D=\sqrt{(-3.5)^2+4.0^2} \\& D=5.3\end{aligned}Use Equation 3.14
\theta=\tan ^{-1} \frac{A_y}{A_x}\quad \quad (3.14)and a calculator to find the direction.
\tan ^{-1} \frac{4.0}{-3.5}=-49^{\circ}CHECK and THINK
This result does not meet our expectation that 90^{\circ}< \theta <180^{\circ} . So, we add 180° to correct the answer returned by the calculator because \vec{D} should point into quadrant \mathrm{II}.
\theta=-49^{\circ}+180^{\circ}=131^{\circ}