State the component form of each vector. 21. \( |\mathbf{v}|=5 ; \theta=20^{\circ} \) 22. \( |\mathbf{u}|=4 \sqrt{3} ; \theta=\frac{\pi}{6} \) 23. \( |\mathbf{w}|=4 \) in the direction of \( \langle 2,-5\rangle \) 24. \( |\mathbf{w}|=9 \) in the direction of \( \langle-7,15\rangle \)
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The component form of a vector can be found using the magnitude and the angle. For vector \( \mathbf{v} \), we convert \( \theta \) to radians and compute the components using: \(\mathbf{v} = \langle |\mathbf{v}| \cos(\theta), |\mathbf{v}| \sin(\theta) \rangle\). Thus, \( \mathbf{v} = \langle 5 \cos(20^\circ), 5 \sin(20^\circ) \rangle \approx \langle 4.7, 1.7 \rangle \). For vector \( \mathbf{u} \), since \(\theta = \frac{\pi}{6}\) (which is 30 degrees), we calculate similarly: \( \mathbf{u} = \langle 4\sqrt{3} \cos\left(\frac{\pi}{6}\right), 4\sqrt{3} \sin\left(\frac{\pi}{6}\right) \rangle \approx \langle 6.93, 4 \rangle \). For vector \( \mathbf{w} \) with a magnitude of 4 in the direction of \( \langle 2, -5 \rangle \), we first normalize the direction vector to find the unit vector \( \mathbf{d} \): \(\mathbf{d} = \frac{\langle 2, -5 \rangle}{\sqrt{2^2 + (-5)^2}} = \langle \frac{2}{\sqrt{29}}, \frac{-5}{\sqrt{29}} \rangle\). Hence, \( \mathbf{w} = 4 \cdot \mathbf{d} \approx \langle \frac{8}{\sqrt{29}}, \frac{-20}{\sqrt{29}} \rangle \). Lastly, for vector \( \mathbf{w} \) with a magnitude of 9 in the direction of \( \langle -7, 15 \rangle \), normalizing gives: \(\mathbf{d} = \frac{\langle -7, 15 \rangle}{\sqrt{(-7)^2 + 15^2}} = \langle \frac{-7}{\sqrt{274}}, \frac{15}{\sqrt{274}} \rangle\). Therefore, \( \mathbf{w} = 9 \cdot \mathbf{d} \approx \langle \frac{-63}{\sqrt{274}}, \frac{135}{\sqrt{274}} \rangle \).
