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 \( \langle-7,15\rangle \)

Sure, let's find the component form (i.e., the \(x\) and \(y\) components) of each vector based on the given magnitude and direction. ### 21. \( |\mathbf{v}| = 5 ; \theta = 20^{\circ} \) A vector with magnitude \(5\) and angle \(20^\circ\) from the positive \(x\)-axis can be expressed using cosine and sine for the \(x\) and \(y\) components, respectively. \[ \mathbf{v} = \langle 5 \cos 20^\circ,\ 5 \sin 20^\circ \rangle \approx \langle 4.70,\ 1.71 \rangle \] ### 22. \( |\mathbf{u}| = 4\sqrt{3} ; \theta = \frac{\pi}{6} \) Here, the angle \(\frac{\pi}{6}\) radians is equivalent to \(30^\circ\). Using the exact values for cosine and sine at \(30^\circ\): \[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2} \] Thus, \[ \mathbf{u} = \langle 4\sqrt{3} \cdot \frac{\sqrt{3}}{2},\ 4\sqrt{3} \cdot \frac{1}{2} \rangle = \langle 6,\ 2\sqrt{3} \rangle \] ### 23. \( |\mathbf{w}| = 4 \) in the direction of \( \langle 2, -5 \rangle \) First, find the unit vector in the direction of \(\langle 2, -5 \rangle\): \[ \text{Magnitude of } \langle 2, -5 \rangle = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \] \[ \text{Unit vector} = \left\langle \frac{2}{\sqrt{29}},\ \frac{-5}{\sqrt{29}} \right\rangle \] Now, multiply by the magnitude \(4\) to get \(\mathbf{w}\): \[ \mathbf{w} = 4 \cdot \left\langle \frac{2}{\sqrt{29}},\ \frac{-5}{\sqrt{29}} \right\rangle = \left\langle \frac{8}{\sqrt{29}},\ \frac{-20}{\sqrt{29}} \right\rangle \] For rationalized components: \[ \mathbf{w} = \left\langle \frac{8\sqrt{29}}{29},\ \frac{-20\sqrt{29}}{29} \right\rangle \] ### 24. \( |\mathbf{w}| = 9 \) in the direction of \( \langle -7, 15 \rangle \) Similarly, find the unit vector in the direction of \(\langle -7, 15 \rangle\): \[ \text{Magnitude of } \langle -7, 15 \rangle = \sqrt{(-7)^2 + 15^2} = \sqrt{49 + 225} = \sqrt{274} \] \[ \text{Unit vector} = \left\langle \frac{-7}{\sqrt{274}},\ \frac{15}{\sqrt{274}} \right\rangle \] Multiply by the magnitude \(9\) to get \(\mathbf{w}\): \[ \mathbf{w} = 9 \cdot \left\langle \frac{-7}{\sqrt{274}},\ \frac{15}{\sqrt{274}} \right\rangle = \left\langle \frac{-63}{\sqrt{274}},\ \frac{135}{\sqrt{274}} \right\rangle \] For rationalized components: \[ \mathbf{w} = \left\langle \frac{-63\sqrt{274}}{274},\ \frac{135\sqrt{274}}{274} \right\rangle \] --- **Summary of Component Forms:** 21. \(\mathbf{v} = \langle 5 \cos 20^\circ,\ 5 \sin 20^\circ \rangle \approx \langle 4.70,\ 1.71 \rangle\) 22. \(\mathbf{u} = \langle 6,\ 2\sqrt{3} \rangle\) 23. \(\mathbf{w} = \left\langle \dfrac{8}{\sqrt{29}},\ \dfrac{-20}{\sqrt{29}} \right\rangle\) 24. \(\mathbf{w} = \left\langle \dfrac{-63}{\sqrt{274}},\ \dfrac{135}{\sqrt{274}} \right\rangle\)