Complete parts (a) through (c) below. a. Find two unit vectors parallel to \( v=-24 i-7 j \). b. Find b if \( \mathrm{v}=\left\langle\frac{1}{6}, \mathrm{~b}\right) \) is a unit vector. c. Find all values of a such that \( \mathrm{w}=\mathrm{ai}-\frac{\mathrm{a}}{10} \mathrm{j} \) is a unit vector. a. The parallel unit vector with the same direction is

To find the unit vectors parallel to \( v = -24i - 7j \), we first calculate the magnitude of \( v \): \[ \|v\| = \sqrt{(-24)^2 + (-7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25. \] The unit vector in the same direction is: \[ u_1 = \frac{1}{\|v\|} v = \frac{1}{25}(-24i - 7j) = -\frac{24}{25}i - \frac{7}{25}j. \] For the unit vector in the opposite direction: \[ u_2 = -u_1 = \frac{24}{25}i + \frac{7}{25}j. \] So the two unit vectors parallel to \( v \) are: \[ -\frac{24}{25}i - \frac{7}{25}j \quad \text{and} \quad \frac{24}{25}i + \frac{7}{25}j. \] Next, let’s find \( b \) for \( v = \left\langle \frac{1}{6}, b \right\rangle \) to be a unit vector: The magnitude of this vector must equal 1: \[ \sqrt{\left(\frac{1}{6}\right)^2 + b^2} = 1. \] Squaring both sides: \[ \frac{1}{36} + b^2 = 1 \implies b^2 = 1 - \frac{1}{36} = \frac{36}{36} - \frac{1}{36} = \frac{35}{36}. \] Thus, \( b = \pm \sqrt{\frac{35}{36}} = \pm \frac{\sqrt{35}}{6} \). Now, for \( w = ai - \frac{a}{10}j \) to be a unit vector: Calculate the magnitude: \[ \|w\| = \sqrt{a^2 + \left(-\frac{a}{10}\right)^2} = \sqrt{a^2 + \frac{a^2}{100}} = \sqrt{a^2\left(1 + \frac{1}{100}\right)} = \sqrt{a^2 \frac{101}{100}} = |a| \sqrt{\frac{101}{100}}. \] Setting \( |a| \sqrt{\frac{101}{100}} = 1 \) gives: \[ |a| = \frac{1}{\sqrt{\frac{101}{100}}} = \frac{10}{\sqrt{101}}. \] Thus, \( a = \pm \frac{10}{\sqrt{101}} \). In summary: - Two unit vectors parallel to \( v \) are \( -\frac{24}{25}i - \frac{7}{25}j \) and \( \frac{24}{25}i + \frac{7}{25}j \). - The values for \( b \) are \( \pm \frac{\sqrt{35}}{6} \). - The values for \( a \) are \( \pm \frac{10}{\sqrt{101}} \).