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

To find two unit vectors parallel to \( v = -24 \mathbf{i} - 7 \mathbf{j} \), we first calculate its magnitude. The magnitude \( ||v|| \) is given by: \[ ||v|| = \sqrt{(-24)^2 + (-7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25. \] Now, the unit vector in the same direction is: \[ u = \frac{v}{||v||} = \frac{-24 \mathbf{i} - 7 \mathbf{j}}{25} = -\frac{24}{25} \mathbf{i} - \frac{7}{25} \mathbf{j}. \] To find another unit vector parallel but in the opposite direction, we simply negate this result: \[ -u = \frac{24}{25} \mathbf{i} + \frac{7}{25} \mathbf{j}. \] So, the two unit vectors parallel to \( v \) are: \[ -\frac{24}{25} \mathbf{i} - \frac{7}{25} \mathbf{j} \quad \text{and} \quad \frac{24}{25} \mathbf{i} + \frac{7}{25} \mathbf{j}. \] For part (b), we need \( v = \langle \frac{1}{6}, b \rangle \) to be a unit vector, which means its magnitude must equal 1: \[ ||v|| = \sqrt{\left(\frac{1}{6}\right)^2 + b^2} = 1. \] Squaring both sides gives: \[ \left(\frac{1}{6}\right)^2 + b^2 = 1 \implies \frac{1}{36} + b^2 = 1 \implies b^2 = 1 - \frac{1}{36} = \frac{36}{36} - \frac{1}{36} = \frac{35}{36}. \] Taking the square root results in two possible values for \( b \): \[ b = \pm \frac{\sqrt{35}}{6}. \] In part (c), to find all values of \( a \) such that \( w = a \mathbf{i} - \frac{a}{10} \mathbf{j} \) is a unit vector, we calculate its magnitude: \[ ||w|| = \sqrt{a^2 + \left(-\frac{a}{10}\right)^2} = \sqrt{a^2 + \frac{a^2}{100}} = \sqrt{a^2\left(1 + \frac{1}{100}\right)} = \sqrt{a^2 \cdot \frac{101}{100}} = |a| \sqrt{\frac{101}{100}}. \] We want this equal to 1, so: \[ |a| \sqrt{\frac{101}{100}} = 1 \implies |a| = \sqrt{\frac{100}{101}}. \] Thus, the values of \( a \) are: \[ a = \pm \sqrt{\frac{100}{101}}. \]