32. Find all possible values of a unit vector \( \boldsymbol{u} \) that will be perpendicular to both \( \boldsymbol{i}+2 \boldsymbol{j}+\boldsymbol{k} \) and \( 3 \boldsymbol{i}-4 \boldsymbol{j}+2 \boldsymbol{k} \).

To find a unit vector \( \boldsymbol{u} \) that is perpendicular to both given vectors, we can utilize the cross product. The cross product of two vectors yields a vector that is orthogonal to both. First, we denote the vectors: \[ \boldsymbol{v_1} = \boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k} \] \[ \boldsymbol{v_2} = 3\boldsymbol{i} - 4\boldsymbol{j} + 2\boldsymbol{k} \] Now, compute the cross product \( \boldsymbol{u} = \boldsymbol{v_1} \times \boldsymbol{v_2} \): \[ \boldsymbol{u} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 1 & 2 & 1 \\ 3 & -4 & 2 \end{vmatrix} \] Calculating this determinant: \[ \boldsymbol{u} = \boldsymbol{i}\begin{vmatrix}2 & 1 \\ -4 & 2\end{vmatrix} - \boldsymbol{j}\begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} + \boldsymbol{k}\begin{vmatrix}1 & 2 \\ 3 & -4\end{vmatrix} \] Calculating each of those 2x2 determinants: \[ \begin{vmatrix}2 & 1 \\ -4 & 2\end{vmatrix} = (2)(2) - (1)(-4) = 4 + 4 = 8 \] \[ \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = (1)(2) - (1)(3) = 2 - 3 = -1 \] \[ \begin{vmatrix}1 & 2 \\ 3 & -4\end{vmatrix} = (1)(-4) - (2)(3) = -4 - 6 = -10 \] Substituting back into the cross product expression: \[ \boldsymbol{u} = 8\boldsymbol{i} + 1\boldsymbol{j} - 10\boldsymbol{k} \] Now, this gives: \[ \boldsymbol{u} = (8, 1, -10) \] To find the unit vector, we need to normalize \( \boldsymbol{u} \): First, compute the magnitude \( ||\boldsymbol{u}|| \): \[ ||\boldsymbol{u}|| = \sqrt{8^2 + 1^2 + (-10)^2} = \sqrt{64 + 1 + 100} = \sqrt{165} \] Thus, the unit vector in the direction of \( \boldsymbol{u} \) is: \[ \hat{\boldsymbol{u}} = \frac{1}{\sqrt{165}}(8, 1, -10) \] Consequently, the possible values for the unit vector that is perpendicular to both original vectors can be: \[ \hat{\boldsymbol{u}} = \frac{8}{\sqrt{165}} \boldsymbol{i} + \frac{1}{\sqrt{165}} \boldsymbol{j} - \frac{10}{\sqrt{165}} \boldsymbol{k} \] or its negative, \[ -\hat{\boldsymbol{u}} = -\frac{8}{\sqrt{165}} \boldsymbol{i} - \frac{1}{\sqrt{165}} \boldsymbol{j} + \frac{10}{\sqrt{165}} \boldsymbol{k} \] To summarize, the two possible values for the unit vector perpendicular to both given vectors are: \[ \hat{\boldsymbol{u}} = \frac{8}{\sqrt{165}} \boldsymbol{i} + \frac{1}{\sqrt{165}} \boldsymbol{j} - \frac{10}{\sqrt{165}} \boldsymbol{k} \quad \text{and} \quad -\hat{\boldsymbol{u}} = -\frac{8}{\sqrt{165}} \boldsymbol{i} - \frac{1}{\sqrt{165}} \boldsymbol{j} + \frac{10}{\sqrt{165}} \boldsymbol{k} \]