19. Find the value(s) of \( b \) such that the vectors are orthogonal. \( \begin{array}{ll}\text { (a) }\left(\begin{array}{l}b \\ 3 \\ 2\end{array}\right) \text { and }\left(\begin{array}{l}1 \\ b \\ 1\end{array}\right) & \text { (b) }\left(\begin{array}{c}4 \\ -2 \\ 7\end{array}\right) \text { and }\left(\begin{array}{c}b^{2} \\ b \\ 0\end{array}\right) \\ \text { (c) }\left(\begin{array}{c}b \\ 11 \\ -3\end{array}\right) \text { and }\left(\begin{array}{c}2 b \\ -b \\ -5\end{array}\right) & \text { (d) }\left(\begin{array}{c}2 \\ 5 \\ 2 b\end{array}\right) \text { and }\left(\begin{array}{c}6 \\ 4 \\ -b\end{array}\right)\end{array} \)

To find the values of \( b \) such that the vectors are orthogonal, we need to set the dot product of each pair of vectors equal to zero. (a) For the vectors \( \begin{pmatrix} b \\ 3 \\ 2 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ b \\ 1 \end{pmatrix} \): \[ b \cdot 1 + 3 \cdot b + 2 \cdot 1 = 0 \implies b + 3b + 2 = 0 \implies 4b + 2 = 0 \implies b = -\frac{1}{2}. \] (b) For the vectors \( \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix} \) and \( \begin{pmatrix} b^2 \\ b \\ 0 \end{pmatrix} \): \[ 4b^2 - 2b + 7 \cdot 0 = 0 \implies 4b^2 - 2b = 0 \implies 2b(2b - 1) = 0 \implies b = 0 \text{ or } b = \frac{1}{2}. \] (c) For the vectors \( \begin{pmatrix} b \\ 11 \\ -3 \end{pmatrix} \) and \( \begin{pmatrix} 2b \\ -b \\ -5 \end{pmatrix} \): \[ b(2b) + 11(-b) + (-3)(-5) = 0 \implies 2b^2 - 11b + 15 = 0. \] Using the quadratic formula, we find \( b = 5 \) or \( b = 1.5 \). (d) For the vectors \( \begin{pmatrix} 2 \\ 5 \\ 2b \end{pmatrix} \) and \( \begin{pmatrix} 6 \\ 4 \\ -b \end{pmatrix} \): \[ 2(6) + 5(4) + 2b(-b) = 0 \implies 12 + 20 - 2b^2 = 0 \implies 2b^2 = 32 \implies b^2 = 16 \implies b = 4 \text{ or } b = -4. \] In summary, the values of \( b \) are: (a) \( -\frac{1}{2} \) (b) \( 0 \text{ or } \frac{1}{2} \) (c) \( 5 \text{ or } 1.5 \) (d) \( 4 \text{ or } -4 \)