## Precalculus (6th Edition) Blitzer

The value of $b$ for which $15\mathbf{i}-3\mathbf{j}$ and $-4\mathbf{i}+b\mathbf{j}$ are orthogonal is $-20$.
For two vectors to be orthogonal, their dot product must be zero.Therefore, $\mathbf{u}\cdot \mathbf{v}=0$ Substituting the values of $\mathbf{u}$ and $\mathbf{v}$ from (1) and (2), we get $\left( 15\mathbf{i}-3\mathbf{j} \right)\centerdot \left( -4\mathbf{i}+b\mathbf{j} \right)=0$ …… (3) Solving equation (3) for the value of $b$, we get \begin{align} & 15\times \left( -4 \right)+\left( -3 \right)b=0 \\ & -60-3b=0 \\ & b=-20 \end{align}