Answer
The value of $b$ for which $15\mathbf{i}-3\mathbf{j}$ and $-4\mathbf{i}+b\mathbf{j}$ are orthogonal is $-20$.
Work Step by Step
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
$\begin{align}
& 15\times \left( -4 \right)+\left( -3 \right)b=0 \\
& -60-3b=0 \\
& b=-20
\end{align}$
