Answer
(b) 2.68 units
$(a)$
3.7 units
Work Step by Step
We know that $\frac{\|A \vec{B} \times A \vec{B}\|}{\|A \vec{p}\|}=d$
$(a)$
\[
\begin{array}{c}
-4 \hat{\imath}+2 \hat{k}=\overrightarrow{A P} \\
-(3 \hat{\imath}-2 \hat{\jmath}+4 \hat{k})= \overrightarrow{A B}\\
=\frac{\|(3 \hat{\imath}-2 \hat{\jmath}+4 \hat{\kappa})\times( -4 \hat{\imath}+2 \hat{R}) \|}{\sqrt{29}} \\
= \sqrt{\frac{141}{29}}.2=4.4 \text { units }
\end{array}
\]
(b)
\[
\begin{array}{c}
2 \hat{\imath}+2 \hat{\jmath}=\overrightarrow{A P} \\
-2 \hat{\imath}+\hat{\jmath}= \overrightarrow{A B}\\
=\frac{\|(-2 \hat{\imath}+\hat{\jmath})\times \
(2 \hat{\imath}+2 \hat{\jmath}) |}{\sqrt{5}} \\
=116 \hat{k} \| / \sqrt{5} \\
=\frac{6}{\sqrt{5}}=2.68 \mathrm{units}
\end{array}
\]
