Problems from Distance from a point to a straight line in space

Development:

To find the distance to the straight line, we consider the point $Q=(1,3,-2)$ and the governing vector $\overrightarrow{v}=(1,0,1)$. We calculate now the vector product of $\overrightarrow{QP}$ times $\overrightarrow{v}$.

$\overrightarrow{QP}=(0,-1,5)$

$\begin{array}{rl}|\overrightarrow{QP}\times \overrightarrow{v}|=& \left|\left|\begin{array}{ccc}and& j& k\\ 0& -1& 5\\ 1& 0& 1\end{array}\right|\right|=|-i+5j+k|=|(-3,2,-1)|\\ =& \sqrt{1+25+1}=\sqrt{27}\end{array}$

And we can already apply the formula: $$\text{d}(P,r)={\displaystyle \frac{|\overrightarrow{QP}\times \overrightarrow{v}|}{|\overrightarrow{v}|}}={\displaystyle \frac{\sqrt{27}}{\sqrt{7}}}=\sqrt{{\displaystyle \frac{27}{2}}}$$

Solution:

$\text{d}(P,r)=\sqrt{{\displaystyle \frac{27}{2}}}$

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