The Shortest Distance between a Line and a Point

First the line joining $P$ and the line l must be defined using vector subtraction. Let this line be defined as $\vec{PN}$ .

$\vec{PN} = \binom{6 - 4 \lambda }{k + 2\lambda} - \binom{2}{4} = \binom{4 - 4 \lambda}{2\lambda + k - 4}$

When this vector $\vec{PN}$ is normal to the direction vector of line l , $\binom{-4}{2}$ , the vector $\vec{PN}$ is at its shortest length. When dotting two orthogonal vectors the output should be 0. Using this definition, one can determine the value of $\lambda$ which minimizes the length of the vector $\vec{PN}$ .

$\binom{-4}{2} \cdot \binom{4 - 4\lambda}{2\lambda + k - 4} = 0$

$-4(4 - 4\lambda) + 2(2\lambda + k - 4) = 0$

$\lambda = \frac{12 - k}{10}$

Inputting this value of lambda into $\vec{PN}$ puts $\vec{PN}$ in terms of $k$ .

$\vec{PN} = \binom{4 - 4(\frac{12-k}{10})}{2(\frac{12-k}{10}) + k - 4} = (\frac{2k-4}{5})$ i $+ (\frac{4k-8}{5})$ j

To determine the length of the vector $\vec{PN}$ , one must add the squares of each component of the vector followed by square rooting the result. The length is indicated by $|\vec{PN}|$ .

$|\vec{PN}|^{2} = (\frac{2k-4}{5})^{2} + (\frac{4k-8}{5})^{2}$

$\quad \quad \quad \! \, \!= \frac{4k^{2}-16k+16}{5} = \frac{4(k-2)^{2}}{5}$

$\therefore |\vec{PN}| = \sqrt{\frac{4(k-2)^{2}}{5}} = \frac{2k-4}{\sqrt{5}}$

Equating this to the value of 7 provided in the question gives us the value of $k$ required.

$7 = \frac{2k-4}{\sqrt{5}}$

$k = \frac{7}{2}\sqrt{5} + 2$

Comparing the value of $k$ to $\frac{a}{b} \sqrt{c} + b$ we determine

$a = 7 \; \; \quad b = 2 \; \; \quad c = 5$

Inputting the values of $a$ , $b$ and $c$ into $3a + 2b + c$ yields

$3(7) + 2(2) + (5) = 30$