The length of the perpendicular from the point $(0,2,3)$ on the line…
The length of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$
$\sqrt{15}$ units
$\sqrt{21}$ units
$\sqrt{33}$ units
$\sqrt{11}$ units
Solution
Any point on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$ can be taken as
$(5 \lambda-3,2 \lambda+1,3 \lambda-4)$
$\begin{aligned}
& \text { for foot of perpendicular } \lambda=\frac{a\left(\alpha-x_1\right)+b\left(\beta-y_1\right)+c\left(\gamma-z_1\right)}{a^2+b^2+c^2} \\
& =\frac{5(0+3)+2(2-1)+3(3+4)}{5^2+2^2+3^2} \\
& \text { i.e., } \lambda=1 \\
& \Rightarrow \text { foot of perpendicular }(2,3,-1) \\
& \Rightarrow \text { Length of the perpendicular }=\sqrt{(0-2)^2+(2-3)^2+(3+1)^2} \\
& =\sqrt{21}
\end{aligned}$