Question

If the projections of the line segment $AB$ on the coordinate axes are $12,3,k$ such that $k\in R^{+}$ and $AB=13,\text{then}{k}^2-2k+3=$

Answer 1

Let $a,b,c$ be the projection of a line on the coordinate axes.
Let $a=$ projection on $x-$axis,
$b=$ projection on $y-$axis,
$c=$ projection on $z-$axis
Then the length of the line given by $A{B}^2=a^2+b^2+c^2$
$\Rightarrow 169=144+9+k^2$
$\Rightarrow k^2=16$
$\therefore k=4$
Hence, $k^2-2k+3=16-8+3=11$