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 d.r's of line is $(p,q,r)=(12,3,k)\text{Using}A{B}^2=p^2+q^2+r^2$
We have, $169=144+9+k^2$
$\Rightarrow 16=k^2$
$\Rightarrow k=4,-4\left(\text{rejected}\right)$
Hence we have $k^2-2k+3=11$