Question

If the projections of the line segment $AB$ on the coordinate axes are $2,k,4$ and $AB=\frac{7}{\sqrt{2}},$ then $2k^2-\sqrt{2}k+1$ is equal to

• A. $0$
• B. $3$
• C. $7$
• D. $13$

Answer 1

The correct option is C $7$

Let $a,b,c$ be the projection of a line on the coordinate axes.
Then the length of the line given by $A{B}^2=a^2+b^2+c^2$
$\frac{49}{2}=4+k^2+16$
$\frac{49}{2}=20+k^2$
$k^2=\frac{9}{2}$
$k=\frac{3}{\sqrt{2}}$
$\therefore 2k^2-\sqrt{2}k+1=9-3+1=7$