Question

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

• A. $7$
• B. $13$
• C. $19$
• D. $0$

Answer 1

The correct option is A $7$

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 $A{B}^2=a^2+b^2+c^2$
$\frac{49}{2}=4+k^2+16$
$\frac{49}{2}=20+k^2$
$k^2=\frac{9}{2}$
$k=\frac{3}{\sqrt{2}}$
$\therefore 2k^2-\sqrt{2}k+1=9-3+1=7$