Question

If $f^2\left(x\right)+g^2\left(x\right)+h^2\left(x\right)\le 2$ and $u\left(x\right)=2f\left(x\right)+3g\left(x\right)+4h\left(x\right),$ then the maximum value of $u^2\left(x\right)$ is equal to

Answer 1

Let two vectors $\overrightarrow{P}=f\left(x\right)\hat{i}+g\left(x\right)\hat{j}+h\left(x\right)\hat{k}$ and $\overrightarrow{Q}=2\hat{i}+3\hat{j}+4\hat{k}$
if $\theta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q},$
then
$\cos \theta =\frac{\overrightarrow{P}\cdot \overrightarrow{Q}}{|\overrightarrow{P}||\overrightarrow{Q}|}=\frac{2f\left(x\right)+3g\left(x\right)+4h\left(x\right)}{\sqrt{f^2\left(x\right)+g^2\left(x\right)+h^2\left(x\right)}\sqrt{29}}\Rightarrow \frac{\left[2f\right(x)+3g(x)+4h(x){]}^2}{29\left(f^2\right(x)+g^2(x)+h^2(x\left)\right)}\le 1[\because {\cos }^2\theta \le 1]\Rightarrow u^2\left(x\right)\le 29\left(f^2\right(x)+g^2(x)+h^2(x\left)\right)\Rightarrow u^2\left(x\right)\le 29\times 2\Rightarrow u^2\left(x\right)\le 58$