Question

If $a+2b+3c=4$ and $k=a^2+b^2+c^2$, then the least possible value of $14k$ is

Answer 1

Consider vectors $\overrightarrow{p}=a\hat{i}+b\hat{j}+c\hat{k}$ and $\overrightarrow{q}=\hat{i}+2\hat{j}+3\hat{k}$
If $\theta$ is the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$, then
$\cos \theta =\frac{a+2b+3c}{\sqrt{a^2+b^2+c^2}\sqrt{1^2+2^2+3^2}}$
$\Rightarrow {\cos }^2\theta =\frac{(a+2b+3c{)}^2}{14(a^2+b^2+c^2)}\le 1$
$\Rightarrow a^2+b^2+c^2\ge \frac{8}{7}$
Hence, the least value of $a^2+b^2+c^2$ is $\frac{8}{7}$.