Question

If $a+2b+3c=4$, then find the least value (to the nearest integer) of $a^2+b^2+c^2$ is

Answer 1

Given, $a+2b+3b=4$

or, $(a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})=4$

Now taking modulus both sides we get,

$\sqrt{a^2+b^2+c^2}.\sqrt{1^2+2^2+3^2}=4$

or, $\sqrt{a^2+b^2+c^2}=\frac{4}{\sqrt{14}}$

or, $a^2+b^2+c^2=\frac{16}{14}=\frac{8}{7}=1+\frac{1}{7}$.

So least value (to the nearest integer) of $(a^2+b^2+c^2)$ is $1$.