If $|\vec{a}|=4,|\vec{b}|=5$, then the values of $k$ for which $\vec{a}+k \vec{b}$ is perpendicular to $\vec{a}-k \vec{b}$ are

Question

Aicharya · Accepted Answer

$\begin{aligned} & (\vec{a}+k \vec{b}) \cdot(\vec{a}-k \vec{b})=0 \ & 	herefore|a|^2-k^2|b|^2=0 \ & 	herefore(4)^2-k^2(5)^2=0 \Rightarrow k= \pm \frac{4}{5}\end{aligned}$

If $|\vec{a}|=4,|\vec{b}|=5$, then the values of $k$ for which $\vec{a}+k \vec{b}$ is perpendicular to…

Solution