If $|\vec{a}|=3$ and $|\vec{b}|=4$, then a value of $\lambda$ for which $\vec{a}+\lambda \vec{b}$ and $\vec{a}-\lambda \vec{b}$ are perpendicular is :
Answer & explanation
Correct answer: option 2
$|\vec{a}|=3 |\vec{b}|=4$
$\vec{v}_1=(\vec{a}+\lambda \vec{b})$
$\vec{v}_2=(\vec{a}-\lambda \vec{b})$
$\vec{v}_1⊥\vec{v}_2 \Rightarrow \vec{v}_1 . \vec{v}_2=0$
$\Rightarrow (\vec{a}+\lambda \vec{b})(\vec{a}-\lambda \vec{b})=0$
$=|\vec{a}|^2-\lambda^2|\vec{b}|^2=0 $
so $|\vec{a}|^2=\lambda^2|\vec{b}|^2$
so $\lambda=\left|\frac{\vec{a}}{\vec{b}}\right|$
$\lambda=\frac{3}{4}$