Practicing Success
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 : |
$\frac{9}{16}$ $\frac{3}{4}$ $\frac{3}{2}$ $\frac{4}{3}$ |
$\frac{3}{4}$ |
$|\vec{a}|=3 |\vec{b}|=4$ $\vec{v}_1=(\vec{a}+\lambda \vec{b})$ $\vec{v}_2=(\vec{a}-\lambda \vec{b})$ $\vec{v}_1⊥\vec{v}_2 \Rightarrow \vec{v}_1 . \vec{v}_2=0$ $\Rightarrow (\vec{a}+\lambda \vec{b})(\vec{a}-\lambda \vec{b})=0$ $=|\vec{a}|^2-\lambda^2|\vec{b}|^2=0 $ so $|\vec{a}|^2=\lambda^2|\vec{b}|^2$ so $\lambda=\left|\frac{\vec{a}}{\vec{b}}\right|$ $\lambda=\frac{3}{4}$ |