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  4. If | veca| = 3, | vecb| = 4, then the value of λ for which veca +λ vecb is perpendicular to veca -λ vecb, is

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Q. If $|\vec{a}| = 3$, $|\vec{b}| = 4$, then the value of $\lambda$ for which $\vec{a} +\lambda \vec{b}$ is perpendicular to $\vec{a} -\lambda \vec{b}$, is

Vector Algebra

Solution:

Given that, $\left|\vec{a} \right|= 3$, $\left|\vec{b}\right| = 4$, and $\vec{a} +\lambda \vec{b}$ is perpendicular to $\vec{a} -\lambda \vec{b}$.
$\therefore \left(\vec{a} +\lambda \vec{b}\right)\cdot\left(\vec{a} -\lambda \vec{b}\right) = 0$
$\Rightarrow \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b}\lambda + \lambda \vec{b} \cdot\vec{a} - \lambda^{2} \vec{b}\cdot \vec{b} = 0$
$\Rightarrow \left|\vec{a} \right|^{2} - \lambda^{2}\left|b \right|^{2} = 0$
$\Rightarrow \lambda ^{2} = \frac{\left|\vec{a} \right|^{2}}{\left|\vec{b} \right|^{2}}$
$\Rightarrow \lambda = \frac{\left|\vec{a} \right|}{\left|\vec{b} \right|} = \frac{3}{4}$