Question

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

• A. $\frac{9}{16}$
• B. $\frac{3}{4}$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

Answer: (B)

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