If |a|=3,|b|=4, then the value of for which a+ b is perpendicular…

MCQ

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
$\frac{9}{16}$
✓
$\frac{3}{4}$
C
$\frac{3}{2}$
D
$\frac{4}{3}$

✓

Answer

Correct option: B.

$\frac{3}{4}$

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

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