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Vector Algebra — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVector Algebra1 Mark
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
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Answer

$(b) :$ 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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