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Question: Answered & Verified by Expert
If $|\overrightarrow{\mathbf{a}}|=3,|\overrightarrow{\mathbf{b}}|=4$, then for what value of 1 is $(\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{b}})$
perpendicular to $(\overrightarrow{\mathbf{a}}-\lambda \overrightarrow{\mathbf{b}})$ ?
MathematicsVector AlgebraNDANDA 2007 (Phase 1)
Options:
  • A $\frac{3}{4}$
  • B $\frac{4}{3}$
  • C $\frac{9}{16}$
  • D $\frac{3}{5}$
Solution:
2589 Upvotes Verified Answer
The correct answer is: $\frac{3}{4}$
$\because(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}})$ is perpendicular to $(\overrightarrow{\mathrm{a}}-\lambda \overrightarrow{\mathrm{b}})$, their dot
product is zero, so, $(\vec{a}+\lambda \vec{b}) \cdot(\vec{a}-\lambda \vec{b})=0$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}-\lambda^{2}|\overrightarrow{\mathrm{b}}|^{2}-\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=0$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}-\lambda^{2}|\overrightarrow{\mathrm{b}}|^{2}=0 \quad(\because \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b} \cdot \mathrm{a}})$
$\Rightarrow 9-16 \lambda^{2}=0$
$\Rightarrow \lambda=\pm \frac{3}{4} \quad \lambda=\frac{3}{4}$ matches with the given option.

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