Question

If $\left|a\right|=3,\left|b\right|=4$ then find the value of $\lambda$ for which $a+\lambda b$ is perpendicular to $a-\lambda b$

• A. $\frac{3}{4}$
• B. $\frac{-3}{4}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{-3}{4}$
B $\frac{3}{4}$

$a+\lambda b$ is perpendicular to $a-\lambda b$
$\therefore (a+\lambda b).(a-\lambda b)=0$
${\left|a\right|}^2-\lambda \times (a.b)+\lambda \times (b.a)-{\lambda }^2\times {\left|b\right|}^2=0$
$\Rightarrow {\left|a\right|}^2-{\lambda }^2\times {\left|b\right|}^2=0$
$\lambda =\pm \frac{\left|a\right|}{\left|b\right|}=\pm \frac{3}{4}$