Vector Algebra Question 344
Question: If $ |\mathbf{a}|,=3,|\mathbf{b}|,=4 $ then a value of l for which $ \mathbf{a}+\lambda \mathbf{b} $ is perpendicular to $ \mathbf{a}-\lambda \mathbf{b} $ is
[Karnataka CET 2004]
Options:
A) $ \mathbf{a}=2,\mathbf{i}+2,\mathbf{j}+3,\mathbf{k},\mathbf{b}=-\mathbf{i}+2,\mathbf{j}+\mathbf{k} $
B) $ \frac{3}{4} $
C) $ \frac{3}{2} $
D) $ \frac{4}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Since $ \mathbf{a}+\lambda \mathbf{b} $ is perpendicular to $ \mathbf{a}-\lambda \mathbf{b} $ , then their product will be zero. So, $ (\mathbf{a}+\lambda \mathbf{b}).(\mathbf{a}-\lambda \mathbf{b})=0 $
Þ $ |\mathbf{a}{{|}^{2}}-{{\lambda }^{2}}|\mathbf{b}{{|}^{2}}=0 $ or $ {{\lambda }^{2}}=\frac{|\mathbf{a}{{|}^{2}}}{|\mathbf{b}{{|}^{2}}}\Rightarrow {{\lambda }^{2}}=\frac{9}{16} $ or $ \lambda =\pm \frac{3}{4} $ , $ [\because ,|\mathbf{a}|=3,|\mathbf{b}|=4] $ .
BETA
