Question

If $a=2i+3j+\alpha k$ and$b=3i-\alpha j+2k$ then the angle between $a+b$ and $a-b$ is equal to

• A. $0$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$
• E. $\frac{\pi }{2}$

Answer 1

Answer: (E)

Given, $a=2i+3j+\alpha k$ and $b=3i-\alpha j+2k$
$\therefore a+b=5i+(3-\alpha )j+(\alpha +2)k$
and $a-b=-i+(3+\alpha )j+(\alpha -2)k$
$\therefore \cos \theta =\frac{(a+b)\cdot (a-b)}{|a+b||a-b|}$
$=\frac{\left[\{5i+(3-\alpha )j+(\alpha +2)k\}-i+(3+\alpha )j+(\alpha -2)k.\right]}{\left[\sqrt{5^2+(3-\alpha {)}^2+(\alpha +2{)}^2}\times \sqrt{1^2+(3+\alpha {)}^2+(\alpha -2{)}^2}\right]}$
$=\frac{-5+9-{\alpha }^2+{\alpha }^2-4}{[\sqrt{5^2+(3-\alpha {)}^2+(\alpha +2{)}^2}\times \sqrt{1+(3+\alpha {)}^2+(\alpha -2{)}^2}]}=0$
$\Rightarrow \theta =\frac{\pi }{2}$