MCQ
The angle between the vector $a = 2i + 3j + k$ and $b = 2i - j - k$ is
  • $\pi /2$
  • B
    $\pi /4$
  • C
    $\pi /3$
  • D
    $0$

Answer

Correct option: A.
$\pi /2$
a
(a) Let $a = 2i + 3j + k$ and $b = 2i - j - k$

Since $\cos \,\theta = \frac{{a\,.b}}{{|a||b|}}$

$ = \frac{{(2i + 3j + k)\,.\,(2i - j - k)\,}}{{\sqrt {{{(2)}^2} + {{(3)}^2} + {{(1)}^2}} \sqrt {{{(2)}^2} + {{( - 1)}^2} + {{( - 1)}^2}} }}$

$ = \frac{{4 - 3 - 1}}{{\sqrt {(4 + 9 + 1)} \sqrt {(4 + 1 + 1)} }} = 0$

$\therefore \theta = \frac{\pi }{2}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free