MCQ
A unit vector perpendicular to the vector $4i - j + 3k$ and $ - 2i + j - 2k$ is
- A$\frac{1}{3}\,(i - 2j + 2k)$
- ✓$\frac{1}{3}\,( - i + 2j + 2k)$
- C$\frac{1}{3}\,(2i + j + 2k)$
- D$\frac{1}{3}\,(2i - 2j + 2k)$
Unit vector perpendicular to $a$ and $b$ is $\frac{{a \times b}}{{|a \times b|}}$
But $a \times b = \left| {\begin{array}{*{20}{c}}i&j&k\\4&{ - 1}&3\\{ - 2}&1&{ - 2}\end{array}} \right|$
$ = i(2 - 3) - j( - 8 + 6) + k(4 - 2) = - i + 2j + 2k$
$\therefore \,\frac{{a \times b}}{{|a \times b|}} = \frac{{ - i + 2j + 2k}}{{\sqrt {1 + 4 + 4} }} = \frac{{ - i + 2j + 2k}}{3}.$
Trick : Check it with the options. Since the vector $\frac{{ - i + 2j + 2k}}{3}$ is unit and perpendicular to both the given vectors.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x) = \sqrt {\left| {{{\sin }^{ - 1}}\left| {\sin x} \right|} \right| - {{\cos }^{ - 1}}\left| {\cos x} \right|} $ is
$\frac{14}{29}$
$\frac{16}{29}$
$\frac{15}{29}$
$\frac{10}{29}$
| $\text{X}:$ | $2$ | $3$ | $4$ | $5$ |
| $\text{P}(\text{X}):$ | $\frac{5}{\text{k}}$ | $\frac{7}{\text{k}}$ | $\frac{9}{\text{k}}$ | $\frac{11}{\text{k}}$ |
The value of k is: