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.
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$\left|\begin{array}{lll}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right|=-648 \alpha$ ?
$(A)$ $-4$ $(B)$ $9$ $(C)$ $-9$ $(D)$ $4$