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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
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
    $\frac{1}{3}\,(2i - 2j + 2k)$
✓

Answer

Correct option: B.
$\frac{1}{3}\,( - i + 2j + 2k)$
b
(b) Let $a = 4i - j + 3k$ and $b = - 2i + j - 2k$

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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