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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Vector coplanar with vectors $i + j $ and $ j + k$  and parallel to the vector $2i -2j -4k$ , is
  • A
    $i -k$
  • ✓
    $i -j -2k$
  • C
    $i + j -k$
  • D
    $3i + 3j -6k$
✓

Answer

Correct option: B.
$i -j -2k$
b
(b) Let vector be $ai + bj + ck$.

$\because ai + bj + ck\,,\,i + j\,,\,\,j + k$ are coplanar.

$\therefore \left| {\,\begin{array}{*{20}{c}}a&b&c\\1&1&0\\0&1&1\end{array}\,} \right| = 0$ $ \Rightarrow a - b + c = 0$

Also, since $(ai + bj + ck\,)|\,|\,\,(2i - 2j - 4k)$

 $\therefore (ai + bj + ck) \times (2i - 2j - 4k) = 0$

i.e., $\left| {\,\begin{array}{*{20}{c}}i&j&k\\a&b&c\\2&{ - 2\,}&{ - \,4\,\,}\end{array}\,} \right| = 0$

==> $i( - \,4b + 2c) - j( - \,4a - 2c) + k( - \,2a - 2b) = 0$

==> $ - \,4b + 2c = 0,\,\,4a + 2c = 0,\,\,\,2a + 2b = 0$

==> $\frac{c}{2} = \frac{b}{1},$ $\frac{c}{2} = \frac{a}{{ - 1}},$ $\frac{a}{{ - 1}} = \frac{b}{1}$

i.e., $\frac{a}{{ - 1}} = \frac{b}{1} = \frac{c}{2}$ or $\frac{a}{1} = \frac{b}{{ - 1}} = \frac{c}{{ - 2}}$

$\therefore $ Required vector is $i - j - 2k.$

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