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$3i + 3j -6k$
$\because ai + bj + ck\,,\,i + j\,,\,\,j + k$ 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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Statement $-1 :$$S=\{x:f(x)=f^{-1}(x)\}=$$\left\{ {1,2} \right\}$
Statement $-2 :$ $f $ is a bijection and ${f^{ - 1}}\left( x \right) = 1 + \sqrt {x - 1} \;,x \ge 1$