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Let $$\overline { a } = \overline { i } + \overline { j } + \overline { k } , \overline { b } = \overline { i } - \overline { j } + 2 \overline { k }$$ and $$\overline { c } = x \overline { i } + ( x - 2 ) \overline { j } + \overline { k }$$ . If the vector $$\overline { c }$$ liesin the plane of $$\overline { a }$$ and $$\overline { b }$$ then $$x$$ equals


A
0
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B
1
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C
-4
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D
-2
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Solution

The correct option is A 0
If $$\vec{a}, \vec{b}, \vec{c}$$ all lie in same plane, then $$(\vec{a} \times \vec{b}) . \vec{c} = 0$$
Now. $$\vec{a} \times \vec{b} = \begin{vmatrix} i&i&k \\ 1&1 &1\\1&-1&2  \end{vmatrix}= 3j - j - 2k $$
$$(3i - j - 2k) . (x\vec{i} + (x-2)\hat{i} + \hat{k}) = 3x - x + 2 - 2 = 0$$
$$\Rightarrow x=0$$

Mathematics

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