If $\overline{a} = - \overline{i} + \overline{j} + \overline{k}$ and $\overline{b} = 2 \overline{i} + \overline{k}$ , then the vector c satisfying the conditions that (i) it is coplanar with a and b (ii) it is perpendicular to b (iii) a .c = 7 is

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$- \overline{i} + 2 \overline{j} + 2 \overline{k}$

$(- 3 / 2) \overline{i} + (5 / 2) \overline{j} + 3 \overline{k}$

$- 6 \overline{i} + \overline{k}$

$- 3 \overline{i} + 5 \overline{j} + 6 \overline{k}$

answer is B.

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Detailed Solution

$\vec{L e t c} = x \hat{i} + y \hat{j} + z \hat{k} since \vec{c} is coplanar with \vec{a} and \vec{b} \begin{aligned} \Rightarrow \vec{c} . (\vec{a} x \vec{b}) = 0 \\ \Rightarrow \vec{c} . [(- \hat{i} + \hat{j} + \hat{k}) x (2 \hat{i} + \hat{k})] = 0 \\ \Rightarrow \vec{c .} [\hat{i} + 3 \hat{j} - 2 \hat{k}] = 0 \\ \Rightarrow (x i + y j + z k) (i + 3 j - 2 k) = 0 \\ \Rightarrow x + 3 y - 2 z = 0 \dots \dots \dots (i) \end{aligned} c is perpendicular to b \Rightarrow c \cdot b = 0 \Rightarrow 2 x + z = 0 F r o m t h e a b o v e 2 e q u a t i o n s x + 3 y - 2 (- 2 x) = 0 \Rightarrow 5 x + 3 y = 0 \Rightarrow y = \frac{- 5 x}{3} and z = - 2 x \vec{c} = x i + (\frac{- 5 x}{3}) j + (- 2 x) k = \frac{x}{3} (3 \hat{i} - 5 \hat{j} - 6 \hat{k}) = \frac{- x}{3} (- 3 \hat{i} + 5 \hat{j} + 6 \hat{k}) \vec{c} = (- 3 \hat{i} + 5 \hat{j} + 6 \hat{k})$