A unit vector parallel to the xz– plane and perpendicular to the vector $4 i + \bar{j} - 3 \bar{k}$ is

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$\frac{- 3 \bar{i}}{5} + \frac{4}{3} \bar{k}$

$\frac{3}{5} \bar{i} + \frac{4}{5} \bar{k}$

$\frac{4}{5} \bar{i} + \frac{3}{5} \bar{k}$

$\frac{4}{5} \bar{i} - \frac{3}{5} \bar{k}$

answer is B.

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Detailed Solution

$l e t \overset{⇀}{a} = x \overset{⏜}{i} + z \overset{⏜}{k} i s a u n i t v e c t o r p a r a l l e l t o x z - p l a n e, w h i c h i s p e r p e n d i c u l a r t o \overset{⇀}{b} = 4 \hat{i} + \hat{j} - 3 \hat{k} g i v e n |\overset{⇀}{a}| = 1 \Rightarrow x^{2} + z^{2} = 1 a n d \overset{⇀}{a} \cdot \overset{⇀}{b} = 0 \Rightarrow 4 x - 3 z = 0 \Rightarrow x = \frac{3 z}{4} \Rightarrow {(\frac{3 z}{4})}^{2} + z^{2} = 1 \Rightarrow z^{2} = \frac{16}{25} \Rightarrow z = \frac{4}{5} \Rightarrow x = \frac{3}{5}$