Let a→=2i +j^+k^,b→=i^+2j^−k^ and a unit vector c→ be coplanar. If c→ perpendicular to a→, then c→ =

Let c→=xi^+yj^+zk^ and a→,b→,c→are coplanar xyz21112−1=0⇒−x+y+z=0 …(i)c→⋅a→=0⇒2x+y+z=0 …(ii)By cross multiplication rule from (i), (ii), x0=y−3=z3⇒x=0,y=−zc→=λ(j^−k^),|c→|=1⇒c→=±(j^−k^)2

Let a→=2i +j^+k^,b→=i^+2j^−k^ and a unit vector c→ be coplanar. If c→ perpendicular to a→, then c→ =

Let c→=xi^+yj^+zk^ and a→,b→,c→are coplanar xyz21112−1=0⇒−x+y+z=0 …(i)c→⋅a→=0⇒2x+y+z=0 …(ii)By cross multiplication rule from (i), (ii), x0=y−3=z3⇒x=0,y=−zc→=λ(j^−k^),|c→|=1⇒c→=±(j^−k^)2

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Let $\vec{a} = 2 \overset{}{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$ and a unit vector $\vec{c}$ be coplanar. If $\vec{c}$ perpendicular to $\vec{a},$ then $\vec{c}$ =

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$\frac{- \hat{j} + \hat{k}}{\sqrt{2}}$

$\frac{\hat{i} - 2 \hat{j}}{\sqrt{5}}$

$\frac{- \hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

$\frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

answer is A.

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Detailed Solution

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k} and \vec{a}, \vec{b}, \vec{c}$
are coplanar
$|\begin{array}{ccc} x & y & z \\ 2 & 1 & 1 \\ 1 & 2 & - 1 \end{array}| = 0 \Rightarrow - x + y + z = 0$ …(i)
$\vec{c} \cdot \vec{a} = 0 \Rightarrow 2 x + y + z = 0$ …(ii)
By cross multiplication rule from (i), (ii),
$\begin{array}{l} \frac{x}{0} = \frac{y}{- 3} = \frac{z}{3} \Rightarrow x = 0, y = - z \\ \vec{c} = λ (\hat{j} - \hat{k}), | \vec{c} | = 1 \Rightarrow \vec{c} = \pm \frac{(\hat{j} - \hat{k})}{\sqrt{2}} \end{array}$