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Linear Combination of Vectors

Let a =2 i +ĵ...

Question

Let $\to a = 2 \to i +^j +^k, \to b =^i + 2^j -^k$ and a unit vector $\to c$ be coplanar. If $\to c$ is perpendicular to $\to a,$ then $\to c$ is equal to

A

\frac{1}{\sqrt{2}} (-^j +^k)

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B

\frac{1}{\sqrt{3}} (-^i -^j -^k)

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C

\frac{1}{\sqrt{5}} (-^i - 2^j)

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D

\frac{1}{\sqrt{5}} (^i -^j -^k)

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Solution

The correct option is A $\frac{1}{\sqrt{2}} (-^j +^k)$
It is given that $\to c$ is coplanar with $\to a$ and $\to b,$ we take $\to c = p \to a + q \to b (i)$
where, p and q are scalars.
Since, $\to c ⊥ \to a \Rightarrow \to c . \to a = 0$
Taking dot product of $\to a$ in Eq. (i), we get
$\to c . \to a = p \to a . \to a + q \to b . \to a \Rightarrow 0 = p | \to a |^{2} + q | \to b . \to a | ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} ∵ \to a = 2^i +^j +^k | \to a | = \sqrt{2^{2} + 1 + 1} = \sqrt{6} . \to a . \to b = (2^i +^j +^k) . (^i + 2^j -^k) = 2 + 2 - 1 = 3 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow 0 = p . 6 + q .3 \Rightarrow q = - 2 p$
On putting in Eq. (i), we get
$\to c = p \to a + \to b (- 2 p) \Rightarrow \to c = p \to a - 2 p \to b \Rightarrow \to c = p (\to a - 2 \to b) \Rightarrow \to c = p [(2^i +^j +^k) - 2 (^i + 2^j -^k)] \Rightarrow \to c = p (- 3^j + 3^k) \Rightarrow | \to c | = p \sqrt{(- 3)^{2} + 3^{2}} \Rightarrow | \to c |^{2} = p^{2} (\sqrt{18})^{2} \Rightarrow | \to c |^{2} = p^{2} .18 \Rightarrow 1 = p^{2} .18 \Rightarrow 1 = p^{2} .18 [∵ | \to c | = 1] \Rightarrow p^{2} = \frac{1}{18} \Rightarrow p = \pm \frac{1}{3 \sqrt{2}} ∴ \to c = \pm \frac{1 (-^j +^k}{\sqrt{2}}$

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