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Question

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A. $ \dfrac{{2\widehat j + \widehat k}}{{\sqrt 5 }} $

B. $ \dfrac{{3\widehat i + 2\widehat j - 2\widehat k}}{{\sqrt {17} }} $

C. $ \dfrac{{3\widehat i + 2\widehat j + 2\widehat k}}{{\sqrt {17} }} $

D. $ \dfrac{{2\widehat i + 2\widehat j - \widehat k}}{3} $

Answer

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Let us assume that the unit vector be $ x\widehat i + y\widehat j + j\widehat k $

Unit vector can be expressed as

$ \sqrt {{x^2} + {y^2} + {j^2}} = 1 $

Squaring both the sides of the equation –

$ \Rightarrow {x^2} + {y^2} + {j^2} = 1 $ .... (A)

A unit vector which is perpendicular to the vector $ 2\widehat i - \widehat j + 2\widehat k $ and is coplanar with the vectors $ \widehat i + \widehat j - \widehat k $ and $ 2\widehat i + 2\widehat j - \widehat k $

$ 2x - y + 2j = 0 $ .... (B)

$ \left| {\begin{array}{*{20}{c}}

x&y&z \\

1&1&{ - 1} \\

2&1&{ - 1}

\end{array}} \right| $

Expand the above determinant –

$ = x( - 1 + 1) - y( - 2 + 1) + z(1 - 2) $

Simplify the above equation –

$ \Rightarrow y = - j $

Place the above value in equation (B)

$ 2x + 3j = 0 $

Make “X” the subject-

$ 2x = - 3j $

When the term multiplicative on one side is moved to the opposite side then it goes to denominator.

$ \Rightarrow x = \dfrac{{ - 3j}}{2} $

Place in equation (A)

$ \dfrac{{9{j^2}}}{4} + {j^2} + {j^2} = 1 $

Take LCM (Least common multiple) and simplify the above equation-

$ \Rightarrow \dfrac{{9{j^2}}}{4} + \dfrac{{4{j^2}}}{4} + \dfrac{{4{j^2}}}{4} = 1 $

When denominators are the same, numerators are added.

$

\Rightarrow \dfrac{{9{j^2} + 4{j^2} + 4{j^2}}}{4} = 1 \\

\Rightarrow \dfrac{{17{j^2}}}{4} = 1 \;

$

Do cross multiplication and simplify the above equation –

$

\Rightarrow 17{j^2} = 4 \\

\Rightarrow {j^2} = \dfrac{4}{{17}} \;

$

Take square-root on both the sides –

$ \Rightarrow \sqrt {{j^2}} = \sqrt {\dfrac{4}{{17}}} $

Square and square-root cancel each other on left hand side of the equation-

$ \Rightarrow j = \pm \dfrac{2}{{\sqrt {17} }} $

So, the required vector is – $ \dfrac{{3\widehat i + 2\widehat j - 2\widehat k}}{{\sqrt {17} }} $