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Question: A unit vector in the plane of the vectors \(2 \mathbf { i } + \mathbf { j } + \mathbf { k }\) , \(...

A unit vector in the plane of the vectors 2i+j+k2 \mathbf { i } + \mathbf { j } + \mathbf { k } , i−j+k\mathbf { i } - \mathbf { j } + \mathbf { k } and orthogonal to 5i+2j+6k5 \mathbf { i } + 2 \mathbf { j } + 6 \mathbf { k } is

A
B
C

2i−5j29\frac { 2 \mathbf { i } - 5 \mathbf { j } } { \sqrt { 29 } }

D

2i+j−2k3\frac { 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } } { 3 }

Answer
Explanation

Solution

Let a unit vector in the plane of 2i+j+k2 \mathbf { i } + \mathbf { j } + \mathbf { k } and i−j+k\mathbf { i } - \mathbf { j } + \mathbf { k } be

a^=α(2i+j+k)+β(i−j+k)\hat { \mathbf { a } } = \alpha ( 2 \mathbf { i } + \mathbf { j } + \mathbf { k } ) + \beta ( \mathbf { i } - \mathbf { j } + \mathbf { k } )= (2α+β)i+(α−β)j+(α+β)k( 2 \alpha + \beta ) \mathbf { i } + ( \alpha - \beta ) \mathbf { j } + ( \alpha + \beta ) \mathbf { k }

As a^\hat { \mathbf { a } } is unit vector, we have

= (2α+β)2+(α−β)2( 2 \alpha + \beta ) ^ { 2 } + ( \alpha - \beta ) ^ { 2 }+(α+β)2=1( \alpha + \beta ) ^ { 2 } = 1

̃ 6α2+4αβ+3β2=16 \alpha ^ { 2 } + 4 \alpha \beta + 3 \beta ^ { 2 } = 1....(i)

As is orthogonal to 5i+2j+6k5 \mathbf { i } + 2 \mathbf { j } + 6 \mathbf { k } , we get

5(2α+β)+2(α−β)+6(α+β)=05 ( 2 \alpha + \beta ) + 2 ( \alpha - \beta ) + 6 ( \alpha + \beta ) = 0

̃ 18α+9β=018 \alpha + 9 \beta = 0

̃ β=−2α\beta = - 2 \alpha

From (i), we get 6α2−8α2+12α2=16 \alpha ^ { 2 } - 8 \alpha ^ { 2 } + 12 \alpha ^ { 2 } = 1 ̃ α=±110\alpha = \pm \frac { 1 } { \sqrt { 10 } }

̃ β=∓210\beta = \mp \frac { 2 } { \sqrt { 10 } }. Thus a^=±(310j−110k)\hat { \mathbf { a } } = \pm \left( \frac { 3 } { \sqrt { 10 } } \mathbf { j } - \frac { 1 } { \sqrt { 10 } } \mathbf { k } \right)