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Question: If **a**, **b**, **c** are mutually perpendicular vectors of equal magnitudes, then the angle betwee...

If a, b, c are mutually perpendicular vectors of equal magnitudes, then the angle between the vectors a and a+b+c\mathbf { a } + \mathbf { b } + \mathbf { c } is

A

π3\frac { \pi } { 3 }

B

π6\frac { \pi } { 6 }

C

cos⁡−113\cos ^ { - 1 } \frac { 1 } { \sqrt { 3 } }

D

π2\frac { \pi } { 2 }

Answer

cos⁡−113\cos ^ { - 1 } \frac { 1 } { \sqrt { 3 } }

Explanation

Solution

Since a,b\mathbf { a } , \mathbf { b } and are mutually perpendicular, so a⋅b=b⋅c=c⋅a=0\mathbf { a } \cdot \mathbf { b } = \mathbf { b } \cdot \mathbf { c } = \mathbf { c } \cdot \mathbf { a } = 0

Angle between a\mathbf { a } and a+b+c\mathbf { a } + \mathbf { b } + \mathbf { c } is cos⁡θ=a⋅(a+b+c)∣a∥a+b+c∣\cos \theta = \frac { \mathbf { a } \cdot ( \mathbf { a } + \mathbf { b } + \mathbf { c } ) } { | \mathbf { a } \| \mathbf { a } + \mathbf { b } + \mathbf { c } | } .....(i)

Now ∣a∣≠b∣≠c∣=a| \mathbf { a } | \neq \mathbf { b } | \neq \mathbf { c } | = a

∣a+b+c∣2=a2+b2+c2+2a⋅b+2b⋅c+2c⋅a| \mathbf { a } + \mathbf { b } + \mathbf { c } | ^ { 2 } = \mathbf { a } ^ { 2 } + \mathbf { b } ^ { 2 } + \mathbf { c } ^ { 2 } + 2 \mathbf { a } \cdot \mathbf { b } + 2 \mathbf { b } \cdot \mathbf { c } + 2 \mathbf { c } \cdot \mathbf { a }

=a2+a2+a2+0+0+0= a ^ { 2 } + a ^ { 2 } + a ^ { 2 } + 0 + 0 + 0

⇒ ∣a+b+c∣2=3a2⇒a+b+c∣=3a| \mathbf { a } + \mathbf { b } + \mathbf { c } | ^ { 2 } = 3 a ^ { 2 } \Rightarrow \mathbf { a } + \mathbf { b } + \mathbf { c } \mid = \sqrt { 3 } a

Putting this value in (i), we get θ=cos⁡−113\theta = \cos ^ { - 1 } \frac { 1 } { \sqrt { 3 } }.