If a,b,c are unit vectors such that (a,b) =π6, (b,c)=π4, and (c,d)=π4, then |a + b+c| =

a=b=c=1 a.b=1×1×cos π6=32 b.c=1×1×cos π3=12 c.a=1×1×cos π4=12 a+b+c2=1+1+1+232+12+12 =3+3+1+2=4+2+3

If a,b,c are unit vectors such that (a,b) =π6, (b,c)=π4, and (c,d)=π4, then |a + b+c| =

a=b=c=1 a.b=1×1×cos π6=32 b.c=1×1×cos π3=12 c.a=1×1×cos π4=12 a+b+c2=1+1+1+232+12+12 =3+3+1+2=4+2+3

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If $a, b, c$ are unit vectors such that $(a, b) = \frac{π}{6}, (b, c) = \frac{π}{4},$ and $(c, d) = \frac{π}{4},$ then $| a + b + c | =$

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$\sqrt{4 + \sqrt{3} + \sqrt{2}}$

$4 + \sqrt{3} + \sqrt{2}$

answer is C.

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Detailed Solution

$|a| = |b| = |c| = 1 a . b = 1 \times 1 \times \cos \frac{π}{6} = \frac{\sqrt{3}}{2} b . c = 1 \times 1 \times \cos \frac{π}{3} = \frac{1}{2} c . a = 1 \times 1 \times \cos \frac{π}{4} = \frac{1}{2} {|a + b + c|}^{2} = 1 + 1 + 1 + 2 (\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}) = 3 + \sqrt{3} + 1 + \sqrt{2} = 4 + \sqrt{2} + \sqrt{3}$