If - -b- - are unit vector such that - -b- -0- then - -b- -b- - - -

Let $^a,^b a n d^c$ be three unit vectors such that
$^a \times (^b \times^c) = \frac{\sqrt{3}}{2} (^b +^c) .$ If
$^b$ is not parallel to $^c$ , then the angle between $^a$ and $^b$ is

Q. Let

^a,^b,^c

are three unit vectors such that

^a +^b +^c

is also a unit vector. If pairwise angles

^a,^b,^c

are

θ_{1}, θ_{2}

and

θ_{3}

respectively then

cos θ_{1} + cos θ_{2} + cos θ_{3}

equals :

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If ^a,^b,^c are unit vector such that ^a+^b+^c=0, then ^a.^b+^b.^c+^c.^a=

The correct option is B −32Since (^a+^b+^c)2=0⇒(→a)2+(→b)2+(→c)2=2(→a.→b+→b.→c+→c.→a)=0∴→a.→b+→b.→c+→c.→a=−12(1+1+1)=−32

If $^a,^b,^c$ are unit vector such that $^a +^b +^c = 0$ , then $^a .^b +^b .^c +^c .^a =$

The correct option is B $- \frac{3}{2}$
Since $(^a +^b +^c)^{2} = 0$
$\Rightarrow {(\to a)}^{2} + {(\to b)}^{2} + {(\to c)}^{2} = 2 (\to a . \to b + \to b . \to c + \to c . \to a) = 0$
$∴ \to a . \to b + \to b . \to c + \to c . \to a = - \frac{1}{2} (1 + 1 + 1) = - \frac{3}{2}$