If a-b- c- are unit vectors and b-c- are non-collinear vectors satisfying a-b- a-c- and a-b-c-b-c-2 then cos-

The correct option is D 1 Given : (a⃗,b⃗)=α, (a⃗,c⃗)=β and a⃗×(b⃗×c⃗)=b⃗+c⃗2 (a⃗·c⃗)b⃗ (a⃗·b⃗)c⃗=b⃗2+c⃗2 from given ( |a⃗ | |c⃗ |cosβ) ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If a⃗,b⃗, c...

Question

If $\to a, \to b, \to c$ are unit vectors and $\to b, \to c$ are non-collinear vectors satisfying $(\to a, \to b) = α, (\to a, \to c) = β$ and $\to a \times (\to b \times \to c) = \frac{\to b + \to c}{2}$ then $c o s (α + β) =$

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Solution

The correct option is D $- 1$
Given :-
$(\to a, \to b) = α, (\to a, \to c) = β a n d \to a \times (\to b \times \to c) = \frac{\to b + \to c}{2}$
$(\to a \cdot \to c) \to b - (\to a \cdot \to b) \to c = \frac{\to b}{2} + \frac{\to c}{2}$
from given
$(| \to a | | \to c | cos β) \to b - (| \to a | ∣ ∣ \to b ∣ ∣ cos α) \to c = \frac{\to b}{2} + \frac{\to c}{2}$
because vectors $\to a, \to b, \to c$ are unit vectors their magnitude is 1
$(1 \times 1 cos β) \to b - (1 \times 1 cos α) \to c = \frac{\to b}{2} + \frac{\to c}{2}$
$c o s β \to b - cos α \to c = \frac{\to b}{2} + \frac{\to c}{2}$
comparing both sides
$cos β = \frac{1}{2}$
$β = {cos}^{-} 1 (\frac{1}{2})$
$β = 60^{.}$
$cos α = - \frac{1}{2}$
$α = {cos}^{-} 1 (- \frac{1}{2})$
$α = 120^{.}$
$cos (α + β) = cos (120 + 60) = cos 180^{.} = - 1$

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If →a,→b, →c are unit vectors and →b,→c are non-collinear vectors satisfying (→a,→b)=α, (→a,→c)=β and →a×(→b×→c)=→b+→c2 then cos(α+β)=

If $\to a, \to b, \to c$ are unit vectors and $\to b, \to c$ are non-collinear vectors satisfying $(\to a, \to b) = α, (\to a, \to c) = β$ and $\to a \times (\to b \times \to c) = \frac{\to b + \to c}{2}$ then $c o s (α + β) =$