Let a-b-c- are three unit vectors such that a-b-c- is also a unit vector- If pairwise angles between a-b-c- are -1-2 and -3 respectively then cos -1-cos -2-cos -3 equals

The correct option is D 1 ∵ â+b̂+ĉ=1 ∴ â^2+b̂^2+ĉ^2+2[ â.b̂+b̂.ĉ+ĉ.â ]=1 ⇒ 1+1+1+2[ cos θ_1+cos θ_2+cos θ_3 ]=1 ⇒ cosθ_1+cosθ_2+cosθ_ ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

Let a⃗,b⃗,c...

Question

Let $\to a, \to b, \to c$ are three unit vectors such that $\to a + \to b + \to c$ is also a unit vector. If pairwise angles between $\to a, \to b, \to c$ are $θ_{1}, θ_{2} a n d θ_{3}$ respectively then $c o s θ_{1} + c o s θ_{2} + c o s θ_{3}$ equals

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Solution

The correct option is D $- 1$
$∵ ∣ ∣ \begin{matrix} ^a +^b +^c \end{matrix} ∣ ∣ = 1$
$∴ {| \begin{matrix} ^a \end{matrix} |}^{2} + {∣ ∣ \begin{matrix} ^b \end{matrix} ∣ ∣}^{2} + {| \begin{matrix} ^c \end{matrix} |}^{2} + 2 (\begin{matrix} ^a .^b +^b .^c +^c .^a \end{matrix}) = 1$
$\Rightarrow 1 + 1 + 1 + 2 (\begin{matrix} c o s θ_{1} + c o s θ_{2} + c o s θ_{3} \end{matrix}) = 1$
$\Rightarrow c o s θ_{1} + c o s θ_{2} + c o s θ_{3} = - 1$

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Let →a,→b,→c are three unit vectors such that →a+→b+→c is also a unit vector. If pairwise angles between →a,→b,→c are θ1,θ2andθ3 respectively then cosθ1+cosθ2+cosθ3 equals

The correct option is D −1∵∣∣^a+^b+^c∣∣=1∴|^a|2+∣∣^b∣∣2+|^c|2+2(^a.^b+^b.^c+^c.^a)=1⇒1+1+1+2(cosθ1+cosθ2+cosθ3)=1⇒cosθ1+cosθ2+cosθ3=−1

Let $\to a, \to b, \to c$ are three unit vectors such that $\to a + \to b + \to c$ is also a unit vector. If pairwise angles between $\to a, \to b, \to c$ are $θ_{1}, θ_{2} a n d θ_{3}$ respectively then $c o s θ_{1} + c o s θ_{2} + c o s θ_{3}$ equals