Let a- b and c be three unit vectors such that a-b-c-0- If -a-b-b-c-c-a and d-a-b-b-c - c-a- then the ordered pair - d is equal to-

Given, a+b+c=0 nbsp; λ = a·b+b·c+c·a ⇒ |a+b+c|^2 = |0|^2 ⇒ nbsp; |a|^2 + |b|^2 + |c|^2 + 2(a·b+b·c+c·a)=0 ⇒λ= 32 nbsp; Also, d=a×b + b×c + c×a ⇒d=a×b + ...

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Byju's Answer

Standard XIII

Mathematics

Cross Product of Two Vectors

Let a, b and ...

Question

Let $\to a, \to b$ and $\to c$ be three unit vectors such that $\to a + \to b + \to c = \to 0$ . If $λ = \to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a$ and $\to d = \to a \times \to b + \to b \times \to c + \to c \times \to a$ , then the ordered pair $(λ, \to d)$ is equal to:

(\frac{3}{2}, 3 \to a \times \to c)

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(- \frac{3}{2}, 3 \to c \times \to b)

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(- \frac{3}{2}, 3 \to a \times \to b)

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(\frac{3}{2}, 3 \to b \times \to c)

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Solution

The correct option is C $(- \frac{3}{2}, 3 \to a \times \to b)$
Given, $\to a + \to b + \to c = \to 0$
$λ = \to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a$
$\Rightarrow | \to a + \to b + \to c |^{2} = | \to 0 |^{2}$
$\Rightarrow | \to a |^{2} + | \to b |^{2} + | \to c |^{2} + 2 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a) = 0$
$\Rightarrow λ = - \frac{3}{2}$

Also, $\to d = \to a \times \to b + \to b \times \to c + \to c \times \to a$
$\Rightarrow \to d = \to a \times \to b + \to b \times (- \to a - \to b) + (- \to a - \to b) \times \to a$
$\Rightarrow \to d = \to a \times \to b + \to a \times \to b + \to a \times \to b$
$\Rightarrow \to d = 3 (\to a \times \to b) = 3 \to a \times \to b$

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Let →a, →b and →c be three unit vectors such that →a+→b+→c=→0. If λ=→a⋅→b+→b⋅→c+→c⋅→a and →d=→a×→b+→b×→c+→c×→a, then the ordered pair (λ,→d) is equal to:

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