If a-b-c- b-c-a- and a-b-c be the moduli of the vectors a- b- c- respectively- then

Since a⃗×b⃗=c⃗ ∴ ( b⃗×c⃗)×b⃗=c⃗ ⇒ ( b⃗.b⃗)c⃗ (b⃗.c⃗)b⃗=c⃗ ⇒ ( b⃗.b⃗)=1, b⃗.c⃗=0 ⇒ b^2=1 i.e. b=1 Angle between b⃗ and c⃗ is 90^∘ ⇒ |b⃗×c⃗|=|b⃗||c⃗|=|c⃗| (∵ ...

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Byju's Answer

Standard XI

Mathematics

Cross Product of Two Vectors

If a⃗×b⃗=c⃗, ...

Question

If $\to a \times \to b = \to c, \to b \times \to c = \to a$ and a,b,c be the moduli of the vectors $\to a, \to b, \to c$ respectively, then

a=1 ,b=1

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c = 1, a = 1

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$\to a . (\to b \times \to c) = - 1$

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b = 1, c = a

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Solution

The correct option is D
b = 1, c = a

Since $\to a \times \to b = \to c ∴ (\to b \times \to c) \times \to b = \to c \Rightarrow (\to b . \to b) \to c - (\to b . \to c) \to b = \to c \Rightarrow (\to b . \to b) = 1, \to b . \to c = 0 \Rightarrow b^{2} = 1 i . e . b = 1$
Angle between $\to b$ and $\to c$ is $90^{\circ}$
$\Rightarrow | \to b \times \to c | = | \to b | | \to c | = | \to c | (∵ | \to b | = 1)$
$∴ | \to b \times \to c | = | \to c | = | \to a | i . e . c = a .$

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If →a×→b=→c, →b×→c=→a and a,b,c be the moduli of the vectors →a, →b, →c respectively, then

The correct option is D b = 1, c = a Since →a×→b=→c∴(→b×→c)×→b=→c⇒(→b.→b)→c−(→b.→c)→b=→c⇒(→b.→b)=1, →b.→c=0⇒b2=1 i.e. b=1 Angle between →b and →c is 90∘ ⇒|→b×→c|=|→b||→c|=|→c| (∵|→b|=1) ∴|→b×→c|=|→c|=|→a| i.e. c=a.

If $\to a \times \to b = \to c, \to b \times \to c = \to a$ and a,b,c be the moduli of the vectors $\to a, \to b, \to c$ respectively, then