Let - a- -1- - b- -2- - c- -3- and a- b-c- - b- c-a- and c- a-b- - then - a-b-c- - is

The correct option is C √(6) Now a⃗ is perpendicular to (b⃗+c⃗) Hence a⃗.(b⃗+c⃗)=0 a⃗.b⃗+a⃗.c⃗=0 Similarly b⃗.c⃗+b⃗.a⃗=0 a⃗ ...

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Definition of Functions

Let | a⃗ |=...

Question

Let $| \to a | = 1, ∣ ∣ \to b ∣ ∣ = \sqrt{2}$ , $| \to c | = \sqrt{3}$ , and $\to a ⊥ (\to b + \to c)$ , $\to b ⊥ (\to c + \to a)$ and $\to c ⊥ (\to a + \to b)$ , then $∣ ∣ \to a + \to b + \to c ∣ ∣$ is

\sqrt{6}

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6

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\sqrt{14}

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None of these

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Solution

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Similar questions

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

\to a =^i +^j +^k

\to b = - 3^i +^k

\to c =^i + 2^j + 3^k

A)

[\to a \times \to b

\to b \times \to c

\to c \times \to a]

B)

[\to a + \to b

\to b + \to c

\to c + \to a]

C)

[\to a

\to b

\to c]

[\to a

\to b

\to c]^{- 1}

D)

[\to a - \to b

\to b - \to c

\to c - \to a]

\to a, \to b

\to c

\to b + \to c, \to c + \to a

\to a + \to b

| \to a + \to b | = 6, | \to b + \to c | = 8

| \to c + \to a | = 10

| \to a + \to b + \to c |

| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 4, | \to c | = 5, \to a ⊥ (\to b + \to c), \to b ⊥ (\to c + \to a)

\to c ⊥ (\to a + \to b)

\sqrt{2} ∣ ∣ \to a + \to b + \to c ∣ ∣

[\to a + 2 \to b - \to c, \to a - \to b, \to a - \to b - \to c] =

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