If a- b- and c- are unit vectors then- -a- b-2 - -b- - c-2 - -c- a-2 can take the value of nbsp-

|a⃗ b⃗|^2 + |b⃗ c⃗|^2 + |c⃗ a⃗|^2 = a^2 + b^2 + b^2 + c^2 + c^2 + a^2 2 [a⃗.b⃗+ b⃗. c⃗+ c⃗. a⃗] = 6 2 [a⃗. b⃗+ b⃗.c⃗+ c⃗.a⃗]—(1) Since a⃗, b⃗ & c⃗ ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If a⃗, b⃗ and...

Question

If $\to a, \to b$ and $\to c$ are unit vectors then, $| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2}$ can take the value of

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Solution

The correct options are
A 3
B 5
D 8
$| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = a^{2} + b^{2} + b^{2} + c^{2} + c^{2} + a^{2} - 2 [\to a . \to b + \to b . \to c + \to c . \to a]$
$= 6 - 2 [\to a . \to b + \to b . \to c + \to c . \to a] - - - (1)$
Since $\to a, \to b & \to c$ are unit vectors
$| \to a + \to b + \to c | \geq 0$
$a^{2} + b^{2} + c^{2} + 2 [\to a . \to b + \to b . \to c + \to c - \to a] \geq 0$
$3 + 2 [\to a . \to b + \to b . \to c + \to c . \to a] \geq 0 [\to a . \to b + \to b . \to c + \to c . \to a] \geq - \frac{3}{2} - - - (2)$
By (1) & (2),
$| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} \leq 6 - 2 \frac{- 3}{2} \leq 9$

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Similar questions

Q. If

\to a, \to b

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Q. If

\to a, \to b

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then prove that

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If →a,→b and →c are unit vectors then, |→a−→b|2+|→b−→c|2+|→c−→a|2 can take the value of

If $\to a, \to b$ and $\to c$ are unit vectors then, $| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2}$ can take the value of