Let a- b and c be three unit vectors such that -a - b -2 - -a - c -2 - 8-Then -a - 2b -2 - -a - 2c -2 is equal to-

|a b|^2 + |a c|^2 = 8 (a b ) .( a b ) + ( a c )( a c ) = 8 a^2 +b^2 2a· b + a^2 +c^2 2a· c = 8 2a^2 +b^2 + c^2 2a· b 2a· c = 8 a· b+a· c = 2 N ...

Let a, b and ...

Question

Let $\to a, \to b$ and $\to c$ be three unit vectors such that ${∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to a - \to c ∣ ∣}^{2} = 8.$ Then ${∣ ∣ ∣ \to a + 2 \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to a + 2 \to c ∣ ∣}^{2}$ is equal to:

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Solution

$| \to a - \to b |^{2} + | \to a - \to c |^{2} = 8$
$(\to a - \to b) . (\to a - \to b) + (\to a - \to c) (\to a - \to c) = 8$
$a^{2} + b^{2} - 2 a \cdot b + a^{2} + c^{2} - 2 a \cdot c = 8$

$2 a^{2} + b^{2} + c^{2} - 2 a \cdot b - 2 a \cdot c = 8$
$a \cdot b + a \cdot c = - 2$

Now $| \to a + 2 \to b |^{2} + | \to a + 2 \to c |^{2}$
$= 2 a^{2} + 4 b^{2} + 4 c^{2} + 4 a \cdot b + 4 a \cdot c$
$= 2 + 4 + 4 + 4 (- 2) = 2$

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