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Condition for Coplanarity of Four Points

Let a, b, c a...

Question

Let $\to a, \to b, \to c$ are vectors such that $| \to a | = 1, | \to b | = 6, | \to c | = 2 \sqrt{3}$ and $(\to a + 2 \to b)$ is perpendicular to $\to c, (\to b + 2 \to c)$ is perpendicular to $\to a$ and $(\to c + 2 \to a)$ is perpendicular to $\to b .$ Then the value of $| \to a + \to b + \to c | =$

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Solution

Given :
$(\to a + 2 \to b) \cdot \to c = 0 \Rightarrow \to a \cdot \to c + 2 \to b \cdot \to c = 0 \dots (i)$ ,
$(\to b + 2 \to c) \cdot \to a = 0 \Rightarrow \to a \cdot \to b + 2 \to a \cdot \to c = 0 \dots (i i)$
and
$(\to c + 2 \to a) \cdot \to b = 0 \Rightarrow \to b \cdot \to c + 2 \to a \cdot \to b = 0 \dots (i i i)$
Adding $(i), (i i)$ and $(i i i) :$ we get,
$3 (\to a \cdot \to b + \to b \cdot \to c + \to a \cdot \to c) = 0 \Rightarrow (\to a \cdot \to b + \to b \cdot \to c + \to a \cdot \to c = 0) \dots (A)$
Also,
$| \to a + \to b + \to c |^{2} = | \to a |^{2} + | \to b |^{2} + | \to c |^{2} + 2 (\to a \cdot \to b + \to b \cdot \to c + \to a \cdot \to c) = 1 + 36 + 12 [using (A)] [∵ | \to a | = 1, | \to b | = 6, | \to c | = 2 \sqrt{3}] = 49 \Rightarrow | \to a + \to b + \to c | = 7$

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Condition for Coplanarity of Four Points

Standard XII Mathematics

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