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Vectors and Its Types

If e =lî +m...

Question

If $\to e = l^i + m^j + n^k$ is a unit vector, then the minimum value of $l m + m n + n l$ is

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Solution

It is given that $e = l^i + m^j + n^k$ is a unit vector
As, $| e | = 1$ where $e$ is a unit vector
Therefore $l^{2} + m^{2} + n^{2} = 1 \to (1)$
As $(l + m + n)^{2} = l^{2} + m^{2} + n^{2} + 2 (l m + m n + n l) \Rightarrow (l + m + n)^{2} = 1 + 2 (l m + m n + n l)$ [using $(i)$ ]
As square of any number is always greater than zero
i.e, $a^{2} \geq 0$
So, $(l + m + n)^{2} \geq 0 ∴ 1 + 2 (l m + m n + n l) \geq 0 \Rightarrow 2 (l m + m n + n l) \geq - 1 ∴ l m + m n + n l \geq - \frac{1}{2}$

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\to e = l^i + m^j + n^k

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