If ABC is a triangle, then the vectors $(- 1, cos C, cos B)$ , $(cos C, - 1, cos A)$ and $(cos B, cos C, - 1)$ are

linearly independent for all triangles

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linearly dependent for all triangles

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linearly independent for all isosceles triangles

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none of these

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Solution

The correct option is B linearly dependent for all triangles
Using laws of projection
$∣ ∣ ∣ ∣ \begin{matrix} - 1 & c o s C & c o s B c o s C & - 1 & c o s A c o s B & c o s C & - 1 \end{matrix} ∣ ∣ ∣ ∣ = - 1 (1 - c o s A c o s C) - c o s C (- c o s C - c o s A c o s B) + c o s B ({c o s}^{2} C + c o s B) = 0$

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