In a triangle ABC, if BC→=3, CA→=5, BA→=7,,then the projection of the vector BA→ on BC→ is equal to :

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In a triangle ABC, if $|\vec{BC}| = 3, |\vec{CA}| = 5, |\vec{BA}| = 7,,$ then the projection of the vector $\vec{BA} on \vec{BC}$ is equal to :

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$\frac{13}{2}$

$\frac{11}{2}$

$\frac{15}{2}$

$\frac{19}{2}$

answer is B.

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Detailed Solution

$a = |BC| = 3 b = |CA| = 5 c = |BA| = 7 \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 ac} = \frac{9 + 49 - 25}{2 \times 3 \times 7} = \frac{33}{14 \times 3} = \frac{11}{14} Proj BA or BC = \frac{BA . BC}{|BC|} = |BA| \cos B = 7 \times \frac{11}{14} = \frac{11}{2}$