In a triangle $A B C$, if $|\overrightarrow{B C}|=8,|\overrightarrow{C A}|=7,|\overrightarrow{A B}|=

Question

In a triangle $A B C$, if $|\overrightarrow{B C}|=8,|\overrightarrow{C A}|=7,|\overrightarrow{A B}|=10$, then the projection of the vector $\overrightarrow{A B}$ on $\overrightarrow{A C}$ is equal to\

Professor PC Thomas Classes · Answer

(d) : The projection of $\overrightarrow{A B}$ on\ $\overrightarrow{A C}=|\overrightarrow{A B}| \cos A$\ $=10 \cos A$\ $=10\left(\frac{85}{140}ight)=\frac{85}{14}$\ $[\because$ By cosine rule, $ \left.\cos A=\frac{10^{2}+7^{2}-8^{2}}{2 \cdot 10 \cdot 7}=\frac{85}{140}ight] $