In a triangle ABC,if |overrightarrow{BC}|=3,| | vedclass

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Question

(A) Let $\vec{a} = \overrightarrow{BC}$,$\vec{b} = \overrightarrow{AC}$,and $\vec{c} = \overrightarrow{BA}$.Given $|\vec{a}| = 3$,$|\vec{b}| = 5$,and

Vedclass · Accepted Answer

$\frac{11}{2}$

Vedclass · Answer

(A) Let $\vec{a} = \overrightarrow{BC}$,$\vec{b} = \overrightarrow{AC}$,and $\vec{c} = \overrightarrow{BA}$.Given $|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$.In $	riangle ABC$,by the Law of Cosines at vertex $B$:$|\overrightarrow{AC}|^2 = |\overrightarrow{BA}|^2 + |\overrightarrow{BC}|^2 - 2 |\overrightarrow{BA}| |\overrightarrow{BC}| \cos(\angle ABC)$$5^2 = 7^2 + 3^2 - 2(7)(3) \cos(\angle ABC)$$25 = 49 + 9 - 42 \cos(\angle ABC)$$25 = 58 - 42 \cos(\angle ABC)$$42 \cos(\angle ABC) = 58 - 25 = 33$$\cos(\angle ABC) = \frac{33}{42} = \frac{11}{14}$.The projection of vector $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is given by $|\overrightarrow{BA}| \cos(\angle ABC)$.Projection $= 7 	imes \frac{11}{14} = \frac{11}{2}$.

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

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