Question 10 Review Exercise 3

Solutions of Question 10 of Review Exercise 3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove that in any triangle $A B C$ that $|\vec{a}|^2=|\vec{b}|^2+|\vec{c}|^2 -2|\vec{b}|| \vec{c}| \cos A$.

Let considering a triangle $A B C$ as shown in figure,

whose sides are represented by $\vec{a}, \vec{b}$ and $\vec{c}$.

By head to tail rule. we get $$\begin{align} \vec{b}&=\vec{a}+\vec{c} \\ \Rightarrow \vec{a}&=\vec{b}-\vec{c} \\ \Rightarrow \vec{a} \cdot \vec{a}&=(\vec{b}-\vec{c}) \cdot(\vec{b}-\vec{c}) \\ \Rightarrow|\vec{a}|^2&=\vec{b} \cdot \vec{b}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c} \\ \Rightarrow|\vec{a}|^2&=|\vec{b}|^2+|\vec{c}|^2-2 \vec{b} \cdot \vec{c} \\ \Rightarrow |\vec{a}|^2=|\vec{b}|^2+|\vec{c}|^2 - 2|\vec{b}||\vec{c}| \cos A .\end{align}$$

Prove that in any triangle $A B C$ that $| \vec{a}|= |\vec{b}| \cos C+| \vec{c}| \cos B$

Let consider three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are represented respectively in order by sides $B C, C A$ and $AB$ of $\triangle A B C$.
So,we know that
$$\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\vec{O}$$ $$\Rightarrow \vec{a}+\vec{b}+ \vec{c}=\overrightarrow{0} .$$ Taking dot product with $\vec{a}$, we get
$$\begin{align}\vec{a} \cdot(\vec{a}+\vec{b}+\vec{c})&=0\\ \vec{a}\cdot\vec{a}+\vec{a}\vec{b}+\vec{a}\vec{c}&=0\\ |\vec{a}||\vec{a}|+|\vec{a}||\vec{b}|cos(\pi-C)+|\vec{a}||\vec{c}|cos(\pi-B)&=0\\ |\vec{a}||\vec{a}|-|\vec{a}||\vec{b}|cosC-|\vec{a}||\vec{c}|cosB&=0\\ |\vec{a}|=|\vec{b}|cosC+|\vec{c}|cosB&\end{align}$$ Which is the reguired result.

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