Let $G$ be the centroid of $\triangle A B C$ & $G_1$ be the centroid of $\triangle A_1 B_1 C_1$ Show that $\overrightarrow{A A}_1+\overrightarrow{B B}_1+\overrightarrow{C C}_1=3 \overrightarrow{G G}_1$.

Solution:

$\overrightarrow{A A}_1=\overrightarrow{A G}+\overrightarrow{G G}_1+\vec{G}_1 \bar{A}_1$

$\overrightarrow{B B}_1=\overrightarrow{B G}+\overrightarrow{G G_1}+\overrightarrow{G_1 \vec{B}_1}$

(Triangle law)

$\overrightarrow{A A}_1=\overrightarrow{A G}+\overrightarrow{G G}_1+\vec{G}_1 \vec{A}_1$

$\overrightarrow{B B}_1=\overrightarrow{B G}+\overrightarrow{G G_1}+\overrightarrow{G_1 \vec{B}_1}$

$\text { (Triangle law) }$

$\overrightarrow{C C}_1=\overrightarrow{C G}+\overrightarrow{G G}_1+\vec{G}_1 \vec{C}_1$

$\overrightarrow{A A}_1+\overrightarrow{B B}_1+\overrightarrow{C C}_1=\overrightarrow{A G}+\overrightarrow{G G_1}+\vec{G}_1 \vec{A}_1 +\overrightarrow{B G}+\overrightarrow{G G_1}+\overrightarrow{G_1 \vec{B}_1}$

$+\overrightarrow{C G}+\overrightarrow{G G_1}+\overrightarrow{G_1 C_1} =3 \overrightarrow{G G}_1+(\overrightarrow{A G}+\overrightarrow{B G}+\overrightarrow{C G}) +(\overrightarrow{G_1 A_1}+\overrightarrow{G_1 B_1}+ \overrightarrow{G_1 C_1}) =3 \overrightarrow{G G}_1+\left(\overrightarrow{A G}+\frac{2 \overrightarrow{D G}}{\overrightarrow{G A}}\right)+\left(\overrightarrow{G_1 A_1}+ 2\vec{G}_1 \vec{D}_1\right)$

Let $P, Q, R \& S$ be the points on the plane with position vectors $-2 i-j, 4 i, 3 i+3 j$ & $-3 i+2 j$ respectively. What kind of quadrilateral is $P Q R S$ ?

Solution:

$\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=(4+2) i+j=6 i+j$

$\overrightarrow{Q R}=\overrightarrow{O R}-\overrightarrow{O Q}=(-1) i+3 j=-i+3 j$

$\overrightarrow{R S}=\overrightarrow{O S}-\overrightarrow{O R}=-6 i-j=-\overrightarrow{P Q}$

$\overrightarrow{S P}=\overrightarrow{O P}-\overrightarrow{O S}=i-3 j=-\overrightarrow{Q R}$

$\overrightarrow{P Q}|\overrightarrow{R S} \quad \overrightarrow{Q R}| \overrightarrow{S P}$

$\quad \text { unlike vectors }$

$|\overrightarrow{P Q}|=\sqrt{36+1}=\sqrt{37} \quad|\overrightarrow{P Q}|=|\overrightarrow{R S}|$

$|\overrightarrow{R S}|=\sqrt{36+1} \quad \sqrt{37} \quad|\overrightarrow{Q R}|=|\overrightarrow{S P}|$

$\overrightarrow{P Q} \cdot \overrightarrow{Q R}=(6 i+j)(-i+3 j)$

$=6(-1)+1(3)=-3 \text {. }$

$\overrightarrow{Q R} \cdot \overrightarrow{R S}=3=\overrightarrow{S P} \cdot \overrightarrow{P Q}$

$\overrightarrow{R S} \cdot \overrightarrow{S P}=-3 \text {. }$

$\vec{a} \cdot \vec{b}=0$

Rectangle X

$|\overrightarrow{Q R}|=\sqrt{1+9}$

$=\sqrt{10} \neq \sqrt{37}=|\overrightarrow{P Q}|$ Rhombus X

$\square$ P Q R S must be a $11^m$.

Prove that the vectors

$3 i+5 j+2 k, 2 i-3 j-5 k \& 5 i+2 j-3 k$ form sides of an equilateral triangle.

Solution:

Let

$\begin{aligned} & \vec{a}=3 i+5 j+2 k \ & \vec{b}=2 i-3 j-5 k \end{aligned}$

$\vec{a}+\vec{b}=5 i+2 j-3 k=\vec{c}$

$\vec{a} \cdot \vec{b}=6-15-10=-19$

$\vec{a} \cdot \vec{c}=15+10-6=19$

$\vec{b} \cdot \vec{c}=10-6+15=19$

$|\vec{a}|^2=9+25+4=38$

$|\vec{b}|^2=38=|\vec{c}|^2$

$\rightarrow \cos \theta_1=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{-19}{38}=-\frac{1}{2}=\cos \left(120^{\circ}\right)$

$\left(180-\theta_1\right)=60^{\circ}$

$\cos \theta_2=\frac{1}{2}=\cos \theta_3 \quad \theta_2=60^{\circ}=\theta_3$

This must he an equilateral $\Delta$ .

Show that the points $A(-1,6,6), B(-4,9,6), C(0,7,10)$ form an isosceles right angled triangle.

Solution:

$\overrightarrow{A B}=-3 i+3 j$

$\overrightarrow{A C}=i+j+4 k$

$\overrightarrow{B C}=4 i-2 j+4 k$

$\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \text { (Triangle) }$

$|\overrightarrow{A B}|=\sqrt{9+9}=\sqrt{18}$

$|\overrightarrow{A C}|=\sqrt{1+1+16}=\sqrt{18}$

$|\overrightarrow{A B}|=|\overrightarrow{A C}| \text {. }$

$\text { (isosceles) }$

$\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \quad$ (Triangle)

$\overrightarrow{A B} \cdot \overrightarrow{A C}=-3+3=0 \text {. }$

$\triangle A B C$ is right angled at $A$.

(isosceles)

$\triangle A B C$ is a right angled isosceles triangle.

If $|\vec{a}|=3$,$|\vec{b}|=1$ and $|\vec{c}|=4$ and $\vec{a}+\vec{b}+\vec{c}=0$.

Find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$

Solution:

$0 =(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c})$

$=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c})$

$=9+1+16+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c})$

$=26+2()$

$()=\frac{-26}{2}=-13$

$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-13$