Vector Algebra

  $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  

⇒   $\overrightarrow{a}+5\overrightarrow{b}$ = $\overrightarrow{c}$           …(1)

$\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$  

⇒   $\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}$               …(2)

From (1) and (2)

  $\overrightarrow{b}+6\overrightarrow{c}=\mu \left(\lambda \overrightarrow{c}-5\overrightarrow{b}\right)$          

-> $\left(1+5\mu \right)\overrightarrow{b}+\left(6-\lambda \mu \right)\overrightarrow{c}=0$

$?\overrightarrow{b}$ and $\overrightarrow{c}$  are non-collinear

-> 1 + 5m = 0 $\mu =\frac{-1}{5}$  and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

$\overrightarrow{b}=6\overrightarrow{c}=\frac{-1}{5}\overrightarrow{a}$

$5\overrightarrow{b}+30\overrightarrow{c}=-\overrightarrow{a}$

$\begin{array}{cc}\hfill \overrightarrow{a}+5\overrightarrow{b}+30\overrightarrow{c}=0\hfill & \hfill \overrightarrow{a}+\alpha \overrightarrow{b}+\beta \overrightarrow{c}=0\hfill \end{array}]$

On comparing

α = 5, β = 30 ⇒ α + β = 35

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$

$\overrightarrow{r}*\overrightarrow{a}=\overrightarrow{c}*\overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$

${\overrightarrow{a}}_1=x\widehat{i}-\widehat{j}+\widehat{k}\&{\overrightarrow{a}}_2=\widehat{i}+y\widehat{j}+z\widehat{k}$           

given  ${\overrightarrow{a}}_1\&{\overrightarrow{a}}_2$ are collinear then ${\overrightarrow{a}}_1=\lambda {\overrightarrow{a}}_2$  

$\Rightarrow \left(x\widehat{i}-\widehat{j}+\widehat{k}\right)=\lambda \left(\widehat{i}+y\widehat{j}+z\widehat{k}\right)$         

Since $\widehat{i},\widehat{j}\&\widehat{k}$ are not collinear so

  $Sox\widehat{i}+y\widehat{j}+z\widehat{k}=\lambda \widehat{i}-\frac{1}{\lambda }\widehat{j}+\frac{1}{\lambda }\widehat{k}$         

Hence possible unit vector parallel to it be  $\frac{1}{\sqrt[]{3}}\left(\widehat{i}-\widehat{j}+\widehat{k}\right)$ for $\lambda$ =

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

Mid point of BC is $\frac{1}{2}\left(5\widehat{i}+\left(\alpha -2\right)\widehat{j}+9\widehat{k}\right)$ $\frac{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$\overline{AB}=\widehat{i}+\left(\alpha -4\right)\widehat{j}+\widehat{k}$

$\overline{AC}=\widehat{i}+\left(-2-\alpha \right)\widehat{j}+\widehat{k}$

For = 1, $\overline{AB}$ $\stackrel{}{}$ and $\overline{AC}$ $\stackrel{}{}$ will be collinear. So for non collinearity

= 2

Mid point of BC is $\frac{1}{2}\left(5\widehat{i}+\left(\alpha -2\right)\widehat{j}+9\widehat{k}\right)$  

$\overline{AB}=\widehat{i}+\left(\alpha -4\right)\widehat{j}+\widehat{k}$  

$\overline{AC}=\widehat{i}+\left(-2-\alpha \right)\widehat{j}+\widehat{k}$

For a = 1,   $\overline{AB}$ and $\overline{AC}$  will be collinear. So for non collinearity

a = 2

$\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}$

$\begin{array}{l}\overrightarrow{b}=3\widehat{i}+3\widehat{j}+\widehat{k}\\ \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}\end{array}$

$Coplnanar\Rightarrow \left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right|=0$

$\begin{array}{l}\Rightarrow -8c_1+7c_2+12c_3=0........(i)\\ \overrightarrow{a}.\overrightarrow{c}=5\Rightarrow 2c_1+c_2+3c_3=5........(ii)\\ \overrightarrow{b}.\overrightarrow{c}=0\Rightarrow 3c_1+3c_2+c_3=0........(iii)\end{array}$

Solving (i), (ii), (iii)

$C_1=\frac{10}{122},c_2=\frac{-85}{122},c_3=\frac{225}{122}$

$\therefore 122\left(c_1+c_2+c_3\right)=150$

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

Given : $\overrightarrow{a}=\left(\alpha ,1,-1\right)and\overrightarrow{b}=\left(2,1,-\alpha \right)\overrightarrow{c}=\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -\alpha \hfill \end{array}\right|$  

$=\left(-\alpha +1\right)\widehat{i}+\left({\alpha }^2-2\right)\widehat{j}+\left(\alpha -2\right)\widehat{k}$ Projection of $\overrightarrow{c}$ on $\overrightarrow{d}=-\widehat{i}+2\widehat{j}-2\widehat{k}=\left|\overrightarrow{c}.\frac{\overrightarrow{d}}{\left|\overrightarrow{d}\right|}\right|=30\left\{Given\right\}$

$\Rightarrow \left|\frac{\alpha -1-4+2{\alpha }^2-2\alpha +4}{\sqrt[]{1+4+4}}\right|=30$  

On solving Þ $\alpha =\frac{-13}{2}$  (Rejected as a > 0) and a = 7

$x^2\left(y+z\right)y^2\left(z+x\right)z^2\left(x+y\right)=a^3{b}^3{c}^3=x^3{y}^3{z}^3$

$\Rightarrow \left(x+y\right)\left(y+z\right)\left(z+x\right)=xyz$

$\begin{array}{l}\Rightarrow x^2\left(y+z\right)+y^2\left(z+y\right)+z^2\left(x+y\right)+xyz=0\\ \Rightarrow a^3+b^3+c^3+abc=0\end{array}$