Question

$\overline{\alpha }=a\overline{i}+b\overline{j}+c\overline{k},\overline{\beta }=b\overline{i}+c\overline{j}+a\overline{k}$ and $\overline{\gamma }=c\overline{i}+a\overline{j}+b\overline{k}$ be three coplanar vectors with $\overline{\alpha }\ne \overline{\beta }\ne \overline{\gamma }$ and $\overline{r}=\overline{i}+\overline{j}+\overline{k}$ then $\overline{r}$ is perpendicular to?

• A. $\overline{\alpha }$
• B. $\overline{\beta }$
• C. $\overline{\gamma }$
• D. $\overline{\alpha },\overline{\beta },\overline{\gamma }$

Answer 1

The correct option is D $\overline{\alpha },\overline{\beta },\overline{\gamma }$

Given:

$\overrightarrow{\alpha }=a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}$

$\overrightarrow{\beta }=b\overrightarrow{i}+c\overrightarrow{j}+a\overrightarrow{k}$

$\overrightarrow{\gamma }=c\overrightarrow{i}+a\overrightarrow{j}+b\overrightarrow{k}$

$\overrightarrow{r}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$

and

$\overrightarrow{\alpha }\ne \overrightarrow{\beta }\ne \overrightarrow{\gamma }$

We know that:

$If\overrightarrow{v_1}\perp \overrightarrow{v_2}then$

$\overrightarrow{v_1}\cdot \overrightarrow{v_2}=0$

and,

If 3 vectors are coplanar, then

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

Solution:

Here,

$\vec{\alpha}, \ \vec{\beta}, \ \vec{\gamma}\ are \ coplanar\\&&

$\Rightarrow \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

$\Rightarrow a(cb-a^2)-b(b^2-ac)+c(ba-c^2)=0$

$\Rightarrow abc-a^3-b^3+abc-c^3+abc=0$

$\Rightarrow 3abc-a^3-b^3-c^3=0$

$\Rightarrow a^3+b^3+c^3-3abc=0$

$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

$\Rightarrow (a+b+c)=0$

Now,

$\overrightarrow{r}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$

$\Rightarrow \overrightarrow{r}\cdot \overrightarrow{\alpha }=(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})(a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k})=a+b+c=0$

$\therefore \overrightarrow{r}\perp \overrightarrow{\alpha }$

$\Rightarrow \overrightarrow{r}\cdot \overrightarrow{\beta }=(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})(b\overrightarrow{i}+c\overrightarrow{j}+a\overrightarrow{k})=b+c+a=0$

$\therefore \overrightarrow{r}\perp \overrightarrow{\beta }$

$\Rightarrow \overrightarrow{r}\cdot \overrightarrow{\gamma }=(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})(c\overrightarrow{i}+a\overrightarrow{j}+b\overrightarrow{k})=c+a+b=0$

$\therefore \overrightarrow{r}\perp \overrightarrow{\gamma }$

$\Rightarrow \overrightarrow{r}\perp \overrightarrow{\alpha },\overrightarrow{\beta },\overrightarrow{\gamma }$