Question

If ${{v}_{1}},{{v}_{2}},{{v}_{3}}$ are unit vectors given by
${{v}_{1}}=ai+bj+ck$
${{v}_{2}}=bi+cj+ak$
${{v}_{3}}=ci+aj+bk$
Where $a,b,c$ are non negative real numbers, and ${{v}_{\alpha }}.{{v}_{\beta }}=0$ , for $\alpha \ne \beta $ , then
(A) $\left| \left[ {{v}_{1}},{{v}_{2}},{{v}_{3}} \right] \right|=1$
(B) $a+b+c=1$
(C) ${{v}_{1}}+{{v}_{2}}+{{v}_{3}}=0$
(D) ${{v}_{1}},{{v}_{2}},{{v}_{3}}\text{ are coplanar}\text{.}$

Answer 1

Hint: Since, ${{v}_{1}},{{v}_{2}},{{v}_{3}}$ are unit vector that means the sum of square of components of a vector is equal to unit that is ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ and we have that ${{v}_{\alpha }}.{{v}_{\beta }}=0$ that means the dot product of two vectors is equal to zero that means they are perpendicular to each other. So, we will get from the dot product of two vectors, $ab+bc+ca=0$ . Then, we will use the formula related to these two conditions that is ${{\left( a+b+c \right)}^{2}}$ . It will help us to find the relationship.

Complete step by step solution:
Since, It is given that ${{v}_{1}},{{v}_{2}},{{v}_{3}}$ are unit vectors from the question that means the sum of square of components of a vector is equal to zero. So, we can evaluate the sum of square of every vector as:
For first vector:
$\Rightarrow {{v}_{1}}=ai+bj+ck$
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$
For second vector:
${{v}_{2}}=bi+cj+ak$
$\Rightarrow {{b}^{2}}+{{c}^{2}}+{{a}^{2}}=1$
For third vector:
${{v}_{3}}=ci+aj+bk$
$\Rightarrow {{c}^{2}}+{{a}^{2}}+{{b}^{2}}=1$
Thus, we finally have from all these three vectors:
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ … $\left( i \right)$
Now, we have ${{v}_{\alpha }}.{{v}_{\beta }}=0$ that means the dot product of any two vectors is zero. So, we will calculate the dot product of two vectors as:
For ${{v}_{1}}$ and ${{v}_{2}}$ :
$\Rightarrow {{v}_{1}}.{{v}_{2}}=0$
$\Rightarrow \left( ai+bj+ck \right).\left( bi+cj+ak \right)=0$
$\Rightarrow ab+bc+ca=0$
For ${{v}_{2}}$ and ${{v}_{3}}$ :
$\Rightarrow {{v}_{2}}.{{v}_{3}}=0$
$\Rightarrow \left( bi+cj+ak \right).\left( ci+aj+bk \right)=0$
$\Rightarrow bc+ca+ab=0$
For ${{v}_{3}}$ and ${{v}_{1}}$ :
$\Rightarrow {{v}_{3}}.{{v}_{1}}=0$
$\Rightarrow \left( ci+aj+bk \right).\left( ai+bj+ck \right)=0$
$\Rightarrow bc+ca+ab=0$
Therefore, from the all the three vectors, we got the final conclusion as:
$\Rightarrow ab+bc+ca=0$ … $\left( ii \right)$
Since, we got two relations above. So, we will use the relation of ${{\left( a+b+c \right)}^{2}}$ as:
$\Rightarrow {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2\left( ab+bc+ca \right)$
Now, we will use the relation that we got above that is equation $\left( i \right)$ and equation $\left( ii \right)$ as:
$\Rightarrow {{\left( a+b+c \right)}^{2}}=1-2\times 0$
Here, we will do necessary calculation as:
$\Rightarrow {{\left( a+b+c \right)}^{2}}=1$
Now, we will take square root both sides and will have:
$\Rightarrow a+b+c=1$

Hence, from the given condition of the question, we got the relation $a+b+c=1$.

Note: Here, we need to take care of the calculation of the vector product. In the solution we did the dot product, here I will show you how we calculate dot product as:
$\Rightarrow {{v}_{1}}.{{v}_{2}}=0$
Now, we will put the vectors value as:
$\Rightarrow \left( ai+bj+ck \right).\left( bi+cj+ak \right)=0$
Now, we will open the bracket as:
$\Rightarrow \left( ai.bi+ai.cj+ai.ak \right)+\left( bj.bi+bj.cj+bj.ak \right)+\left( ck.bi+ck.cj+ck.ak \right)=0$
Here, we need to know that $i.i=1$ and the multiplication different vectors as $i.j$ or $j.k$ or $k.i$ will be zero as:
$\Rightarrow \left( ab+0+0 \right)+\left( 0+bc+0 \right)+\left( 0+0+ca \right)=0$
After simplifying it we will get:
$\Rightarrow ab+bc+ca=0$
From this process we got the above relation in the solution.