Let the vectors u_1=i+j+ak,u_2=i+bj+k and u_3=ci+j+k be coplanar.

Question

Let the vectors ${\vec{u}}_{1}=\hat{i}+\hat{j}+a\hat{k},{\vec{u}}_{2}=\hat{i}+b\hat{j}+\hat{k}$ and ${\vec{u}}_{3}=c\hat{i}+\hat{j}+\hat{k}$ be coplanar. If the vectors ${\vec{v}}_{1}=(a+b)\hat{i}+c\hat{j}+c\hat{k},{\vec{v}}_{2}=a\hat{i}+(b+c)\hat{j}+a\hat{k}$ and ${\vec{v}}_{3}=b\hat{i}+b\hat{j}+(c+a)\hat{k}$ are also coplanar, then $6(a+b+c)$ is equal to :

Accepted Answer

Since, ${\vec{u}}_{1},{\vec{u}}_{2},{\vec{u}}_{3}$ are coplanar.

So, $[\begin{gathered} {\vec{u}}_{1} \quad  {\vec{u}}_{2} \quad  {\vec{u}}_{3} \end{gathered}]=0$

$\begin{gathered} \Rightarrow |\begin{array}{ccc}1 \quad  1 \quad  a \ 1 \quad  b \quad  1 \ c \quad  1 \quad  1\end{array}|=0 \ \Rightarrow 1(b-1)-1(1-c)+a(1-bc)=0 \ \Rightarrow b-1-1+c+a-abc=0 \ \Rightarrow a+b+c-2=abc...........(i) \end{gathered}$

Also, $[\begin{gathered} {\vec{v}}_{1} \quad  {\vec{v}}_{2} \quad  {\vec{v}}_{3} \end{gathered}]=0$

$\begin{gathered} \Rightarrow |\begin{array}{ccc}a+b \quad  c \quad  c \ a \quad  b+c \quad  a \ b \quad  b \quad  c+a\end{array}|=0 \ \Rightarrow (a+b)[bc+ba+{c}^{2}+ca-ab]-c[ac+{a}^{2}-ab] \ \, +c[ab-{b}^{2}-bc]=0 \end{gathered}$

$\begin{gathered} \Rightarrow abc+a{c}^{2}+{a}^{2}c+{b}^{2}c+b{c}^{2}+abc-a{c}^{2}-{a}^{2}c \ +abc+abc-{b}^{2}c-b{c}^{2}=0 \ \Rightarrow 4abc=0\Rightarrow abc=0 \ \begin{gathered} a+b+c-2=0[from(i)] \ \Rightarrow a+b+c=2 \end{gathered} \ \Rightarrow 6(a+b+c)=12 \end{gathered}$

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