Question

The vectors $\overline{a},\overline{b},\overline{c}$ are of the same length and taken pair wise, they form equal angles. If $\overline{a}=\overline{i}+\overline{j}$ and $\overline{b}=\overline{j}+\overline{k}$ then the components of $\overline{c}$ are?

• A. $(1,0,1)$
• B. $(1,2,3)$
• C. $(-1,1,2)$
• D. $(-1,4,1)$

Answer 1

Answer: (A)

$(1,0,1)$

$\begin{array}{c}Given,\\ \left|\overline{a}\right|=\left|\overline{b}\right|=\left|\overline{c}\right|=\sqrt{2}\\ \Rightarrow (\overline{a}.\overline{b})=(\overline{b}.\overline{c})=(\overline{c}.\overline{a})\\ \Rightarrow \overline{a}.\overline{b}=\left|\overline{a}\right|.\left|\overline{b}\right|\cos \theta |\overline{b}=xi+yj+zk\\ \Rightarrow \overline{a}.\overline{b}=\overline{b}.\overline{c}=\overline{a}.\overline{c}\\ 1=y+z=x+y\\ y+z=1,x+y=1\\ but,\\ \left|\overline{b}\right|=\sqrt{2}\Rightarrow x^2+y^2+z^2=2------\left(i\right)\\ \therefore x=z\\ then,\\ x+y=1\Rightarrow y=1-x\\ fromeqn\left(i\right)\Rightarrow \\ x^2+(1-x{)}^2+x^2=2\\ \Rightarrow 3x^2-2x-1=0\\ \Rightarrow x=1,x=\frac{-1}{3}\\ ifx=0,y=0,z=0\\ \therefore z=(1,0,1)\\ so,correctoptionisA.\end{array}$