Question

The vector \[a=\alpha i+2j+\beta k\] lies in the plane of the vectors \[b=i+j\] and \[c=j+k\] and bisects the angle between \[b\] and \[c\]. Then, which of the following gives possible values of \[\alpha \] and \[\beta \]
(1) \[\alpha =1,\beta =1\]
(2) \[\alpha =2,\beta =2\]
(3) \[\alpha =1,\beta =2\]
(4) \[\alpha =2,\beta =1\]

Answer 1

Hint: In this type of question we have to use the concept of vectors. We know that if the vector a lies in the plane of the vectors b and c and the vector a bisects the angle between \[b\] and \[c\] then \[a=\lambda \left( b+c \right)\] where \[b=\dfrac{\overline{b}}{\left| \overline{b} \right|}\] and \[c=\dfrac{\overline{c}}{\left| \overline{c} \right|}\]. Also we know that if \[\overline{a}={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k\] then we have \[\left| \overline{a} \right|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}}\]. Here, by using this we express a in the form of b and c and after simplification and comparison we can find the values of \[\alpha \] and \[\beta \]

Complete step-by-step solution:
Now here we have to find the value of \[\alpha \] and \[\beta \] such that the vector \[a=\alpha i+2j+\beta k\] lies in the plane of the vectors \[b=i+j\] and \[c=j+k\] and bisects the angle between \[b\] and \[c\].
We have given that, the vector \[a=\alpha i+2j+\beta k\] lies in the plane of the vectors \[b=i+j\] and \[c=j+k\] and bisects the angle between \[b\] and \[c\], hence we can write
\[\Rightarrow a=\lambda \left( b+c \right)\]
Where \[b=\dfrac{\overline{b}}{\left| \overline{b} \right|}\] and \[c=\dfrac{\overline{c}}{\left| \overline{c} \right|}\]
By substituting the values of a, b and c we can write
\[\Rightarrow \alpha i+2j+\beta k=\lambda \left[ \left( \dfrac{i+j}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right)+\left( \dfrac{j+k}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right) \right]\]
\[\begin{align}
  & \Rightarrow \alpha i+2j+\beta k=\lambda \left( \dfrac{i+2j+k}{\sqrt{2}} \right) \\
 & \Rightarrow \alpha i+2j+\beta k=\lambda \left( \dfrac{1}{\sqrt{2}}i+\sqrt{2}j+\dfrac{1}{\sqrt{2}}k \right) \\
 & \Rightarrow \alpha i+2j+\beta k=\dfrac{\lambda }{\sqrt{2}}i+\sqrt{2}\lambda j+\dfrac{\lambda }{\sqrt{2}}k \\
\end{align}\]
On comparing both the sides, we get,
\[\begin{align}
  & \Rightarrow \alpha =\dfrac{\lambda }{\sqrt{2}},2=\sqrt{2}\lambda ,\beta =\dfrac{\lambda }{\sqrt{2}} \\
 & \Rightarrow \lambda =\sqrt{2}\alpha ,\lambda =\sqrt{2},\lambda =\sqrt{2}\beta \\
 & \Rightarrow \sqrt{2}\alpha =\sqrt{2}=\sqrt{2}\beta \\
 & \Rightarrow \alpha =1,\beta =1 \\
\end{align}\]
Hence, option (1) is the correct option.

Note: In this type of question students have to note that the vector a lies in the plane of the vectors b and c so that the three vectors can be considered as coplanar vectors. Now as the vectors are coplanar then students can write,
\[\begin{align}
  & \Rightarrow \left| \begin{matrix}
   \alpha & 2 & \beta \\
   1 & 1 & 0 \\
   0 & 1 & 1 \\
\end{matrix} \right|=0 \\
 & \Rightarrow \alpha \left( 1-0 \right)-2\left( 1-0 \right)+\beta \left( 1-0 \right)=0 \\
 & \Rightarrow \alpha -2+\beta =0 \\
 & \Rightarrow \alpha +\beta =2 \\
\end{align}\]
Which gets satisfied by the values \[\alpha =1,\beta =1\].