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

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
$\Rightarrow \alpha i+2j+\beta k=\lambda \left[ \left( \dfrac{i+j}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right)+\left( \dfrac{j+k}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right) \right]$

& \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{aligned}$$ On comparing both the sides, we get, $$\begin{aligned} & \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{aligned}$$ **Hence, option (1) is the correct option.** **Note:** In this type of question students have to note that the vector a lies in the plane of the vectors b and c so that the three vectors can be considered as coplanar vectors. Now as the vectors are coplanar then students can write, $$\begin{aligned} & \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{aligned}$$ Which gets satisfied by the values $$\alpha =1,\beta =1$$.