Question

In a four-dimensional space where unit vectors along the axes are $\hat{i},\hat{j},\hat{k}\text{ and }\hat{l}$ , and ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},{{\vec{a}}_{4}}$ are four non-zero vectors such that no vector can be expressed as linear combination of others and $\left( \lambda -1 \right)\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)+\mu \left( {{{\vec{a}}}_{2}}+{{{\vec{a}}}_{3}} \right)+\gamma \left( {{{\vec{a}}}_{3}}+{{{\vec{a}}}_{4}}-2{{{\vec{a}}}_{2}} \right)+{{\vec{a}}_{3}}+\delta {{\vec{a}}_{4}}=0$ , then
(A) $\lambda =1$
(B) $\mu =\dfrac{-2}{3}$
(C) $\gamma =\dfrac{2}{3}$
(D) $\delta =\dfrac{1}{3}$

Answer 1

Hint: As no vector can be expressed as the linear combination of ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},$ and ${{\vec{a}}_{4}}$ . It means the vectors ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},$ and ${{\vec{a}}_{4}}$ are independent of each other. Our given equation is $\left( \lambda -1 \right)\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)+\mu \left( {{{\vec{a}}}_{2}}+{{{\vec{a}}}_{3}} \right)+\gamma \left( {{{\vec{a}}}_{3}}+{{{\vec{a}}}_{4}}-2{{{\vec{a}}}_{2}} \right)+{{\vec{a}}_{3}}+\delta {{\vec{a}}_{4}}=0$ . The RHS of the equation is equal to zero. So, LHS must also be equal to zero. The only way to get the LHS equal to zero is to make the coefficients of the vectors ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},\text{ and }{{\vec{a}}_{4}}$ in the above expression equal to zero. The coefficient of ${{\vec{a}}_{1}}$ is $\left( \lambda -1 \right)$ . The coefficient of ${{\vec{a}}_{2}}$ is $-\left( \lambda -1 \right)+\mu -2\gamma =\left( 1-\lambda \right)+\mu -2\gamma$. The coefficient of ${{\vec{a}}_{3}}$ is $\left( \mu +\gamma +1 \right)$. The coefficient of ${{\vec{a}}_{4}}$ is $\left( \gamma +\delta \right)$. Now, make these coefficients equal to zero and solve it further.

Complete step-by-step answer:
According to the question it is given that the vectors ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},{{\vec{a}}_{4}}$ are four non-zero vectors such that no vector can be expressed as linear combination of others and $\left( \lambda -1 \right)\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)+\mu \left( {{{\vec{a}}}_{2}}+{{{\vec{a}}}_{3}} \right)+\gamma \left( {{{\vec{a}}}_{3}}+{{{\vec{a}}}_{4}}-2{{{\vec{a}}}_{2}} \right)+{{\vec{a}}_{3}}+\delta {{\vec{a}}_{4}}=0$ …………………………(1)
It is given that no any vectors out of ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},\text{ and }{{\vec{a}}_{4}}$ can be expressed as linear combination of others. It means that the vector ${{\vec{a}}_{1}}$ can be expressed without the using the vectors ${{\vec{a}}_{2}}$ , ${{\vec{a}}_{3}}$ , and ${{\vec{a}}_{4}}$ . In the same way, the vector ${{\vec{a}}_{2}}$ can be expressed without the using the vectors ${{\vec{a}}_{1}}$ , ${{\vec{a}}_{3}}$ , and ${{\vec{a}}_{4}}$ . We can also say that the vector ${{\vec{a}}_{3}}$ can be expressed without the using the vectors ${{\vec{a}}_{1}}$ , ${{\vec{a}}_{2}}$ , and ${{\vec{a}}_{4}}$ . Similarly, the vector ${{\vec{a}}_{4}}$ can be expressed without the using the vectors ${{\vec{a}}_{1}}$ , ${{\vec{a}}_{2}}$ , and ${{\vec{a}}_{3}}$ . It means that the vectors ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},{{\vec{a}}_{4}}$ are independent of each other.
From equation (1), we have the expression
$\left( \lambda -1 \right)\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)+\mu \left( {{{\vec{a}}}_{2}}+{{{\vec{a}}}_{3}} \right)+\gamma \left( {{{\vec{a}}}_{3}}+{{{\vec{a}}}_{4}}-2{{{\vec{a}}}_{2}} \right)+{{\vec{a}}_{3}}+\delta {{\vec{a}}_{4}}=0$ .
First of all, we have to convert the above expression in simpler form.

& \left( \lambda -1 \right)\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)+\mu \left( {{{\vec{a}}}_{2}}+{{{\vec{a}}}_{3}} \right)+\gamma \left( {{{\vec{a}}}_{3}}+{{{\vec{a}}}_{4}}-2{{{\vec{a}}}_{2}} \right)+{{{\vec{a}}}_{3}}+\delta {{{\vec{a}}}_{4}}=0 \\\ & \Rightarrow {{{\vec{a}}}_{1}}\left( \lambda -1 \right)-{{{\vec{a}}}_{2}}\left( \lambda -1 \right)+\mu {{{\vec{a}}}_{2}}+\mu {{{\vec{a}}}_{3}}+\gamma {{{\vec{a}}}_{3}}+\gamma {{{\vec{a}}}_{4}}-2\gamma {{{\vec{a}}}_{2}}+{{{\vec{a}}}_{3}}+\delta {{{\vec{a}}}_{4}}=0 \\\ & \Rightarrow {{{\vec{a}}}_{1}}\left( \lambda -1 \right)-{{{\vec{a}}}_{2}}\left( \lambda -1 \right)+\mu {{{\vec{a}}}_{2}}-2\gamma {{{\vec{a}}}_{2}}+\mu {{{\vec{a}}}_{3}}+\gamma {{{\vec{a}}}_{3}}+{{{\vec{a}}}_{3}}+\gamma {{{\vec{a}}}_{4}}+\delta {{{\vec{a}}}_{4}}=0 \\\ & \Rightarrow {{{\vec{a}}}_{1}}\left( \lambda -1 \right)+{{{\vec{a}}}_{2}}\left\\{ -\left( \lambda -1 \right)+\mu -2\gamma \right\\}+{{{\vec{a}}}_{3}}\left( \mu +\gamma +1 \right)+{{{\vec{a}}}_{4}}\left( \gamma +\delta \right)=0 \\\ \end{aligned}$$ In the above equation, we have the RHS equal to zero. So, the LHS of the above equation should also be equal to zero. The only way to get the LHS equal to zero is to make the coefficients of the vectors $${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},\text{ and }{{\vec{a}}_{4}}$$ in the above expression equal to zero. Our equation is, $${{\vec{a}}_{1}}\left( \lambda -1 \right)+{{\vec{a}}_{2}}\left\\{ -\left( \lambda -1 \right)+\mu -2\gamma \right\\}+{{\vec{a}}_{3}}\left( \mu +\gamma +1 \right)+{{\vec{a}}_{4}}\left( \gamma +\delta \right)=0$$ . The coefficient of the vector $${{\vec{a}}_{1}}$$ = $$\left( \lambda -1 \right)$$ ……………….(2) The coefficient of the vector $${{\vec{a}}_{2}}$$ = $$-\left( \lambda -1 \right)+\mu -2\gamma =\left( 1-\lambda \right)+\mu -2\gamma $$ ……………….(3) The coefficient of the vector $${{\vec{a}}_{3}}$$ = $$\left( \mu +\gamma +1 \right)$$ ……………….(4) The coefficient of the vector $${{\vec{a}}_{4}}$$ = $$\left( \gamma +\delta \right)$$ ……………….(5) The coefficient of the vector $${{\vec{a}}_{1}}$$ should be equal to zero. From equation (2), we have the coefficient of the vector $${{\vec{a}}_{1}}$$ . $$\left( \lambda -1 \right)=0$$ $$\Rightarrow \lambda =1$$ …………………(6) The coefficient of the vector $${{\vec{a}}_{2}}$$ should be equal to zero. From equation (3), we have the coefficient of the vector $${{\vec{a}}_{2}}$$ . $$\left( 1-\lambda \right)+\mu -2\gamma =0$$ ……………….(7) Putting $$\lambda =1$$ in equation (7), we get $$\left( 1-1 \right)+\mu -2\gamma =0$$ $$\Rightarrow \mu =2\gamma $$ ………………(8) The coefficient of the vector $${{\vec{a}}_{3}}$$ should be equal to zero. From equation (4), we have the coefficient of the vector $${{\vec{a}}_{2}}$$ . $$\left( \mu +\gamma +1 \right)=0$$ ……………….(9) From equation (8), we have $$\mu =2\gamma $$ . Putting $$\mu =2\gamma $$ in equation (9), we get $$\begin{aligned} & \left( 2\gamma +\gamma +1 \right)=0 \\\ & \Rightarrow 3\gamma =-1 \\\ \end{aligned}$$ $$\Rightarrow \gamma =\dfrac{-1}{3}$$ ………………(10) We have, $$\mu =2\gamma $$ . So, $$\mu =2\gamma =2.\dfrac{-1}{3}=\dfrac{-2}{3}$$ ………………..(11) The coefficient of the vector $${{\vec{a}}_{4}}$$ should be equal to zero. From equation (3), we have the coefficient of the vector $${{\vec{a}}_{4}}$$ . $$\left( \gamma +\delta \right)=0$$ $$\Rightarrow \gamma =-\delta $$ ……………….(12) From equation (10), we have $$\gamma =\dfrac{-1}{3}$$ . Putting $$\gamma =\dfrac{-1}{3}$$ in equation (12), we get $$\begin{aligned} & \Rightarrow \delta =-\gamma \\\ & \Rightarrow \delta =-\dfrac{-1}{3} \\\ \end{aligned}$$ $$\Rightarrow \delta =\dfrac{1}{3}$$ ………………(13) From equation (6), equation (10), equation (11), and equation (13), we have $$\lambda =1$$ , $$\mu =\dfrac{-2}{3}$$ , $$\gamma =\dfrac{-1}{3}$$ , and $$\delta =\dfrac{1}{3}$$ . Hence, option (A), (B) and (D) are correct. Note: In this question, the relation between the vectors $${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},$$ and $${{\vec{a}}_{4}}$$ is not given directly. So, one might get confused how the solve the expression $${{\vec{a}}_{1}}\left( \lambda -1 \right)+{{\vec{a}}_{2}}\left\\{ -\left( \lambda -1 \right)+\mu -2\gamma \right\\}+{{\vec{a}}_{3}}\left( \mu +\gamma +1 \right)+{{\vec{a}}_{4}}\left( \gamma +\delta \right)=0$$ . But the hidden information in this question is that the vectors $${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{a}}_{3}},$$ and $${{\vec{a}}_{4}}$$ are independent of each other. It means no vectors can be expressed in terms of others.