Question

Let $\vec{a}=2\hat{i}+{{\lambda }_{1}}\hat{j}+3\hat{k}$, $\vec{b}=4\hat{i}+\left( 3-{{\lambda }_{2}} \right)\hat{j}+6\hat{k}$ and $\vec{c}=3\hat{i}+6\hat{j}+\left( {{\lambda }_{3}}-1 \right)\hat{k}$ be three vectors such that $\vec{b}=2\vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$, then find the possible value of $\left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)$?
(a) $\left( \dfrac{1}{2},4,-2 \right)$,
(b) $\left( -\dfrac{1}{2},4,0 \right)$,
(c) $\left( 1,3,1 \right)$,
(d) $\left( 1,5,1 \right)$.

Answer 1

We start solving the problem by using the given condition $\vec{b}=2\vec{a}$ to get the relation between ${{\lambda }_{1}}$ and ${{\lambda }_{2}}$. We use the fact that the dot product of product of two perpendicular vectors is zero for the vectors $\vec{a}$ and $\vec{c}$ to get the relation between the ${{\lambda }_{1}}$ and ${{\lambda }_{3}}$. We assume the values for ${{\lambda }_{1}}$ as per the requirement of the problem to get the desired result.

Complete step-by-step answer:

According to the problem, we have three vectors given as $\vec{a}=2\hat{i}+{{\lambda }_{1}}\hat{j}+3\hat{k}$, $\vec{b}=4\hat{i}+\left( 3-{{\lambda }_{2}} \right)\hat{j}+6\hat{k}$ and $\vec{c}=3\hat{i}+6\hat{j}+\left( {{\lambda }_{3}}-1 \right)\hat{k}$. It is also said that $\vec{b}=2\vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$. We need to find the possible value of $\left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)$.
Let us use our first condition $\vec{b}=2\vec{a}$.
$\Rightarrow 4\hat{i}+\left( 3-{{\lambda }_{2}} \right)\hat{j}+6\hat{k}=2\times \left( 2\hat{i}+{{\lambda }_{1}}\hat{j}+3\hat{k} \right)$.
$\Rightarrow 4\hat{i}+\left( 3-{{\lambda }_{2}} \right)\hat{j}+6\hat{k}=4\hat{i}+2{{\lambda }_{1}}\hat{j}+6\hat{k}$.
We compare the coefficients of $\hat{j}$ on both sides.
$\Rightarrow 3-{{\lambda }_{2}}=2{{\lambda }_{1}}$
$\Rightarrow {{\lambda }_{2}}=3-2{{\lambda }_{1}}$ ---(1).
We have our second condition as $\vec{a}$ is perpendicular to $\vec{c}$. We know that the dot product of two perpendicular vectors is zero.
So, we have $\vec{a}\bullet \vec{c}=0$.
$\Rightarrow \left( 2\hat{i}+{{\lambda }_{1}}\hat{j}+3\hat{k} \right)\bullet \left( 3\hat{i}+6\hat{j}+\left( {{\lambda }_{3}}-1 \right)\hat{k} \right)=0$.
We know that dot product of two vectors $a\hat{i}+b\hat{j}+c\hat{k}$ and $d\hat{i}+e\hat{j}+f\hat{k}$ is defined as $\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\bullet \left( d\hat{i}+e\hat{j}+f\hat{k} \right)=ad+be+cf$.
$\Rightarrow \left( 2\times 3 \right)+\left( {{\lambda }_{1}}\times 6 \right)+\left( 3\times \left( {{\lambda }_{3}}-1 \right) \right)=0$.
$\Rightarrow 6+6{{\lambda }_{1}}+3{{\lambda }_{3}}-3=0$.
$\Rightarrow 6{{\lambda }_{1}}+3{{\lambda }_{3}}+3=0$.
$\Rightarrow 2{{\lambda }_{1}}+{{\lambda }_{3}}+1=0$.
$\Rightarrow {{\lambda }_{3}}=-2{{\lambda }_{1}}-1$ ---(2).
From equations (1) and (2), we got $\left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( {{\lambda }_{1}},3-2{{\lambda }_{1}},-2{{\lambda }_{1}}-1 \right)$ ---(3).
Let us assume the value of ${{\lambda }_{1}}$ as $\dfrac{1}{2}$. We substitute in equation (3).
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{1}{2},3-2\left( \dfrac{1}{2} \right),-2\left( \dfrac{1}{2} \right)-1 \right)$.
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{1}{2},3-1,-1-1 \right)$.
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{1}{2},2,-2 \right)$ ---(4).
Let us assume the value of ${{\lambda }_{1}}$ as $\dfrac{-1}{2}$. We substitute in equation (3).
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{-1}{2},3-2\left( \dfrac{-1}{2} \right),-2\left( \dfrac{-1}{2} \right)-1 \right)$.
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{-1}{2},3+1,+1-1 \right)$.
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{-1}{2},4,0 \right)$ ---(5).
Let us assume the value of ${{\lambda }_{1}}$ as 1. We substitute in equation (3).
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( 1,3-2\left( 1 \right),-2\left( 1 \right)-1 \right)$.
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( 1,3-2,-2-1 \right)$.
$\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( 1,1,-3 \right)$ ---(6).
From equations (4), (5) and (6), we can see there is only one option matching our answer i.e., $\Rightarrow \left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)=\left( \dfrac{-1}{2},4,0 \right)$.
We have found one of the possible values of $\left( {{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}} \right)$ as $\left( \dfrac{-1}{2},4,0 \right)$.

So, the correct answer is “Option b”.

Note: We can get other condition as the vector $\vec{b}$ is perpendicular to $\vec{c}$ as vector $\vec{b}$ is parallel to the vector $\vec{a}$. This condition also leads to the same condition as equation (1) which makes our problem not solvable for a single solution. Since we have only two equations for solving three variables, we could not get a single definite solution. Instead, we get infinite no. of solutions for the given problem.