Question

Find the value of $\lambda \And \mu$ if $\left( 2\widehat{i}+6\widehat{j}+27\widehat{k} \right)\times \left( \widehat{i}+\lambda \widehat{j}+\mu \widehat{k} \right)=\overrightarrow{0}$?

Answer 1

First of all, we are going to find the cross product of the two vectors $\left( 2\widehat{i}+6\widehat{j}+27\widehat{k} \right)\And \left( \widehat{i}+\lambda \widehat{j}+\mu \widehat{k} \right)$. We know that cross product of $\left( {{a}_{1}}\widehat{i}+{{b}_{1}}\widehat{j}+{{c}_{1}}\widehat{k} \right)\And \left( {{a}_{2}}\widehat{i}+{{b}_{2}}\widehat{j}+{{c}_{2}}\widehat{k} \right)$ is equal to $\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|$. Now, simplify this determinant, by expanding along the first row. Then as the R.H.S of this equation is equal to 0 so equating the coefficient of $\widehat{i},\widehat{j},\widehat{k}$ to 0 and from this find the value of $\lambda \And \mu$.

Complete step by step answer:
In the above problem, we have given the following vector equation:
$\left( 2\widehat{i}+6\widehat{j}+27\widehat{k} \right)\times \left( \widehat{i}+\lambda \widehat{j}+\mu \widehat{k} \right)=\overrightarrow{0}$
We know the cross product of two vectors $\left( {{a}_{1}}\widehat{i}+{{b}_{1}}\widehat{j}+{{c}_{1}}\widehat{k} \right)\And \left( {{a}_{2}}\widehat{i}+{{b}_{2}}\widehat{j}+{{c}_{2}}\widehat{k} \right)$ is equal to:
$\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|$
Now, finding the cross product of $\left( 2\widehat{i}+6\widehat{j}+27\widehat{k} \right)\And \left( \widehat{i}+\lambda \widehat{j}+\mu \widehat{k} \right)$ using the above determinant form we get,
$\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ 2 & 6 & 27 \\\ 1 & \lambda & \mu \\\ \end{matrix} \right|$
Expanding along first row of the above determinant we get,
$\widehat{i}\left( 6\mu -27\lambda \right)-\widehat{j}\left( 2\mu -27 \right)+\widehat{k}\left( 2\lambda -6 \right)$
As the above vector is equal to 0 so equating each of $\widehat{i},\widehat{j}\And \widehat{k}$ to 0 we get,
$\begin{aligned} & 6\mu -27\lambda =0......Eq.(1) \\\ & 2\mu -27=0.........Eq.(2) \\\ & 2\lambda -6=0............Eq.(3) \\\ \end{aligned}$
Solving eq. (2) we get,
$2\mu -27=0$
Adding 27 on both the sides of the above equation we get,
$2\mu =27$
Dividing 2 on both the sides of the above equation we get,
$\mu =\dfrac{27}{2}$
From the above, we have calculated the value of $\mu$.
Now, we are going to solve eq. (3) and we get,
$2\lambda -6=0$
Adding 6 on both the sides of the above equation and we get,
$2\lambda =6$
Dividing 2 on both the sides of the above equation and we get,
$\lambda =\dfrac{6}{2}=3$
From the above, we have calculated the value of $\lambda$ as 3.

Hence, we have calculated the value of $\mu \And \lambda$ as $\dfrac{27}{2}\And 3$ respectively.

Note: We can check the values of $\lambda \And \mu$ obtained in the above solution by substituting these values in eq. (1) and see whether these values are satisfying this equation or not.
The values of $\lambda \And \mu$ are as follows:
$\begin{aligned} & \lambda =3; \\\ & \mu =\dfrac{27}{2} \\\ \end{aligned}$
Substituting these values in eq. (1) we get,
$6\left( \dfrac{27}{2} \right)-27\left( 3 \right)=0$
In the L.H.S of the above equation, 6 will get divided by 2 by 3 times and the above equation will look like:
$\begin{aligned} & 3\left( 27 \right)-27\left( 3 \right)=0 \\\ & \Rightarrow 81-81=0 \\\ & \Rightarrow 0=0 \\\ \end{aligned}$
As you can see that L.H.S = R.H.S so $\lambda \And \mu$ are satisfying eq. (1).