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# Find $\lambda \text{ and }\mu$if $(\widehat{i}+3\widehat{j}+9\widehat{k})\times (3\widehat{i}-\lambda \widehat{j}+\mu \widehat{k})=0$  Hint:In this question we have given cross product of two vectors as zero vectors and we have to find the values of $\lambda \text{ and }\mu$. So, in order to find the values of $\lambda \text{ and }\mu$we have to first find the cross product of the vectors. Once we find the cross-product we can equate the resultant vectors as zero vectors. Upon comparing we get the values of $\lambda \text{ and }\mu$. If $\overrightarrow{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}\text{ and }\overrightarrow{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}.$then
We have $\overrightarrow{a}\times \overrightarrow{b}=({{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}})\widehat{i}+({{a}_{3}}{{b}_{1}}-{{a}_{1}}{{b}_{3}})\widehat{j}+({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})\widehat{k}$, and if $\overrightarrow{c}={{c}_{1}}\widehat{i}+{{c}_{2}}\widehat{j}+{{c}_{3}}\widehat{k}\text{ and }d={{d}_{1}}\widehat{i}+{{d}_{2}}\widehat{j}+{{d}_{3}}\widehat{k}$and $\overrightarrow{c}=\overrightarrow{d}$ then ${{c}_{1}}={{d}_{1}},{{c}_{2}}={{d}_{2}},{{c}_{3}}={{d}_{3}}$

Let us assume that
$\overrightarrow{A}=(\widehat{i}+3\widehat{j}+9\widehat{k})$ and
$\overrightarrow{B}=(3\widehat{i}-\lambda \widehat{j}+\mu \widehat{k})$
Now we have to find the cross product of vectors,
As we know from the definition of cross product of two vectors
$\overrightarrow{a}\times \overrightarrow{b}=({{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}})\widehat{i}+({{a}_{3}}{{b}_{1}}-{{a}_{1}}{{b}_{3}})\widehat{j}+({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})\widehat{k}$
Here we have the following values
\begin{align} & {{a}_{1}}=1,{{a}_{2}}=3,{{a}_{3}}=9 \\ & {{b}_{1}}=3,{{b}_{2}}=-\lambda ,{{b}_{3}}=\mu \\ \end{align}
So, we can write
$\overrightarrow{A}\times \overrightarrow{B}=\{(3)(\mu )-(9)(-\lambda )\}\widehat{i}+\{(9)(3)-(1)(\mu )\}\widehat{j}+\{(1)(-\lambda )-(3)(3)\}\widehat{k}$
We can further write
$\overrightarrow{A}\times \overrightarrow{B}=\left( 3\mu +9\lambda \right)\widehat{i}+\left( 27-\mu \right)\widehat{j}+\left( -\lambda -9 \right)\widehat{k}$--------------------------(1)
Now it is given from question that $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{0}$
So, we can apply here the rule of equality, Hence, we can write further
$\left( 3\mu +9\lambda \right)\widehat{i}+\left( 27-\mu \right)\widehat{j}+\left( -\lambda -9 \right)\widehat{k}=\widehat{0}$
Upon comparison we can write further
$\left( 3\mu +9\lambda \right)=0$----------------------------(2)
$\left( 27-\mu \right)=0$--------------------------------(3)
$\left( -\lambda -9 \right)=0$--------------------------------(4)
So from equation (3) we can find the value of $\mu =27$ and from equation (4) we can find the value of $\lambda =-9$.
Hence our solution is
$\mu =27$
$\lambda =-9$
Here we see that if we put the value of $\mu =27$and $\lambda =-9$ in equation (2) it satisfy the equation.

Note:
It should be important to note that if the two vectors represent the side of a parallelogram then cross product of the vectors is the vector area of the parallelogram. If cross product of two vectors is zero it means that the magnitude of the area of parallelogram so formed is zero which means the two vectors are parallel or collinear. Hence, we can conclude that for the value $\mu =27$and $\lambda =-9$, the vectors
$\widehat{i}+3\widehat{j}+9\widehat{k}$and $3\widehat{i}-\lambda \widehat{j}+\mu \widehat{k}$are either parallel or collinear.

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