If the vectors \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] , $\vec{b}=\hat{i}+\lambda \hat{j}+4\hat{k}$ and $\vec{c}=2\hat{i}+4\hat{j}+\left( {{\lambda }^{2}}-1 \right)\hat{k}$ be coplanar vectors then non-zero vector $\vec{a}\times \vec{c}$ is.
$\begin{align}
& a)-14\hat{i}-5\hat{j} \\
& b)-10\hat{i}-5\hat{j} \\
& c)-10\hat{i}+5\hat{j} \\
& d)-14\hat{i}+5\hat{j} \\
\end{align}$
Answer
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Hint: Now the three vectors are given to be coplanar. We know that $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ , $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ are coplanar then $\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|=0$ Hence using this condition we will find the value of $\lambda $ . Now we want to find $\vec{a}\times \vec{c}$. Now if $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ then $\vec{b}\times \vec{c}$ is given by $\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$ . hence we will also find $\vec{b}\times \vec{c}$.
Complete step-by-step solution:
Now we are given that the vectors \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] , $\vec{b}=\hat{i}+\lambda \hat{j}+4\hat{k}$ and $\vec{c}=2\hat{i}+4\hat{j}+\left( {{\lambda }^{2}}-1 \right)\hat{k}$ are coplanar vectors.
Now we know that if the vectors $\vec{a},\vec{b}\And \vec{c}$ defined as $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ , $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ are coplanar then $\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|=0$.
Hence using this condition we get.
$\left| \begin{matrix}
1 & 2 & 4 \\
1 & \lambda & 4 \\
2 & 4 & {{\lambda }^{2}}-1 \\
\end{matrix} \right|=0$
Now opening the determinant we get
$\begin{align}
& 1\left( \lambda \left( {{\lambda }^{2}}-1 \right)-16 \right)-2\left( \left( {{\lambda }^{2}}-1 \right)-8 \right)+4\left( 4-2\lambda \right)=0 \\
& \Rightarrow \lambda \left( {{\lambda }^{2}}-1 \right)-16-2\left( {{\lambda }^{2}}-1 \right)+16+8\left( 2-\lambda \right)=0 \\
& \Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-1 \right)+8\left( 2-\lambda \right)=0 \\
& \Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-1 \right)-8\left( \lambda -2 \right)=0 \\
& \Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-1-8 \right)=0 \\
\end{align}$
$\Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-9 \right)=0$
Now we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ hence we get
$\left( \lambda -2 \right)\left( \lambda -3 \right)\left( \lambda +3 \right)=0$
Hence we have the value of $\lambda $ is 2, 3 or -3.
Now let us check the values of c if $\lambda =3,-3$
We have $\vec{c}=2\hat{i}+4\hat{j}+\left( {{\lambda }^{2}}-1 \right)\hat{k}$ .
Hence for $\lambda =3,-3$ we get $\vec{c}=2\hat{i}+4\hat{j}+\left( 8 \right)\hat{k}$
Now we can write this as $\vec{c}=2\left( \hat{i}+2\hat{j}+4\hat{k} \right)$ .
But we know that \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] hence we get $\vec{c}=2\vec{a}$ .
Now if we have two vectors such that $\vec{a}=\lambda \vec{b}$ then we say that the two vectors are parallel.
Hence we have $\vec{a}$ and $\vec{c}$ are parallel.
Now we know that cross product of parallel vectors is equal to zero. Hence if we take $\lambda =3,-3$ we will get $\vec{a}\times \vec{c}=0$ . and since we need non zero cross product we will take the value of $\lambda =2$ .
Hence using this we get $\vec{c}=2\hat{i}+4\hat{j}+\left( {{2}^{2}}-1 \right)\hat{k}$
Hence we get $\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$
Now we have $\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$ , \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] and we want to find $\vec{a}\times \vec{c}$ .
Now we know that if $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ then $\vec{b}\times \vec{c}$ is given by $\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$
Hence we get
$\vec{a}\times \vec{c}=\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
1 & 2 & 4 \\
2 & 4 & 3 \\
\end{matrix} \right|$
Now expanding the determinant we get
$\begin{align}
& \vec{a}\times \vec{c}=\left( 6-16 \right)\hat{i}-\left( 3-8 \right)\hat{j}+\left( 4-4 \right)\hat{k} \\
& \Rightarrow \vec{a}\times \vec{c}=-10\hat{i}+5\hat{j} \\
\end{align}$
Hence option C is the correct option.
Note: Now note that when the vectors are parallel their cross product is zero and when they are perpendicular their dot product is zero. Now also remember that when the vectors are collinear we get the corresponding determinant as 0. This is nothing but a triple product of vectors. Now triple product represents the volume of parallelepiped formed hence if the vectors are collinear we have this volume as zero. Hence we get the condition.
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|=0$ Hence using this condition we will find the value of $\lambda $ . Now we want to find $\vec{a}\times \vec{c}$. Now if $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ then $\vec{b}\times \vec{c}$ is given by $\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$ . hence we will also find $\vec{b}\times \vec{c}$.
Complete step-by-step solution:
Now we are given that the vectors \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] , $\vec{b}=\hat{i}+\lambda \hat{j}+4\hat{k}$ and $\vec{c}=2\hat{i}+4\hat{j}+\left( {{\lambda }^{2}}-1 \right)\hat{k}$ are coplanar vectors.
Now we know that if the vectors $\vec{a},\vec{b}\And \vec{c}$ defined as $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ , $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ are coplanar then $\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|=0$.
Hence using this condition we get.
$\left| \begin{matrix}
1 & 2 & 4 \\
1 & \lambda & 4 \\
2 & 4 & {{\lambda }^{2}}-1 \\
\end{matrix} \right|=0$
Now opening the determinant we get
$\begin{align}
& 1\left( \lambda \left( {{\lambda }^{2}}-1 \right)-16 \right)-2\left( \left( {{\lambda }^{2}}-1 \right)-8 \right)+4\left( 4-2\lambda \right)=0 \\
& \Rightarrow \lambda \left( {{\lambda }^{2}}-1 \right)-16-2\left( {{\lambda }^{2}}-1 \right)+16+8\left( 2-\lambda \right)=0 \\
& \Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-1 \right)+8\left( 2-\lambda \right)=0 \\
& \Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-1 \right)-8\left( \lambda -2 \right)=0 \\
& \Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-1-8 \right)=0 \\
\end{align}$
$\Rightarrow \left( \lambda -2 \right)\left( {{\lambda }^{2}}-9 \right)=0$
Now we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ hence we get
$\left( \lambda -2 \right)\left( \lambda -3 \right)\left( \lambda +3 \right)=0$
Hence we have the value of $\lambda $ is 2, 3 or -3.
Now let us check the values of c if $\lambda =3,-3$
We have $\vec{c}=2\hat{i}+4\hat{j}+\left( {{\lambda }^{2}}-1 \right)\hat{k}$ .
Hence for $\lambda =3,-3$ we get $\vec{c}=2\hat{i}+4\hat{j}+\left( 8 \right)\hat{k}$
Now we can write this as $\vec{c}=2\left( \hat{i}+2\hat{j}+4\hat{k} \right)$ .
But we know that \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] hence we get $\vec{c}=2\vec{a}$ .
Now if we have two vectors such that $\vec{a}=\lambda \vec{b}$ then we say that the two vectors are parallel.
Hence we have $\vec{a}$ and $\vec{c}$ are parallel.
Now we know that cross product of parallel vectors is equal to zero. Hence if we take $\lambda =3,-3$ we will get $\vec{a}\times \vec{c}=0$ . and since we need non zero cross product we will take the value of $\lambda =2$ .
Hence using this we get $\vec{c}=2\hat{i}+4\hat{j}+\left( {{2}^{2}}-1 \right)\hat{k}$
Hence we get $\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$
Now we have $\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$ , \[\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\] and we want to find $\vec{a}\times \vec{c}$ .
Now we know that if $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\vec{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ then $\vec{b}\times \vec{c}$ is given by $\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{2}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$
Hence we get
$\vec{a}\times \vec{c}=\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
1 & 2 & 4 \\
2 & 4 & 3 \\
\end{matrix} \right|$
Now expanding the determinant we get
$\begin{align}
& \vec{a}\times \vec{c}=\left( 6-16 \right)\hat{i}-\left( 3-8 \right)\hat{j}+\left( 4-4 \right)\hat{k} \\
& \Rightarrow \vec{a}\times \vec{c}=-10\hat{i}+5\hat{j} \\
\end{align}$
Hence option C is the correct option.
Note: Now note that when the vectors are parallel their cross product is zero and when they are perpendicular their dot product is zero. Now also remember that when the vectors are collinear we get the corresponding determinant as 0. This is nothing but a triple product of vectors. Now triple product represents the volume of parallelepiped formed hence if the vectors are collinear we have this volume as zero. Hence we get the condition.
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