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Give answer! - Let, and be coplanar vectors. Then the non-zero vector is : - Vector Algebra - JEE Main

Let \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}+(\lambda ^{2}-1)\hat{k}  be  coplanar vectors.

Then the non-zero vector \vec{a}\times \vec{c}  is :

  • Option 1)

    -10\hat{i}-5\hat{j}

     

  • Option 2)

    -14\hat{i}-5\hat{j}

  • Option 3)

    -10\hat{i}+5\hat{j}

  • Option 4)

    -14\hat{i}+5\hat{j}

Answers (1)
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A admin

 

Coplanar vectors -

A given number of vectors are called coplanar if their line segments are parallel to the same plane.

- wherein

Two vectors are always coplanar.

 

 

Coplanar vectors -

\left [ \vec{a}\;\vec{b}\;\vec{c} \right ]=0

- wherein

\vec{a}\vec{b} and \vec{c} are three vectors.

 

\vec{a}=\hat{i}+2\hat{j}+4\hat{k}

\vec{b}=\hat{i}+\lambda \hat{j}+4\hat{k}

\vec{c}=2\hat{i}+4 \hat{j}+(\lambda ^{2}-1)\hat{k}

[\vec{a}\vec{b}\vec{c}]=\left | \begin{matrix} 1 & 2& 4\\ 1& \lambda &4 \\ 2 & 4 & \lambda ^{2}-1 \end{matrix}\right |

=\lambda (\lambda ^{2}-1)-16-2(\lambda ^{2}-9)+4(4-2\lambda )

\Rightarrow [\vec{a} \vec{b}\vec{c}]=(\lambda -3)(\lambda +3)(\lambda -2)

for \lambda =\pm 3  {\vec{c}}=2\vec{a}\Rightarrow \vec{a}\times \vec{c}=0

for \lambda =2,

\vec{a}\times \vec{c}=\left | \begin{matrix} \hat{i} &\hat{j}& \hat{k}\\ 1 &2 &4 \\ 2& 4& 3 \end{matrix} \right |=-10\hat{i}+5\hat{j}

=-5(2\hat{i}-\hat{j})

 

 

 


Option 1)

-10\hat{i}-5\hat{j}

 

Option 2)

-14\hat{i}-5\hat{j}

Option 3)

-10\hat{i}+5\hat{j}

Option 4)

-14\hat{i}+5\hat{j}

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