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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)

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

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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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