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Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2\hat{k}\; \; and\; \;\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}.$ If the vectors $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$ , then $x$ equals

Option 1)
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–2

Option 3)
0

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1

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As we learnt in

Plane passing through a point and parallel to two given vectors (Cartesian form) -

Let the plane passes through $A(x_{1},y_{1},z_{1})$ and parallel to vectors having DR's $(a_{1},b_{1},c_{1})\: and \: (a_{2},b_{2},c_{2})$ , then the plane is given by

$\begin{vmatrix} x-x_{1} & y-y_{1} &z-z_{1} \\ a_{1} & b_{1} &c_{1} \\ a_{2}& b_{2} & c_{2} \end{vmatrix}=0$

- wherein

$\vec{n}=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ a_{1} & b_{1} &c_{1} \\ a_{2}& b_{2} & c_{2} \end{vmatrix}$

$\left ( \vec{r}-\vec{a} \right )\cdot \vec{n}= 0$

Normal vector of plane with $\underset{a}{\rightarrow}\ and\ \underset{b}{\rightarrow}$ is $=\begin{vmatrix} \hat{i}& \hat{j} & \hat{k}\\ 1&1&1\\ 1&-1&2 \end{vmatrix}=3 \hat{i} - \hat{j}-2 \hat{k}$

$Equation\, \, \, of \, \, \, plane\, \, \, is\, \, \, 3x-y-2z=0$

$Put\, \, \, \left ( x,\left ( x-2 \right )-1 \right )$

$We\, \, \, get \, \, \,3x-\left ( x-2 \right )+2= 0$

$2x+4=0$

$x=-2$

Option 1)

– 4

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Option 2)

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Correct Option

Option 3)

0

Incorrect Option

Option 4)

1

Incorrect Option

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