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)

    – 4

  • Option 2)

    –2

  • Option 3)

    0

  • Option 4)

    1

 

Answers (1)
A Aadil Khan

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

Incorrect Option

 

Option 2)

–2

Correct Option

 

Option 3)

0

Incorrect Option

 

Option 4)

1

Incorrect Option

 

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