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  • Help me understand! - The plane passing through the point (4,-1,2) and parallel to the lines and also passes through th - Three Dimensional Geometry - JEE Main

The plane passing through the point (4,-1,2) and parallel to the lines

 \frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2 } and \frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3 }

also passes through the point:

  • Option 1)

     

    (1,1,-1)

  • Option 2)

     

    (1,1,1)

     

  • Option 3)

     

    (-1,-1,-1)

  • Option 4)

     

    (-1,-1,1)

Answers (1)

best_answer

 

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

 

 

Condition for line to be parallel to plane -

\vec{b}\cdot \vec{n}= 0 or a_{1}a+b_{1}b+c_{1}c= 0

-

 

From the concept

Let \vec{n} be the normal vector to the plane passing through (4, -1 , 2 ) and parallel to the lines L1 and L2

then ,

\vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 3 & -1 &2 \\ 1 &2 & 3 \end{vmatrix} = -7 \hat{i} - 7 \hat{j} + 7\hat{k}

\therefore Equation of plane is

-1(x-4) -1 (y+1) + 1(z-2) = 0

\therefore x + y - z -1 = 0

(1,1,1) satisfies the plane.

 


Option 1)

 

(1,1,-1)

Option 2)

 

(1,1,1)

 

Option 3)

 

(-1,-1,-1)

Option 4)

 

(-1,-1,1)

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