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  • Stuck here, help me understand: - Three Dimensional Geometry - JEE Main-2

The plane containing the line  \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} and parallel to the line   \frac{x}{1}=\frac{y}{1}=\frac{z}{4}  passes through the point :

  • Option 1)

    (1, -2, 5)

  • Option 2)

    (1, 0, 5)

  • Option 3)

    (0, 3, -5)

  • Option 4)

    (-1, -3, 0)

 

Answers (1)

best_answer

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

 

 Line is \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}

The normal vector of plane is 

\begin{vmatrix} \hat{i} & \hat{j}& \hat{k}\\ 1& 2 &3 \\ 1 &1 &4 \end{vmatrix}=5\hat{i}-\hat{j}-\hat{k}

So equation is 5x-y-z = 5-2-3=0

5x-y-z=0

It passes through (1,0,5)

 


Option 1)

(1, -2, 5)

This option is incorrect

Option 2)

(1, 0, 5)

This option is correct

Option 3)

(0, 3, -5)

This option is incorrect

Option 4)

(-1, -3, 0)

This option is incorrect

Posted by

prateek

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