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Equation of plane passing through point (1,2,3) and line\frac{x}{2}=\frac{y}{3}=\frac{z}{4} is

  • Option 1)

    4x+y-2z=0

  • Option 2)

    x-2y+z=0

  • Option 3)

    x+y-z=0

  • Option 4)

    2x+2y-z=0

 

Answers (1)

best_answer

As we have learned

Plane passing through a point and a line (Cartesian form) -

Let the plane passes through A(x_{1},y_{1},z_{1}) and a line \frac{x-x_{2}}{a}=\frac{y-y_{2}}{b}=\frac{z-z_{2}}{c},

then the plane is given by \begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1}&y_{2}-y_{1} &z_{2}-z_{1} \\ a_{2}& b_{2} & c_{2} \end{vmatrix}= 0

- wherein

\vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_{2}-x_{1}&y_{2}-y_{1} &z_{2}-z_{1} \\ a& b & c \end{vmatrix}

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

 

 

 Equation of plane will be \begin{vmatrix} x-1&y-2 &z-3 \\ 0-1& 0-2 &0-3 \\ 2&3 &4 \end{vmatrix}=0

\Rightarrow\begin{vmatrix} x-1&y-2 &z-3 \\ -1& -2 &-3 \\ 2&3 &4 \end{vmatrix}=0

\Rightarrow(x-1)(1)-(y-2)(2)+(z-3)(1)=0

\Rightarrow x-1-2y+4+z-3=0

\Rightarrow x-2y+z=0


Option 1)

4x+y-2z=0

Option 2)

x-2y+z=0

Option 3)

x+y-z=0

Option 4)

2x+2y-z=0

Posted by

Himanshu

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