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Can someone explain - The equationof the line passing throughparallel to the planeand intersecting the lineis : - Three Dimensional Geometry - JEE Main

The equation of the line passing through (-4, 3, 1), parallel to the plane x +2y -z -5 = 0 and intersecting the line \frac{x+1}{-3} = \frac{y-3}{2} = \frac{z-2}{-1} is :

  • Option 1)

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

  • Option 2)

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

  • Option 3)

  • Option 4)

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Cartesian eqution of a line -

The equation of a line passing through two points A\left ( x_{0},y_{0},z_{0} \right )and parallel to vector having direction ratios as \left ( a,b,c \right )is given by

\frac{x-x_{0}}{a}= \frac{y-y_{0}}{b}= \frac{z-z_{0}}{c}

The equation of a line passing through two points A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right ) is given by

\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

- wherein

 

 

Angle between line and Plane (vector form ) -

The angle between a line \vec{r}= \vec{a}+\lambda \vec{b}and the plane \vec{r}.\vec{n}= d is given by

\sin \Theta = \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |}

- wherein

From the concept we have learnt 

Normal vector of plane containing two intersecting  lines is parallel to vector 

\vec{v_{1}}=\begin{vmatrix} \hat{i} & \hat{\hat{j}} &\hat{k} \\ 3 &0 & 1\\ -3&2 & -1 \end{vmatrix}=-2\hat{i}+6\hat{k}

\therefore Required line is parallel to vector 

\left ( \vec{v_{2}} \right )=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1& 2 &-1 \\ -2&0 &6 \end{vmatrix}=3\hat{i}-\hat{j}+\hat{k}

\therefore  Required equation of line is 

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

 


Option 1)

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

Option 2)

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

Option 3)

Option 4)

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