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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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