3D Geometry 3 Question 28
####28. The equation of the line passing through $(-4,3,1)$, parallel to the plane $x+2 y-z-5=0$ and intersecting the line
(2019 Main, 9 Jan I) $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$ is
(a) $\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$
(b) $\frac{x+4}{-1}=\frac{y-3}{1}=\frac{z-1}{1}$
(c) $\frac{x+4}{1}=\frac{y-3}{1}=\frac{z-1}{3}$
(d) $\frac{x-4}{2}=\frac{y+3}{1}=\frac{z+1}{4}$
Show Answer
Answer:
(a)
Solution:
- Any point on the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$ is of the form $(-3 \lambda-1,2 \lambda+3,-\lambda+2)$
$ \begin{array}{r} {\left[\operatorname{take} \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}=\lambda \Rightarrow x=-3 \lambda-1,\right.} \\ y=2 \lambda+3 \text { and } z=-\lambda+2] \end{array} $
So, the coordinates of point of intersection of two lines will be $(-3 \lambda-1,2 \lambda+3,-\lambda+2)$ for some $\lambda \in R$.
Let the point $A \equiv(-3 \lambda-1,2 \lambda+3,-\lambda+2)$ and $B \equiv(-4,3,1)$
$ \begin{aligned} & \text { Then, } \quad \mathbf{A B}=\mathbf{O B}-\mathbf{O A} \\ & =(-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})-{(-3 \lambda-1) \hat{\mathbf{i}}+(2 \lambda+3) \hat{\mathbf{j}}+(-\lambda+2) \hat{\mathbf{k}}} \\ & =(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}} \end{aligned} $
Now, as the line is parallel to the given plane, therefore $\mathbf{A B}$ will be parallel to the given plane and so $\mathbf{A B}$ will be perpendicular to the normal of plane.
$\Rightarrow \mathbf{A B} \cdot \lambda=0$, where $\mathbf{n}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ is normal to the plane.
$ \begin{aligned} & \Rightarrow \quad((3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\boldsymbol{\lambda}-1) \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=0 \\ & \Rightarrow 3(\boldsymbol{\lambda}-1)-4 \lambda+(-1)(\lambda-1)=0 \\ & \quad\left[\because \text { If } \mathbf{a}=a_{1} \hat{\mathbf{i}}+a_{2} \hat{\mathbf{j}}+a_{3} \hat{\mathbf{k}} \text { and } \mathbf{b}=b_{1} \hat{\mathbf{i}}+b_{2} \hat{\mathbf{j}}+b_{3} \hat{\mathbf{k}},\right. \\ & \left.\quad \text { then } \mathbf{a} \cdot \mathbf{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right] \\ & \Rightarrow \quad 3 \lambda-3-4 \lambda-\lambda+1=0 \\ & \Rightarrow \quad-2 \lambda=2 \Rightarrow \lambda=-1 \end{aligned} $
Now, the required equation is the equation of line joining $A(2,1,3)$ and $B(-4,3,1)$, which is
$ \frac{x-(-4)}{2-(-4)}=\frac{y-3}{1-3}=\frac{z-1}{3-1} $
$\left[\because\right.$ Equation of line joining $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is
$ \begin{aligned} & \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ \Rightarrow \quad \frac{x+4}{6} & =\frac{y-3}{-2}=\frac{z-1}{2} \\ \text { or } \quad & \frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1} \quad \text { [multiplying by } 2 \text { ] } \end{aligned} $
