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Find the meeting point of the line $\overrightarrow{r}(\overrightarrow{2i}+\overrightarrow{j}-\overrightarrow{3k}) + t(\overrightarrow{2i}-\overrightarrow{j}-\overrightarrow{k}) $ and the plane. $x-2y+3z+7=0$

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  • Equation of a straight line passing through a given point and parallel to a given vector $ \overrightarrow r = \overrightarrow a+t\overrightarrow v$ ( vector equation ) where $\overrightarrow a$ is the pv of the point and $ \overrightarrow v$ the vector parallel to the line, a scalar $ \large\frac{x-x_1}{l} = \large\frac{y-y_1}{m} = \large\frac{z-z_1}{n}$ ( cartesian form) where $(x_1, y_1, z_1) $ is the point on the line and $ l, m, n$ are the d.c.s of the vector parallel to the line $ l, m, n$ can also be replaced by the d.r.s $ a, b, c$.
Step 1
The line $ \overrightarrow r =(2\overrightarrow i-\overrightarrow j+3\overrightarrow k)+t(2\overrightarrow i-\overrightarrow j-\overrightarrow k)$ passes through $(2, -1, 3)$ and is parallel to $ \overrightarrow u=2\overrightarrow i-\overrightarrow j-\overrightarrow k$
The cartesian equation of the line is
$ \large\frac{ x-2}{2}=\large\frac{y+1}{-1}=\large\frac{z-3}{-1}$ (i)
Step 2
Let $ \large\frac{x-2}{2}=\large\frac{y+1}{-1}=\large\frac{z-3}{-1}=\mu$ at the point of intersection of the line (i) and the plane $ x-2y+3z+7=0$ (ii)
$ \therefore x=2\mu+2, \: y= - \mu-1, \: z=-\mu+3$
Substitute the above in (ii)
$ 2 \mu+2-2(-\mu-1)+3(-\mu+3)+7=0$
$ 2\mu+2+2\mu+2-3\mu+9+7=0$
$ \mu+20=0 \Rightarrow \mu=-20$
Step 3
$ \therefore x = -40+2=-38\:, y=20-1=19, \: z=20+3=23$
The point of intersection of (i) with (ii) is $(-38, 19, 23)$


answered Jun 17, 2013 by thanvigandhi_1
edited Jul 18, 2013 by vijayalakshmi_ramakrishnans

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