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Home  >>  TN XII Math  >>  Vector Algebra
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Find the point of intersection of the line $\overrightarrow{r}=(\overrightarrow{j}-\overrightarrow{k})+s(\overrightarrow{2i}-\overrightarrow{j}+\overrightarrow{k})$ and $x z$ - plane.

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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$.
  • Equation of the $xy$ plane is $ z=0$, equation of the $yz$ plane is $ x=0$ and equation of the $xz$ plane is $y=0$
Step 1
The line $ \overrightarrow r =(\overrightarrow j-\overrightarrow k)+s(2\overrightarrow i-\overrightarrow j+\overrightarrow k)$ passes through the point $(0, 1, -1)$ and is parallel to $ \overrightarrow u=2\overrightarrow i-\overrightarrow j+\overrightarrow k$.
The cartesian equation of the line is
$ \large\frac{x}{2}=\large\frac{y-1}{-1}=\large\frac{z+1}{1}$ (i)
The equation of the $xz$ plane is $ y=0$ (ii)
Step 2
Subset $ y=0$ in (i)
$ \large\frac{x}{2}=1=z+1 \Rightarrow x = 2, \: y=0, \: z=0$
The point of intersection is $(2, 0, 0)$
answered Jun 17, 2013 by thanvigandhi_1
 

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