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Find the vector and cartesian equation of the line through the point $(3 , -4 , -2 )$ and parallel to vector $\overrightarrow{9i} +\overrightarrow{6j} +\overrightarrow{2k}$.

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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$.
Vector equation : The line passes through the point $ A(3, -4, -2)$ i.e., $ \overrightarrow a = 3\overrightarrow i-4\overrightarrow j-2\overrightarrow k$ and is parallel to the vector $ \overrightarrow u = 9\overrightarrow i+6\overrightarrow j+2\overrightarrow k \therefore $ the vector equation is $ \overrightarrow r = \overrightarrow a+t\overrightarrow u$ $ \overrightarrow r = 3\overrightarrow i-4\overrightarrow j-2\overrightarrow k + t(9\overrightarrow i+6\overrightarrow j+2\overrightarrow k), t$ is a scalar.
Cartesian equation : $(x, y, z) = (3, -4, -2)$ and the d.r.s of the parallel vector are $ (9, 6, 2) = (a, b, c).$
The equation is $ \large\frac{x-x_1}{a} = \large\frac{y-y_1}{b} = \large\frac{z-z_1}{c}$
$ \large\frac{x-3}{9} = \large\frac{y+4}{6} = \large\frac{z+2}{2}$
answered Jun 11, 2013 by thanvigandhi_1

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