Find the cartesian equation of the line that passes through the point(-2,4,-5)and parallel to the line given by \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Find the cartesian equation of the line that passes through the point(-2,4,-5)and parallel to the line given by \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Solution and Explanation
It is given that the line passes through the point (-2,4,-5) and is parallel to \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
The direction ratios of the line, \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\), are 3, 5, and 6.
The required line is parallel to \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Therefore, its direction ratios are 3k, 5k, and 6k, where k≠0
It is known that the equation of the line through the point (x1,y1,z1) and with direction ratios, a,b,c is given by \(\frac{x-x_1}{a}\)=\(\frac{y-y_1}{b}\)=\(\frac{z-z_1}{c}\)
Therefore, the equation of the required line is
\(\Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k\)
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)