Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).
Solution and Explanation
The required line passes through the origin.
Therefore, its position vector is given by, \(\vec a\)=\(\vec 0\)…………....(1)
The direction ratios of the line through the origin and (5,-2,3) are (5-0)=5, (-2-0)=-2, (3-0)=3
The lines are parallel to the vector given by the equation, b→5\(\hat i\)-2\(\hat j\)+3\(\hat k\)
The equation of the line in vector form through a point with position vector a and parallel to \(\vec b\) is,
\(\vec r = \vec a+\lambda \vec b\), λ∈R
⇒ \(\vec r=\vec 0+\lambda(5\vec i-2\vec j+3\vec k)\)
⇒ \(\vec r=\lambda(5\vec i-2\vec j+3\vec k)\)
The equation of the line through the point (x1,y1,z1) and 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 in the cartesian form is
\(\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\)
⇒\(\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\)
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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\(\hat{k}\)