The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
Solution and Explanation
The Cartesian equation of the line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)……….....(1)
The given line passes through the point (5,-4,6).
The position vector of this point is \(\vec a = 5\hat i-4\hat j +6 \hat k\)
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of the vector, \(\vec b\)=3\(\hat i\)+7\(\hat j\)+2\(\hat k\)
It is known that the line through position vector \(\vec a\) and in the direction of the vector \(\vec b\) is given by the equation,
\(\vec r\)=\(\vec a\)+λ\(\vec b\), λ∈R
⇒ \(\vec r\)=(5\(\vec i\)-4\(\vec j\)+6\(\vec k\))+λ(3\(\vec i\)+7\(\vec j\)+2\(\vec k\))
This is the required equation of the given line in vector form.
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Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
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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\(\hat{k}\)