Find the vector and the cartesian equations of the line that passes through the points(3,-2,-5),(3,-2,6).
Solution and Explanation
Let the line passing through the points P (3,-2,-5) and Q(3,-2,6), be PQ.
Since PQ passes through P (3,-2,-5), its position vector is given by,
\(\overrightarrow a=3\widehat{i}-2\widehat j-5\widehat k\)
The direction ratios of PQ are given by, (3-3)=0, (-2+2)=0, (6+5)=11
The equation of the vector in the direction of PQ is
\(\overrightarrow b=0 \widehat i-0. \widehat j+11 \widehat k=11 \widehat k\)
The equation of PQ in vector form is given by, \(\overrightarrow r= \overrightarrow a+ \lambda \overrightarrow b, \lambda \in R\)
\(\Rightarrow \overrightarrow r=(3 \widehat i-2\widehat j-5\widehat k)+11 \lambda \widehat k\)
The equation of PQ in cartesian form is
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} i.e.,\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\)
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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\(\hat{k}\)