Show that the line through the points(4,7,8)(2,3,4)is parallel to the line through the points(-1,-2,1),(1,2,5).
Solution and Explanation
Let AB be the line through the points (4,7,8), and (2,3,4), and CD be the line through the points (-1,-2,1), and (1,2,5).
The direction ratios, a1, b1, c1 of AC are (2-4), (3-7), and (4-8) i.e., -2, -4, and -4.
The direction ratios, a2, b2, c2 of CD are (1-(-1)), (2-(-2)), and (5-1) i.e., 2, 4, and 4.
AB will parellel to CD, if \(\frac{a_1}{a_2}\)=\(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\)
=\(-\frac{2}{2}\)
=-1
\(\frac{b_1}{b_2}\)
=\(-\frac{4}{4}\)
=-1
\(\frac{c_1}{c_2}\)
=\(-\frac{4}{4}\)
=-1
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
Thus, AB is parallel to CD.
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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\(\hat{k}\)