Show that the lines through the points(1,-1,2)(3,4,-2)is perpendicular to the line through the points(0,3,2)and(3,5,6).
Solution and Explanation
Let AB be the lines joining the points (1,-1,2) and (3,4,-2), and CD be the lines joining the points, (0,3,2), (3,5,6).
The direction ratios: a1,b1,c1 of AB are (3-1), (4-(-1)), and (-2-2) i.e., 2, 5, and -4.
The direction ratios: a2,b2,c2 of CD are (3-0), (5-3), and (6-2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other if a1a2+b1b2+c1c2=0
a1a2+b1b2+c1c2
=2×3+5×2+(-4)×4
=6+10-16
=0
Therefore, AB and CD are perpendicular to each other.
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What is the Planning Process?
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Notes on Three Dimensional Geometry
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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\(\hat{k}\)