Find the values of P so the line\(\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}\) and \(\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Find the values of P so the line\(\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}\) and \(\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Solution and Explanation
The given equation can be written in the standard form as
\(\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}\) and \(\frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}=\frac{6-z}{-5}\)
The direction ratios of the lines are -3 ,\(\frac{2p}{7}\), 2, and \(\frac{-3p}{7}\), 1, -5 respectively.
Two lines with direction ratios, a1, b1, c1, and a2, b2, c2, are perpendicular to each other, if a1a2+b1b2+c1c2=0
∴(-3)\(\bigg(\frac{-3p}{7}\bigg)+\bigg(\frac{2p}{7}\bigg)\)(1)+2.(-5)=0
\(\Rightarrow \frac{9p}{7}+\frac{2p}{7}=10\)
\(\Rightarrow \) 11p=70
\(\Rightarrow \) p=\(\frac{70}{11}\)
Thus, the value of P is \(\frac{70}{11}\).
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Concepts Used:
Angle between Two Lines
The two straight lines, whenever intersects, form two sets of angles. The angles so formed after the intersection are;
- A pair of acute angle
- Another pair of an obtuse angle
The absolute values of angles created depend on the slopes of the intersecting lines.
It is also worth taking note, that the angle so formed by the intersection of two lines cannot be calculated if any of the lines is parallel to the y-axis as the slope of a line parallel to the y-axis is an indeterminate.
Read More: Angle Between Two Lines