Question:
The angle between the lines $2x = 3y = - z$ and $6x = - y = - 4z$ is
The angle between the lines $2x = 3y = - z$ and $6x = - y = - 4z$ is
Updated On: May 19, 2024
- $0^{\circ}$
- $45^{\circ}$
- $90^{\circ}$
- $30^{\circ}$
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The Correct Option is C
Solution and Explanation
Given $L_1 : 2x = 3y = - z$ and $L_2 : 6x = - y = -4x $
$\Rightarrow \, L_1 : \frac{x}{\frac{1}{2}} = \frac{y}{\frac{1}{3}} = \frac{z}{-1}$ and $L_2 : \frac{x}{\frac{1}{16}} = \frac{y}{-1} = \frac{z}{ - \frac{1}{4}}$
Now, $a_1 a_2 + b_1 b_2 + c_1 c_2 = \left(\frac{1}{2} \times\frac{1}{6}\right) + \left(\frac{1}{3} \times-1\right) + \left(-1 \times- \frac{1}{4}\right) = \frac{1}{12} - \frac{1}{3} + \frac{1}{4}= 0 $
$\therefore$ Angle between $L_1$ and $L_2$ is $90^{\circ}$
$\Rightarrow \, L_1 : \frac{x}{\frac{1}{2}} = \frac{y}{\frac{1}{3}} = \frac{z}{-1}$ and $L_2 : \frac{x}{\frac{1}{16}} = \frac{y}{-1} = \frac{z}{ - \frac{1}{4}}$
Now, $a_1 a_2 + b_1 b_2 + c_1 c_2 = \left(\frac{1}{2} \times\frac{1}{6}\right) + \left(\frac{1}{3} \times-1\right) + \left(-1 \times- \frac{1}{4}\right) = \frac{1}{12} - \frac{1}{3} + \frac{1}{4}= 0 $
$\therefore$ Angle between $L_1$ and $L_2$ is $90^{\circ}$
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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