Question

The angle between the line $2x = 3y = - z$ and $6x = - y = - 4z$ is
A. $90^\circ$
B. $0^\circ$
C. $30^\circ$
D. $45^\circ$

Answer 1

First we will first divide the equation of the line $2x = 3y = - z$ by 6 on each side and divide the equation of the line $6x = - y = - 4z$ by 12 on each side. Then we will use formula of the angle between the lines $\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$ and $\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$is given by $\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}$. Then we will find the value of ${a_1}$, ${b_1}$, ${c_1}$, ${a_2}$, ${b_2}$, and ${c_2}$ from the obtained equations and then substitute these values in the above formula of angle between lines. Then we will take the ${\cos ^{ - 1}}$ on both sides in the above equation.

Complete step by step answer:

We are given that the equation of the lines are $2x = 3y = - z$ and $6x = - y = - 4z$.

Dividing the equation of the line $2x = 3y = - z$ by 6 on each side, we get

$$\Rightarrow \dfrac{{2x}}{6} = \dfrac{{3y}}{6} = \dfrac{{ - z}}{6} \\\ \Rightarrow \dfrac{x}{3} = \dfrac{y}{2} = \dfrac{z}{{ - 6}}{\text{ ......eq.(1)}} \\\$$

Dividing the equation of the line $6x = - y = - 4z$ by 12 on each side, we get

$$\Rightarrow \dfrac{{6x}}{{12}} = \dfrac{{ - y}}{{12}} = \dfrac{{ - 4z}}{{12}} \\\ \Rightarrow \dfrac{x}{2} = \dfrac{y}{{ - 12}} = \dfrac{z}{{ - 3}}{\text{ ......eq.(2)}} \\\$$

We know that angle between the lines $\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$ and $\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$is given by $\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}$.

Let us assume that $\theta$ be the angle between the given lines.

Finding the value of ${a_1}$, ${b_1}$, ${c_1}$, ${a_2}$, ${b_2}$, and ${c_2}$ from the equation (1) and equation (2), we get

$\Rightarrow {a_1} = 3$
$\Rightarrow {b_1} = 2$
$\Rightarrow {c_1} = - 6$
$\Rightarrow {a_2} = 2$
$\Rightarrow {b_2} = - 12$

$\Rightarrow {c_2} = - 3$
Substituting the value of ${a_1}$, ${b_1}$, ${c_1}$, ${a_2}$, ${b_2}$, and ${c_2}$ from the above formula of angle between lines, we get

$$\Rightarrow \cos \theta = \dfrac{{3 \times 2 + 2 \times \left( { - 12} \right) + \left( { - 6} \right) \times \left( { - 3} \right)}}{{\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }} \\\ \Rightarrow \cos \theta = \dfrac{{6 - 24 + 18}}{{\sqrt {9 + 4 + 36} \sqrt {4 + 144 + 9} }} \\\ \Rightarrow \cos \theta = \dfrac{0}{{\sqrt {49} \sqrt {157} }} \\\ \Rightarrow \cos \theta = 0 \\\$$

Taking the ${\cos ^{ - 1}}$ on both sides in the above equation, we get

$$\Rightarrow {\cos ^{ - 1}}\cos \theta = {\cos ^{ - 1}}0 \\\ \Rightarrow \theta = \dfrac{\pi }{2} \\\$$

So, the angle between the lines is $\dfrac{\pi }{2}$.

Hence, option A is correct.

Note: In solving this question, students need to know about the basic formulas of the angles between two lines. One should know that the inverse of cosine will only lead to the principal value of $\theta$ not the general value or else the problem can be wrong.