Question

The angle between the lines 2x = 3y = – z and 6x = -y = -4z is:
A.0°
B.90°
C.45°
D.30°

Answer 1

We will simplify the given equation of lines in the standard form of $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ where a, b and c are the direction ratios of the line. We will calculate the angle between these lines using the formula cos$\theta$= $\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}$ .

Complete step-by-step answer:
we are given two lines as 2x = 3y = – z and 6x = -y = -4z
We can write them in standard form $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$as:
For 2x = 3y = – z $\Rightarrow \dfrac{x}{{1/2}} = \dfrac{y}{{1/3}} = \dfrac{z}{{ - 1}}$
Comparing this equation with the standard equation, we get the direction ratios of this line as:
$a_1$ = $\dfrac{1}{2}$, $b_1$ = $\dfrac{1}{3}$, $c_1$ = -1
For 6x = -y = -4z $\Rightarrow \dfrac{x}{{1/6}} = \dfrac{y}{{ - 1}} = \dfrac{z}{{ - 1/4}}$
Comparing this equation with the standard equation, we get the direction ratios of this line as:
$a_2$ = $\dfrac{1}{6}$ , $b_2$ = -1, $c_2$ = $\dfrac{{ - 1}}{4}$
Now, putting these values of $a_1$, $a_2$, $b_1$, $b_2$, $c_1$, and $c_2$ in the equation of cos$\theta$, we get
$\Rightarrow \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} }} \\\ \Rightarrow \cos \theta = \dfrac{{\dfrac{1}{2} \cdot \dfrac{1}{6} + \dfrac{1}{3} \cdot ( - 1) + ( - 1)\dfrac{{\left( { - 1} \right)}}{4}}}{{\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( { - 1} \right)}^2}} \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {\dfrac{{ - 1}}{4}} \right)}^2}} }} \\\$
Upon simplifying this expression, we get
$\Rightarrow \cos \theta = \dfrac{{\dfrac{1}{{12}} - \dfrac{1}{3} + \dfrac{1}{4}}}{{\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( { - 1} \right)}^2}} \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {\dfrac{{ - 1}}{4}} \right)}^2}} }} = \dfrac{{\dfrac{1}{{12}} - \dfrac{1}{{12}}}}{{\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( { - 1} \right)}^2}} \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {\dfrac{{ - 1}}{4}} \right)}^2}} }} = 0$
$\Rightarrow \cos \theta = 0$
Operating ${\cos ^{ - 1}}$both sides, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \theta } \right) = {\cos ^{ - 1}}\left( 0 \right) \\\ \Rightarrow \theta = {90^ \circ } \\\$
($\because {\cos ^{ - 1}}$ 0 = 90$^ \circ$)
Therefore, the angle between the lines 2x = 3y = – z and 6x = -y = -4z is 90$^ \circ$.
Hence, option (B) is correct.

Note: In such questions, you may get confused in the selection of the formula for calculating the angle between the given pair of the lines. Be careful in determining the direction ratios of the lines and after that, in the simplification of the expression of cos$\theta$ by putting the values of the obtained direction ratios.