Question

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

Answer 1

Assume $\theta$ as the angle between the lines. Then we will write the equation of the lines in standard form and the vectors parallel to the given lines are $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$. Then the angle between two lines is equal to the angle between $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$.

Complete step by step solution:
The given lines are
$2x = 3y = - z$ … (1)
And $6x = - y = - 4z$ … (2)
Now, we will write the given equation of lines in standard form as:
Since, we can write 2x as $\dfrac{x}{{\dfrac{1}{2}}}$.
Therefore, $2x = \dfrac{x}{{\dfrac{1}{2}}}$
And we can write 3y as $\dfrac{y}{{\dfrac{1}{3}}}$.
Therefore, $3y = \dfrac{y}{{\dfrac{1}{3}}}$
Also, we can write – z as $\dfrac{z}{{ - 1}}$.
Therefore, $- z = \dfrac{z}{{ - 1}}$
Therefore, the standard form of the equation $(1)$ is given by
$\Rightarrow \dfrac{x}{{\dfrac{1}{2}}} = \dfrac{y}{{\dfrac{1}{3}}} = \dfrac{z}{{ - 1}}$ … (3)
Similarly, we can write 6x as $\dfrac{x}{{\dfrac{1}{6}}}$ .
Therefore, $6x = \dfrac{x}{{\dfrac{1}{6}}}$
-Y can be written as $\dfrac{y}{{ - 1}}$
Therefore, $- y = \dfrac{y}{{ - 1}}$
And – 4z can be written as $\dfrac{z}{{ - 4}}$
Therefore, $- 4z = \dfrac{z}{{ - 4}}$
Therefore, the standard form of the equation $(2)$ can be written as
$\Rightarrow \dfrac{x}{{\dfrac{1}{6}}} = \dfrac{y}{{ - 1}} = \dfrac{z}{{ - 4}}$ … (4)
Since, the equation $(3)$ is
$\Rightarrow \dfrac{x}{{\dfrac{1}{2}}} = \dfrac{y}{{\dfrac{1}{3}}} = \dfrac{z}{{ - 1}}$
Let $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$ be vectors parallel to equations $(3)$and $(4)$respectively.
First, we will write $\overrightarrow {{b_1}}$ . The denominator of x in equation (3) will be the coefficient of $\mathop i\limits^ \wedge$ , the denominator of y in equation (3) will be the coefficient of $\mathop j\limits^ \wedge$ and the denominator of x in equation (3) will be the coefficient of $\mathop k\limits^ \wedge$
$\Rightarrow \overrightarrow {{b_1}} = \dfrac{1}{2}\mathop i\limits^ \wedge + \dfrac{1}{3}\mathop j\limits^ \wedge - \mathop k\limits^ \wedge$ … (5)
Similarly, we can write $\overrightarrow {{b_2}}$
$\Rightarrow \overrightarrow {{b_2}} = \dfrac{1}{6}\mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \dfrac{1}{4}\mathop k\limits^ \wedge$ … (6)
If $\theta$ is the angle between the given lines, then by using the following formula we can determine the value of $\theta$.
If $\theta$ is the angle between the given lines, then
$\cos \theta = \dfrac{{\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} }}{{\left| {\overrightarrow {{b_1}} } \right|\left| {\overrightarrow {{b_2}} } \right|}}$ … (7)
Here $\overrightarrow {{b_1}} = \dfrac{1}{2}\mathop i\limits^ \wedge + \dfrac{1}{3}\mathop j\limits^ \wedge - \mathop k\limits^ \wedge$ , $\overrightarrow {{b_2}} = \dfrac{1}{6}\mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \dfrac{1}{4}\mathop k\limits^ \wedge$
And $\left| {\overrightarrow {{b_1}} } \right|$ can be determined by squaring and adding the coefficients of $\mathop i\limits^ \wedge$ , $\mathop j\limits^ \wedge$ and $\mathop k\limits^ \wedge$ and then we will take its square root
$\Rightarrow \left| {\overrightarrow {{b_1}} } \right| = \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( { - 1} \right)}^2}}$ … (8)
Similarly, we can determine $\left| {\overrightarrow {{b_2}} } \right|$
$\Rightarrow \left| {\overrightarrow {{b_2}} } \right| = \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{( - 1)}^2} + {{\left( { - \dfrac{1}{4}} \right)}^2}}$ … (9)
Putting the value of $\overrightarrow {{b_1}}$ , $\overrightarrow {{b_2}}$ , $\left| {\overrightarrow {{b_1}} } \right|$ and $\left| {\overrightarrow {{b_2}} } \right|$ from equations (6) , (7) , (8) and (9) respectively, we have
$\Rightarrow \cos \theta = \dfrac{{\left( {\dfrac{1}{2}\mathop i\limits^ \wedge + \dfrac{1}{3}\mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right)\left( {\dfrac{1}{6}\mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \dfrac{1}{4}\mathop k\limits^ \wedge } \right)}}{{\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( { - 1} \right)}^2}} \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - \dfrac{1}{4}} \right)}^2}} }}$
On simplification, we get
$\Rightarrow \cos \theta = \dfrac{{\dfrac{1}{{12}} - \dfrac{1}{3} + \dfrac{1}{4}}}{{\sqrt {\dfrac{1}{4} + \dfrac{1}{9} + 1} \sqrt {\dfrac{1}{{36}} + 1 + \dfrac{1}{{16}}} }}$
Now, we take LCM and simplify
$\Rightarrow \cos \theta = \dfrac{{\dfrac{{1 - 4 + 3}}{{12}}}}{{\sqrt {\dfrac{{9 + 4 + 36}}{{36}}} \sqrt {\dfrac{{4 + 144 + 9}}{{144}}} }}$
On solving further, we get
$\Rightarrow \cos \theta = \dfrac{0}{{\sqrt {\dfrac{{49}}{{36}}} \sqrt {\dfrac{{157}}{{144}}} }}$
So, we have
$\Rightarrow \cos \theta = \dfrac{0}{{\dfrac{7}{6}\sqrt {\dfrac{{157}}{{144}}} }}$
When 0 is divided by any non-zero value, the result will always be zero.
$\Rightarrow \cos \theta = 0$
Since $\cos \theta = 0$ at $\theta = {90^ \circ }$.
$\Rightarrow \theta = {90^ \circ }$
Therefore, the angle between the lines $2x = 3y = - z$ and $6x = - y = - 4z$ is ${90^ \circ }$.

Hence option A is correct.

Note:
Let the Cartesian equation of two lines be
$\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$ … (a)
and $\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$ … (b)
Therefore, vector parallel to line (a) is
${\vec m_1} = {a_1}\mathop i\limits^ \wedge + {b_1}\mathop j\limits^ \wedge + {c_1}\mathop k\limits^ \wedge$
and vector parallel to line (b) is
${\vec m_2} = {a_2}\mathop i\limits^ \wedge + {b_2}\mathop j\limits^ \wedge + {c_2}\mathop k\limits^ \wedge$
Let $\theta$ is the angle between the lines (a) and (b). Then, $\theta$ is also the angle between ${\vec m_1}$ and ${\vec m_2}$.
Therefore, $\cos \theta = \dfrac{{\overrightarrow {{m_1}} \cdot \overrightarrow {{m_2}} }}{{\left| {\overrightarrow {{m_1}} } \right|\left| {\overrightarrow {{m_2}} } \right|}}$