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

Hint:
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, \[\cos \theta = \dfrac{{\overrightarrow {{m_1}} \cdot \overrightarrow {{m_2}} }}{{\left| {\overrightarrow {{m_1}} } \right|\left| {\overrightarrow {{m_2}} } \right|}}\]