Question

The point of intersection of the lines \[\dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}}\] and \[\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}}\] is
A. \[\left( {0,0, - 4} \right)\]
B. \[\left( {1,0,0} \right)\]
C. \[\left( {0,2,0} \right)\]
D. \[\left( {1,2,0} \right)\]

Answer 1

Hint: At first we will find the coordinate of any point lying on both the lines by equating the line equation with a constant to find the values of x, y and, z in the form of that constant.
Now we will equate the coordinate of a general point on both the lines with each other to get the coordinate of the intersection point.

Complete step-by-step answer:
Given data: The lines \[\dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}}\] and \[\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}}\]
Let the line ${L_1} = a$
 \[ \Rightarrow \dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}} = a\]
Now equating x term with a
 \[ \Rightarrow \dfrac{{x - 6}}{{ - 6}} = a\]
On cross multiplication
 \[ \Rightarrow x - 6 = - 6a\]
Adding 6 on both sides
 \[ \Rightarrow x = - 6a + 6\]
Similarly, equating y term with a
 \[ \Rightarrow \dfrac{{y + 4}}{4} = a\]
On cross multiplication
 \[ \Rightarrow y + 4 = 4a\]
Adding -4 on both sides
 \[ \Rightarrow y = 4a - 4\]
Similarly, equating z term with a
 \[ \Rightarrow \dfrac{{z - 4}}{{ - 8}} = a\]
On cross multiplication
 \[ \Rightarrow z - 4 = - 8a\]
Adding 4 on both sides
 \[ \Rightarrow z = - 8a + 4\]
Now we have a general point on line 1 i.e. $\left( { - 6a + 6,4a - 4, - 8a + 4} \right)$
Similarly, finding a general point on line 2 i.e. \[\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}}\]
Let ${L_2} = b$
 \[ \Rightarrow \dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}} = b\]
Now equating x term with b
 \[ \Rightarrow \dfrac{{x + 1}}{2} = b\]
On cross multiplication
 \[ \Rightarrow x + 1 = 2b\]
Adding -1 on both sides
 \[ \Rightarrow x = 2b - 1\]
Similarly, equating y term with b
 \[ \Rightarrow \dfrac{{y + 2}}{4} = b\]
On cross multiplication
 \[ \Rightarrow y + 2 = 4b\]
Adding -2 on both sides
 \[ \Rightarrow y = 4b - 2\]
Similarly, equating z term with b
 \[ \Rightarrow \dfrac{{z + 3}}{{ - 2}} = b\]
On cross multiplication
 \[ \Rightarrow z + 3 = - 2b\]
Adding -3 on both sides
 \[ \Rightarrow z = - 2b - 3\]
Now we have a general point on line1 i.e. $\left( {2b - 1,4b - 2, - 2b - 3} \right)$
Now, to get the intersection point of both the lines we will equate the general form of the point lying on the lines i.e.
  $ \Rightarrow \left( { - 6a + 6,4a - 4, - 8a + 4} \right) = \left( {2b - 1,4b - 2, - 2b - 3} \right)$
Now equating the x-terms
 $ \Rightarrow - 6a + 6 = 2b - 1$
Solving by combining the like terms
 \[ \Rightarrow - 6a + 6 + 1 = 2b\]
 \[ \Rightarrow 2b = 7 - 6a..........(i)\]
Similarly equating the y-terms
 $ \Rightarrow 4a - 4 = 4b - 2$
Dividing the whole equation by 2
 $ \Rightarrow 2a - 2 = 2b - 1$
Solving by combining the like terms
 $ \Rightarrow 2b = 2a - 1............(ii)$
By equating the value of 2b from equation(i) and equation(ii)
 $ \Rightarrow 7 - 6a = 2a - 1$
Solving by combining the like terms
 $ \Rightarrow 8 = 8a$
Dividing both sides by 8 we get,
 $\therefore a = 1$
Substituting the value of ‘a’ in equation(i)
 $ \Rightarrow 2b = 7 - 6(1)$
 $ \Rightarrow 2b = 1$
Dividing both sides by 2
 $ \Rightarrow b = \dfrac{1}{2}$
Now, substituting the value of ‘a’ in the general form of the point of line 1
We get a point $\left( {0,0, - 4} \right)$
Therefore the intersection point of both the lines is $\left( {0,0, - 4} \right)$
Option(A) is correct.

Note: We equated the x and y part of the general form of point of both the line, but we can also equate it with the z-term as it will give the same result
Equating the z-terms
 $ \Rightarrow - 8a + 4 = - 2b - 3$
Solving by combining the like terms
 \[ \Rightarrow 8a - 4 - 3 = 2b\]
 \[ \Rightarrow 2b = 8a - 7\]
Substituting the value of 2b from equation(i)
 \[ \Rightarrow 7 - 6a = 8a - 7\]
 \[ \Rightarrow 14 = 14a\]
Dividing both sides by 14
 \[\therefore a = 1\] , since the value of a is similar to the value of ‘a’ in the above solution the point will also be equivalent giving us the same intersection point.