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Question: The point of intersection of the lines \[\dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - ...

The point of intersection of the lines x−6−6=y+44=z−4−8\dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}} and x+12=y+24=z+3−2\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}} is
A. (0,0,−4)\left( {0,0, - 4} \right)
B. (1,0,0)\left( {1,0,0} \right)
C. (0,2,0)\left( {0,2,0} \right)
D. (1,2,0)\left( {1,2,0} \right)

Explanation

Solution

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 x−6−6=y+44=z−4−8\dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}} and x+12=y+24=z+3−2\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}}
Let the line L1=a{L_1} = a
⇒x−6−6=y+44=z−4−8=a\Rightarrow \dfrac{{x - 6}}{{ - 6}} = \dfrac{{y + 4}}{4} = \dfrac{{z - 4}}{{ - 8}} = a
Now equating x term with a
⇒x−6−6=a\Rightarrow \dfrac{{x - 6}}{{ - 6}} = a
On cross multiplication
⇒x−6=−6a\Rightarrow x - 6 = - 6a
Adding 6 on both sides
⇒x=−6a+6\Rightarrow x = - 6a + 6
Similarly, equating y term with a
⇒y+44=a\Rightarrow \dfrac{{y + 4}}{4} = a
On cross multiplication
⇒y+4=4a\Rightarrow y + 4 = 4a
Adding -4 on both sides
⇒y=4a−4\Rightarrow y = 4a - 4
Similarly, equating z term with a
⇒z−4−8=a\Rightarrow \dfrac{{z - 4}}{{ - 8}} = a
On cross multiplication
⇒z−4=−8a\Rightarrow z - 4 = - 8a
Adding 4 on both sides
⇒z=−8a+4\Rightarrow z = - 8a + 4
Now we have a general point on line 1 i.e. (−6a+6,4a−4,−8a+4)\left( { - 6a + 6,4a - 4, - 8a + 4} \right)
Similarly, finding a general point on line 2 i.e. x+12=y+24=z+3−2\dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}}
Let L2=b{L_2} = b
⇒x+12=y+24=z+3−2=b\Rightarrow \dfrac{{x + 1}}{2} = \dfrac{{y + 2}}{4} = \dfrac{{z + 3}}{{ - 2}} = b
Now equating x term with b
⇒x+12=b\Rightarrow \dfrac{{x + 1}}{2} = b
On cross multiplication
⇒x+1=2b\Rightarrow x + 1 = 2b
Adding -1 on both sides
⇒x=2b−1\Rightarrow x = 2b - 1
Similarly, equating y term with b
⇒y+24=b\Rightarrow \dfrac{{y + 2}}{4} = b
On cross multiplication
⇒y+2=4b\Rightarrow y + 2 = 4b
Adding -2 on both sides
⇒y=4b−2\Rightarrow y = 4b - 2
Similarly, equating z term with b
⇒z+3−2=b\Rightarrow \dfrac{{z + 3}}{{ - 2}} = b
On cross multiplication
⇒z+3=−2b\Rightarrow z + 3 = - 2b
Adding -3 on both sides
⇒z=−2b−3\Rightarrow z = - 2b - 3
Now we have a general point on line1 i.e. (2b−1,4b−2,−2b−3)\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.
⇒(−6a+6,4a−4,−8a+4)=(2b−1,4b−2,−2b−3)\Rightarrow \left( { - 6a + 6,4a - 4, - 8a + 4} \right) = \left( {2b - 1,4b - 2, - 2b - 3} \right)
Now equating the x-terms
⇒−6a+6=2b−1\Rightarrow - 6a + 6 = 2b - 1
Solving by combining the like terms
⇒−6a+6+1=2b\Rightarrow - 6a + 6 + 1 = 2b
⇒2b=7−6a..........(i)\Rightarrow 2b = 7 - 6a..........(i)
Similarly equating the y-terms
⇒4a−4=4b−2\Rightarrow 4a - 4 = 4b - 2
Dividing the whole equation by 2
⇒2a−2=2b−1\Rightarrow 2a - 2 = 2b - 1
Solving by combining the like terms
⇒2b=2a−1............(ii)\Rightarrow 2b = 2a - 1............(ii)
By equating the value of 2b from equation(i) and equation(ii)
⇒7−6a=2a−1\Rightarrow 7 - 6a = 2a - 1
Solving by combining the like terms
⇒8=8a\Rightarrow 8 = 8a
Dividing both sides by 8 we get,
∴a=1\therefore a = 1
Substituting the value of ‘a’ in equation(i)
⇒2b=7−6(1)\Rightarrow 2b = 7 - 6(1)
⇒2b=1\Rightarrow 2b = 1
Dividing both sides by 2
⇒b=12\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 (0,0,−4)\left( {0,0, - 4} \right)
Therefore the intersection point of both the lines is (0,0,−4)\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
⇒−8a+4=−2b−3\Rightarrow - 8a + 4 = - 2b - 3
Solving by combining the like terms
⇒8a−4−3=2b\Rightarrow 8a - 4 - 3 = 2b
⇒2b=8a−7\Rightarrow 2b = 8a - 7
Substituting the value of 2b from equation(i)
⇒7−6a=8a−7\Rightarrow 7 - 6a = 8a - 7
⇒14=14a\Rightarrow 14 = 14a
Dividing both sides by 14
∴a=1\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.