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Question: If the lines \(\dfrac{{x - 1}}{2} = \dfrac{{y + 1}}{3} = \dfrac{{z - 1}}{4}\) and \(\dfrac{{x - 3}}{...

If the lines x−12=y+13=z−14\dfrac{{x - 1}}{2} = \dfrac{{y + 1}}{3} = \dfrac{{z - 1}}{4} and x−31=y−k2=z1\dfrac{{x - 3}}{1} = \dfrac{{y - k}}{2} = \dfrac{z}{1} intersect, then k is equal to:
(a)−1\left( a \right) - 1
(b)29\left( b \right)\dfrac{2}{9}
(c)92\left( c \right)\dfrac{9}{2}
(d)0\left( d \right)0

Explanation

Solution

In this particular question use the concept that equate the given equation of lines with any different variables and calculate the value of x, y and z for both of the lines and equate x to x, y to y and z to z, so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given lines
x−12=y+13=z−14\dfrac{{x - 1}}{2} = \dfrac{{y + 1}}{3} = \dfrac{{z - 1}}{4}
x−31=y−k2=z1\dfrac{{x - 3}}{1} = \dfrac{{y - k}}{2} = \dfrac{z}{1}
Now let, x−12=y+13=z−14\dfrac{{x - 1}}{2} = \dfrac{{y + 1}}{3} = \dfrac{{z - 1}}{4} = p................ (1)
And let, x−31=y−k2=z1\dfrac{{x - 3}}{1} = \dfrac{{y - k}}{2} = \dfrac{z}{1} = q................... (2)
Now solve equation (1) for x, y and z we have,
⇒x−12=p\Rightarrow \dfrac{{x - 1}}{2} = p, y+13=p\dfrac{{y + 1}}{3} = p, z−14=p\dfrac{{z - 1}}{4} = p
⇒x=2p+1\Rightarrow x = 2p + 1, y = 3p – 1, z = 4p + 1
Now solve equation (2) for x, y and z we have,
⇒x−31=q\Rightarrow \dfrac{{x - 3}}{1} = q, y−k2=q\dfrac{{y - k}}{2} = q, z1=q\dfrac{z}{1} = q
⇒x=q+3\Rightarrow x = q + 3, y = 2q + k, z = q.
Now it is given that both of the lines are intersect each other so, equate x to x, y to y and z to z of both the equations we have,
⇒2p+1=q+3\Rightarrow 2p + 1 = q + 3.............. (3)
⇒3p−1=2q+k\Rightarrow 3p - 1 = 2q + k.............. (4)
⇒4p+1=q\Rightarrow 4p + 1 = q............... (5)
Now substitute the value of q from equation (5) in equation (3) we have,
⇒2p+1=4p+1+3\Rightarrow 2p + 1 = 4p + 1 + 3
⇒0=2p+3\Rightarrow 0 = 2p + 3
⇒p=−32\Rightarrow p = \dfrac{{ - 3}}{2}
Now substitute the value of q from equation (5) in equation (4) we have,
⇒3p−1=2(4p+1)+k\Rightarrow 3p - 1 = 2\left( {4p + 1} \right) + k
⇒3p−1=8p+2+k\Rightarrow 3p - 1 = 8p + 2 + k
⇒0=5p+3+k\Rightarrow 0 = 5p + 3 + k
Now substitute the value of p in the above equation we have,
⇒0=5(−32)+3+k\Rightarrow 0 = 5\left( {\dfrac{{ - 3}}{2}} \right) + 3 + k
⇒k=5(32)−3=15−62=92\Rightarrow k = 5\left( {\dfrac{3}{2}} \right) - 3 = \dfrac{{15 - 6}}{2} = \dfrac{9}{2}
So this is the required value of k.
Hence option (c) is the correct answer.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall that if both the equations intersect each other than the solutions of the equations i.e. the values of x, y and z should be equal to the values of x, y and z of the other equation respectively so simply equate them as above and simplify we will get the required value of k.