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Show that the given lines, \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and \frac{x-4}{5}=\frac{y-1}{2}=z intersect.  

Also, find the point of intersection of the lines.

Answers (1)

We have the lines,



Let us denote these lines as L1and L2, such that



where \lambda ,\mu \epsilon \mathbb{R}

We must show that the lines L1and L2 intersect.

To show this, let us first find any point on line L1 and line L2

For L1:


\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda

\Rightarrow \frac{x-1}{2}=\lambda,\frac{y-2}{3}=\lambda,\frac{z-3}{4}=\lambda

We must find the values of x, y, and z. Therefore, let us take \Rightarrow \frac{x-1}{2}=\lambda

\Rightarrow x-1=2\lambda

\Rightarrow x=2\lambda+1

Take \frac{y-2}{3}=\lambda

\Rightarrow y-2=3\lambda

\Rightarrow y=3 \lambda+2

Take \frac{z-3}{4}=\lambda

\Rightarrow z-3=4\lambda

\Rightarrow z=4\lambda+3...(i)

Therefore, any point on L1 can be represented as (2\lambda + 1, 3\lambda + 2, 4\lambda + 3).


For L2:


\Rightarrow \frac{x-4}{5}=\frac{y-1}{2}=z=\mu

\Rightarrow \frac{x-4}{5}=\mu,\frac{y-1}{2}=\mu,z=\mu

We must find the values of x, y, and z. Therefore,

Take \frac{x-4}{5}=\mu

\Rightarrow x-4=5\mu

\Rightarrow x=5\mu+4

Take \frac{y-1}{2}=\mu

\Rightarrow y-1=2\mu

\Rightarrow y=2\mu+1

Take z=\mu

\Rightarrow z=\mu........(ii)

Hence, any point on line L? can be represented as (5μ + 4, 2μ + 1, μ).

If lines L1 and L2 intersect, then there exist λ and μ such that

\left ( 2\lambda+1,3\lambda+3,4\lambda+3 \right )\equiv \left ( 5\mu+4,2\mu+1,\mu \right )

\Rightarrow 2\lambda+1= 5\mu+4......(iii)



Substituting the value of μ from equation (v) into equation (iv),

3\lambda+2=2\left ( 4\lambda+3 \right )+1

\Rightarrow 3\lambda+2=8\lambda+6+1

\Rightarrow 3\lambda+2=8\lambda+7

\Rightarrow 8\lambda-3\lambda=2-7

\Rightarrow 5\lambda=-5

\Rightarrow \lambda=-\frac{5}{5}

\Rightarrow \lambda=-1

Putting this value of \lambda in eq (v),

4\left ( -1 \right )+3=\mu

\Rightarrow \mu=-4+3

\Rightarrow \mu=-1

To check, we can substitute the values of \lambda and \mu in equation (iii), giving us:

2(-1) + 1 = 5(-1) + 4

\Rightarrow -2 + 1 = -5 + 4

\Rightarrow -1 = -1

Therefore \lambda and \mu also satisfy equation (iii).

So, the z-coordinate from equation (i),

z=4\lambda +3

\Rightarrow z=4\left ( -1 \right )+3 \left [ \because \lambda=-1 \right ]

\Rightarrow z=-4+3

\Rightarrow z=-1

And the z-coordinate from equation (ii),


z=-1\left [ \because \mu=-1 \right ]
So, the lines intersect at the point

(5\mu + 4, 2\mu + 1, \mu) = (5(-1) + 4, 2(-1) + 1, -1).\\ Or, (5\mu + 4, 2\mu + 1, \mu) = (-5 + 4, -2 + 1, -1)\\ Or (5\mu + 4, 2\mu + 1, \mu) = (-1, -1, -1)

Therefore the lines intersect at the point (-1, -1, -1).

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