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Show that the lines `(x+1)/3=(y+3)/5=(z+5)/7 and (x−2)/1=(y−4)/3=(z−6)/5` intersect. Also find their point of intersection

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#### Solution

We have:

`(x+1)/3=(y+3)/5=(z+5)/7=λ (say)`

x=3λ−1, y=5λ−3 and z=7λ−5

So, the coordinates of a general point on this line are (3λ−1, 5λ−3, 7λ−5).

The equation of the second line is given below:

`(x−2)/1=(y−4)/3=(z−6)/5=μ (say)`

x=μ+2, y=3μ+4 and z=5μ+6

So, the coordinates of a general point on this line are (μ+2, 3μ+4, 5μ+6).

If the lines intersect, then they have a common point.

So, for some values of λ and μ, we must have:

3λ−1=μ+2, 5λ−3=3μ+4 and 7λ−5=5μ+6

⇒ 3λ−μ=3, 5λ−3μ=7 and 7λ−5μ=11

Solving the first two equations, 3λ−μ=3 and 5λ−3μ=7, we get:

λ=1/2 and μ=−3/2

Since λ=1/2 and μ=−3/2 satisfy the third equation, 7λ−5μ=11, the given lines intersect each other.

When λ=1/2 in (3λ−1, 5λ−3, 7λ−5), the coordinates of the required point of intersection are (1/2, −1/2, −3/2)

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