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Find the equation of the line of interse...

Find the equation of the line of intersection of the planes `4x + 4y - 5z = 12, 8x + 12y - 13z = 32` in the symmetric form.

A

`(x-1)/(2)=(y+2)/(-3)=(z)/(4)`

B

`(x-1)/(2)=(y-2)/(3)=(z)/(4)`

C

`(x)/(2)=(y+1)/(3)=(z-2)/(4)`

D

`(x)/(2)=(y)/(3)=(z-2)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given that equation of planes are,
`4x+4y-5z=12` . . .(i)
and `8x+12y-13z=32` . . . (ii)
Let direction ratios of the line are (l,m,n)
`:.` Eqs. (i) and (ii) becomes
`4l+4m-5n=0` . . . (iii)
`and " "8l+12m-13n=0` . .. . (iv)
`rArr" "(l)/(-52+60)=(-m)/(-52+40)=(n)/(48-32)`
`rArr" "(l)/(8)=(-m)/(-12)=(n)/(16)rArr(l)/(2)=(m)/(3)=(n)/(4)`
Now, we take intersection point with z=0 is given by
`4x+4y=12` . . . (v)
`and" "8x+12y=32` . . . (vi)
On solving Eqs. (v) and (vi), we get (1,2,0)
`:.` Required line is `(x-1)/(2)=(y-2)/(3)=(z)/(4)`
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