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Find the shortest distance between the z-axis and the line, `x+y+2z-3=0,2x+3y+4z-4=0.`

Text Solution

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We have line of intersection of planes `x+y+2z-3=0, 2x+3y+4z-4=0`.
Vector along line of intersection is
`|{:(hati,,hatj,,hatk),(1,,1,,2),(2,,3,,4):}|=-2hati+hatk`
For any point on this line let z=0, so we get x=5 and y =-2
Therefore, equation of line in symmetric form is
`" "(x-5)/(-2)=(y+2)/(0)=(z)/(1)`
Equation of z-axis is `(x)/(0)=(y)/(0)=(z)/(1)`
Hence, the shortest distance between these lines is
`" "|(|{:(5-0,,-2-0,,0-0),(-2,,0,,1),(0,,0,,1):}|)/(|{:(hati,,hatj,,hatk),(-2,,0,,1),(0,,0,,1):}|)|=2`
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