Shortest distance between line `2x+3y+4z-4=0=x+y+2z-3` and z-axis is -

1.	Shortest distance between line `2x+3y+4z-4=0=x+y+2z-3` and z-axis is -A. 1B. 2C. 4D. 3
Answer» Correct Answer - B We have `x+y+2z-3=0` and `2x+3y+4z-4=0` Let a,b,c be the direction ratios of the line, then line lies on the both planes. `therefore a+b+2c=0 and 2a+3b+4c=0` On solving these equations by cross-multiplication method. `a/(4-6)=(b)/(4-4)=(c)/(3-2) Rightarrow (a)/(-2)=(b)/(0)=c/1` `therefore` Direction ratios of the line is (-2,0,1) and direction ratios of Z-axis (0,0,1) Hence, distance between Z-axis to the given line is `sqrt((-2-0)(2)+(0)^(2)+(1-1)^(2))=sqrt4=2`

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Shortest distance between line `2x+3y+4z-4=0=x+y+2z-3` and z-axis is -A. 1B. 2C. 4D. 3

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