Determine the distance between x-intercept and z-intercept of the plane where equation is 6x+8y-3z=72.

a)3 b)26.83 c)9 d)25.63

• x-intercept=(a,0,0) let
  so value of a=72/6
  =12
  z-intercept=(0,0,b) let
  value of b=72/(-3)
  =-24
  applying Pythagoras theorem:
  distance between x- and z-intercept=sqrt[sqr(12)+sqr(24)]
  =sqrt(144+576)
  =26.83
• 12 years agoHelpfull: Yes(19) No(0)
• let the equation : ax+by+cz=k
  
  if x-intercept put y=0 & z=0
  if y-intercept put y=0 & x=0
  if z-intercept put y=0 & x=0
  
  now for this question: x=72/6=12
  z=72/-3=-24
  y=72/8=9
  distance=sqrt((x^2)+(z^2))=sqrt(144+576)=26.832
  
• 12 years agoHelpfull: Yes(9) No(0)
• given 6x+8y-3z=72.dividing both sides by 72 we have (x/12)+(y/9)-(z/24)=1.
  So x-intercept=12 and z-intercept=24.differnce is 12
• 12 years agoHelpfull: Yes(1) No(9)