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Numerical Ability
Co-ordinate geometry
Determine the distance between the x-intercept and the z-intercept of the plane whose equation is 2x+9y-3z=18...
Read Solution (Total 8)
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- putting y=z=0
x=9
putting x=y=0
z=-6
distance=(x^2+z^2)^0.5=10.8unit - 12 years agoHelpfull: Yes(53) No(1)
- (9^2+6^2)^0.5
- 12 years agoHelpfull: Yes(4) No(1)
- (2x+9y-3z)/18=1
x/9+y/2+(-z)/6=1
den x-intercept will b 9
z-intercept will b -6
so distance will b 9-(-6)=15. - 12 years agoHelpfull: Yes(3) No(3)
- (2x+9y-3z)/18=1
x/9+y/2+(-z)/6=1
den x-intercept will b 9
z-intercept will b -6
here these 2 forms a right angled triangle
so......distance=sqrt(9^2+6^2)
- 12 years agoHelpfull: Yes(3) No(0)
- (2x+9y-3z)/18=1
x/9+y/2+(-z)/6=1
den x-intercept will b 9
z-intercept will b -6
so distance will b 9-(-6)=15.
- 12 years agoHelpfull: Yes(2) No(10)
- x=9 and z=-6 den by using distance formula sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2) ans is 3sqrt13.
- 12 years agoHelpfull: Yes(1) No(4)
- for ax+by+cz=d
The formula for x and z intercept is
d=sqrt[(d/a)^2+(d/c)^2] - 12 years agoHelpfull: Yes(1) No(0)
- d=sqrt[(d/a)^2+(d/c)^2]
d=18,a=2,c=3
by solving we get 10.8 - 12 years agoHelpfull: Yes(0) No(0)
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