Latest TCS Aptitude Question SOLUTION: Determine the distance between x-intercept and z-intercept of the plane where equation is 6x+8y-3z=72. a)3 b)26.83

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(#M40003075) TCS Numerical Ability question Keep an EYE

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

Asked In TCS irfan (12 years ago)

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alka rathore  (12 years ago)

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

Is this solution Helpfull? Yes (19) | No   

Ashish yadav  (12 years ago)

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

Is this solution Helpfull? Yes (9) | No   

Subha  (12 years ago)

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

Is this solution Helpfull? Yes (1) | No (9)  

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