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A variable plane passes through a fixed point (3, 2, 1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz-plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is :

  • Option 1)

    \frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1

     

     

     

  • Option 2)

    x+y+z=6

  • Option 3)

    \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}

  • Option 4)

    \frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1

 

Answers (1)

best_answer

As we learned

 

Intercept form -

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 1

- wherein

Let the plane cuts the coordinate axis at A\left ( a,0,0 \right ),B\left ( 0,b,0 \right )\, and \: C\left ( 0,0,c \right )

 

 

Let a,b,c be intercepts of plane on axis.

Equation of plane is \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

And point of intersection of plane is (a,b,c)

since (3,2,1) lies on variable plane 

\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1

Locus will be 

\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1

 


Option 1)

\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1

 

 

 

Option 2)

x+y+z=6

Option 3)

\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}

Option 4)

\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1

Posted by

Himanshu

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