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If the plane $\mathrm{ P}$ passes through the intersection of two mutually perpendicular planes $\mathrm{ 2 x+k y-5 z=1}$ and $\mathrm{ 3 k x-k y+z=5,}$ $\mathrm{ k<3}$ and intercepts a unit length on positive $\mathrm{ x}$ -axis, then the intercept made by the plane $\mathrm{ P}$ on the $\mathrm{ y}$ -axis is
Option: 1
$\frac{1}{11}$

Option: 2
$\frac{5}{11}$

Option: 3
$6$

Option: 4
$7$

Answers (1)

best_answer

Two given plane mutually perpendicular

$\mathrm{2(3 k)+k(-k)+(-5) 1=0 }\\$

$\mathrm{k=1,5}$

$\mathrm{but\; k<3 \quad so\; k=1}$

Plane passing through these phase is

$\mathrm{2 x+4=5 z-1+\lambda(3 x-y+z-5)=0} \\$

$\mathrm{\frac{x}{\frac{5 \lambda+1}{2+3 \lambda}}+\frac{y}{\frac{5 \lambda+1}{1-\lambda}}+\frac{z}{\frac{5 \lambda+1}{\lambda-5}}=1 }\\$

$\mathrm{\text { given } \frac{5 \lambda+1}{2+3 \lambda}=1 \Rightarrow \lambda=\frac{1}{2}}$

$\mathrm{So\; intercept\; on\; y-axis =\frac{5 \lambda+1}{1-\lambda}=7}$

Hence correct option is 4

Posted by

Gautam harsolia

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