Let the line \frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2} lie in the plane

x+3y-\alpha z+\beta = 0.Then \left ( \alpha ,\beta \right ) equals to

  • Option 1)

    (-6,7)

  • Option 2)

    (5,-15)

  • Option 3)

    (-5,5)

  • Option 4)

    (6,-17)

 

Answers (1)

As we learnt in 

Condition for line to be lie in plane -

\vec{b}\cdot \vec{n}= 0 and \vec{a}\cdot \vec{n}= d or

a_{1}a+b_{1}b+c_{1}c= 0  and a_{1}x_{1}+b_{1}y_{1}+c_{1}z_{1}+d= 0

-

 

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

(2,1,-2) lies in x+3y-\alpha z+ \beta=0

2+3+2\alpha + \beta=0

2\alpha +\beta =-5

Also, (\hat{i}+3\hat{j} -\hat{k}). (3\hat{i}-5\hat{j}+2\hat{k})=0

Since line 1 isperpendicular to normal plane, 3-15-2k=0

\Rightarrow    k= -6

 


Option 1)

(-6,7)

Correct answer

Option 2)

(5,-15)

Incorrect answer

Option 3)

(-5,5)

Incorrect answer

Option 4)

(6,-17)

Incorrect answer

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