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Two lines \frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1} and  \frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}  intersect at the point R.

The reflection of R in the xy-plane has coordinates:

  • Option 1)

    (2,-4,7)

  • Option 2)

    (2,4,7)

  • Option 3)

    (-2,4,7)

  • Option 4)

    (2,-4,-7)

Answers (1)

best_answer

 

Image of a point -

Let L' be the image of point P\left ( \alpha ,\beta ,\gamma \right ) in the plane ax+by+cz+d=0

L' will be given by the formula 

\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}= \frac{-2\left ( a\alpha +b\beta +c\gamma +d \right )}{a^{2}+b^{2}+c^{2}}

-

Point  P_{1}  on  line 1  (\lambda +3,3\lambda -1,6-\lambda )

Point  P_{2}  on  line 2  (7\mu -5,2-6\mu ,4\mu+3 )

At point R    P_{1} = P_{2}

=>\lambda +3=7\mu -5....................(1)

=>3\lambda-1=2-6\mu..................(2)

=>6-\lambda=4\mu+3....................(3)

Solving any 2 equations out of (1),(2) and (3)

\lambda=-1 \: ,\: \mu=1

point R (2 , -4 , 7 )

Reflection of R in xy plane is ( 2, -4, -7 ).

 


Option 1)

(2,-4,7)

Option 2)

(2,4,7)

Option 3)

(-2,4,7)

Option 4)

(2,-4,-7)

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