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  • Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 unit

22.   Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

              (A) 2 units         (B) 4 units           (C) 8 units       (D) \frac{2}{\sqrt{29}}unit

Answers (1)

best_answer

Given equations are 
2x+3y+4z= 4 \ \ \ \ \ \ \ \ \ -(i)
and 
4x+6y+8z= 12\\ 2(2x+3y+4z)= 12\\ 2x+3y+4z = 6 \ \ \ \ \ \ \ \ \ \ -(ii)
Now, it is clear from  equation (i) and (ii) that given planes are parallel
We know that the distance between two parallel planes ax+by +cz = d_1 \ and \ ax+by +cz = d_2  is given by 
D= \left | \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}} \right |
Put the values in this equation
we will get,
D= \left | \frac{6-4}{\sqrt{2^2+3^2+4^2}} \right |
D= \left | \frac{2}{\sqrt{4+9+16}} \right |= \left | \frac{2}{\sqrt{29}} \right |
Therefore, the correct answer is (D) 
 

Posted by

Gautam harsolia

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