Find the distance of the point(-2,3,-4) from the line (x+2)/3=(2y+3)/4=(3z+4)/5 measured parallel to the plane r.(4i+12j-3k)=2

Asked by DIVYA GARG | 8th Jun, 2013, 12:05: AM

Expert Answer:

We can first find the equation of line which passes through point P(-2,3,-4) parallel to the plane r.(4i+12j-3k)=2  
Since the line needs to be parallel to the given plane, so its direction cosines would be (4, 12,-3)
Hence, equation of line passing through P and parallel to the given plane would be 
(x+2)/4 = (y-3)/12 = (z+4)/-3
Now we need to find the intersection of this line with the given line (x+2)/3=(2y+3)/4=(3z+4)/5
So let, (x+2)/4 = (y-3)/12 = (z+4)/-3 = m
x = -2+4m, y = 3+12m, z = -4-3m
Also, let (x+2)/3=(2y+3)/4=(3z+4)/5 = n
x = -2+3n, y = (-3+4n)/2, z = (-4+5n)/3
To find the common intersection point, we can equate respective x, y and z coordinates. 
Hence, -2+4m = -2+3n
4m = 3n
Also, 3+12m = (-3+4n)/2
6+24m = -3+4n
6+24(3n/4) = -3+4n
9+18n = 4n
9 = -14n
n = -9/14
So, m = -27/56
Hence, x = 55/14, y = 39/14
But z from both equations is different i.e. one z = 143/56 and other is equal to 101/42. 
So, these lines would actually never intersect. 

Answered by  | 9th Jun, 2013, 05:41: AM

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