Question

Find the distance of the Point 2 3 4 from the line

Answer 1

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.