Question

Find the equation of the plane through the line x/3-1/3 = y/2-4/2 = z/-2-4/-2 and parallel to the line x/2+1/2 = 1/4-y/4 = z/1+2/1. Hence, find the shortest distance between the lines.

Answer 1

Answer:

6x+7y+16z = 98  Distance = 18.466

Step-by-step explanation:

Plane passes through the line :

$\frac{x-1}{3}=\frac{y-4}{2} =\frac{z-4}{-2}$ = K

x = 1+3K , y = 4+2k, z = 4-2K

So the direction ratios are a = (3,2,-2)

Plane is parallel to the line:

$\frac{x+1}{2}=\frac{y-1}{-4} =\frac{z+2}{1}$ = K

x = 2K-1, y = -4K+1 , x= K-2

Direction ratios are : b = (2,-4,1)

Now the normal of the plane must be perpendicular to the direction ratios of both the lines, So will find the direction rations by: a×b

N = $\left[\begin{array}{ccc}i&j&k\\3&2&-2\\2&-4&1\end{array}\right]$

Ai+ Bj +Ck   = -6i-7j-16k

Now, the plane passes through(x_0,y_0,z_0) =  (1,4,4)

Equation of plane is given by:

A(x-x_0)+B(y-y_0)+C(z-z_0) = 0

-6(x-1) -7(y-4) -16(z-4) = 0

-6x-7y-16z+98 = 0

Required equation of plane is 6x+7y+16z = 98

Shortest Distance between the lines is the length of the perpendicular between them

Perpendicular between the line is the normal vector

Shortest distance = √(-6)^{2}+(-7)^{2}+(-16)^{2}

                             = √341

                             = 18.466