Question

If lines `(x−1)/2=(y+1)/3=(z−1)/4 and  (x−3)/1=(y−k)/2=z/1` intersect, then find the value of k and hence find the equation of the plane containing these lines.

Answer 1

he coordinates of any point on the first line are given by

`(x−1)/2=(y+1)/3=(z−1)/4=lambda`

i.e

x=2λ+1y=3λ−1z=4λ+1

Thus, the coordinates of any point on this line are (2λ+1, 3λ−1, 4λ+1).

The coordinates of any point on the second line are given by

`(x−3)/1=(y−k)/2=z/1=μ`

i.e.

x=μ+3

y=2μ+k

z=μ

Thus, the coordinates of any point on this line are (μ+3, 2μ+k, μ).

If these two lines intersect each other, then

2λ+1=μ+3, 3λ−1=2μ+k, 4λ+1=μ

⇒2λ−μ=2, 3λ−2μ=k+1, 4λ−μ=−1

Solving 2λ−μ=2 and 4λ−μ=−1, we get

λ=−3/2 and μ=−5

By substituting the values λ=−3/2 and μ=−5 in 3λ−2μ=k+1, we get

k=`9/2`

Also, we have

`vecb_1=2hati+3hatj+4hatk and vecb_2=hati+2hatj+hatk`

Now, the required plane contains both the given lines.

So, it passes through a point `veca` (1, −1, 1) and perpendicular vector `vecn` , given by

`vecN=vecb_1xxvecb_2`

`therefore vecN=[[hati,hatj,hatk],[2,3,4],[1,2,1]]`

`=>vecN=-5hati+2hatj+hatk`

Therefore, the equation of plane passing through veca and perpendicular to vecN is given by

`(vecr-veca).vecN=0`

`=>[vecr-(hati-hatj+hatk)].(-5hati+2hatj+hatk)=0`

`=>vecr.(-5hati+2hatj+hatk)=(hati-hatj+hatk).(-5hati+2hatj+hatk)`

`=>vecr.(-5hati+2hatj+hatk)=-6`

or

5x − 2y − z − 6 = 0