Question

The two lines `(x - 1)/3` = −y, z + 1 = 0 and `(-x)/2 = (y + 1)/2` = z + 2 intersect at a point whose y-coordinate is 1. Find the coordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.

Answer 1

Given line `(x - 1)/3` = −y, z + 1 = 0

Writing in normal form.

`(x - 1)/3 = y/(-1) = (z + 1)/0`    .....(1)

Second line:

`(-x)/2 = (y + 1)/2 = (z + 2)/1`   .....(2)

To find the point of intersection, we find the general point of both lines and equate them.

If the point is in line (1):

`(x - 1)/3 = y/(-1) = (z + 1)/0`

General point is

`(x - 1)/3 = y/(-1) = (z + 1)/0 = p`

So, x = 3p + 1

y = −p

z = −1

If the point is in line (2):

`(-x)/2 = (y + 1)/2 = (z + 2)/1`

General point is

`(-x)/2 = (y + 1)/2 = (z + 2)/1 = q`

So, x = −2q

y = 2q − 1

z = q − 2

Equating z-coordinates:

−1 = q − 2

−1 + 2 = q

1 = q

q = 1

Putting q = 1, in x, y, z.

⇒ x = −2q

= −2 × 1

= −2

⇒ y = 2q − 1

= 2 × 1 − 1

= 2 − 1

= 1

⇒ z = q − 2

= 1 − 2

= −1

Thus, the point of intersection is (−2, 1, −1).

We need to find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.

The vector equation of a line passing through a point with position vector `veca` and parallel to a vector `vecb` is

`vecr = veca + λvecb`

The line passes through (−2, 1, −1).

So, `veca = -2hati + hatj - hatk`

Given, the line is perpendicular to both lines.

∴ `vecb` is perpendicular to both lines.

We know that `vecx xx vecy` is perpendicular to both `vecx` and `vecy`.

So, `vecb` is the cross product of both lines `(x - 1)/3 = y/(-1) = (z + 1)/0` and `(-x)/2 = (y + 1)/2 = (z + 2)/1`.

Required normal = `|(hati,hatj,hatk),(3,-1,0),(-2,2,1)|`

= `hati [-1(1) - 2(0)] - hatj[3(1) - (-2)(0)] + hatk[3(2) - (-2)(-)]`

= `hati(-1) - hatj(3) + hatk(6 - 2)`

= `-hati - 3hatj + 4hatk`

Thus, `vecb = -hati-3hatj + 4hatk`

Now, putting the value of `veca` and `vecb` in the formula:

`vecr = veca + λvecb`

∴ `vecr = (-2hati + hatj - hatk) + λ (-hati -3hatj + 4hatk)`

Therefore, the equation of the line is `(-2hati + hatj - hatk) + λ (-hati -3hatj + 4hatk)`.