Question

A plane passes through three points A, B and C with position vectors `hati + hatj, hatj + hatk` and `hatk + hati` respectively. The equation of the line passing through the point P with position vector `hati + 2hatj + 2hatk` and normal to the plane is ______.

Options

• `vecr = (hati + 2hatj + 2hatk) + λ(hati + hatj + hatk), λ ∈ R`

• `vecr = (hati + hatj + hatk) + λ(hati + 2hatj + 2hatk), λ ∈ R`

• `vecr * (hati - hatj - hatk) = hati + 2hatj + 2hatk`

• x – 1 = y = z

Answer 1

A plane passes through three points A, B and C with position vectors `hati + hatj, hatj + hatk` and `hatk + hati` respectively. The equation of the line passing through the point P with position vector `hati + 2hatj + 2hatk` and normal to the plane is `underlinebb(vecr = (hati + 2hatj + 2hatk) + λ(hati + hatj + hatk), λ ∈ R)`.

Explanation:

Let the position vectors of the points be:

`vecA = hati + hatj, vecB = hatj + hatk, vecC = hati + hatk`

Find direction vectors of the plane:

`vec(AB) = vecB - vecA`

= `(hatj + hatk) - (hati + hatj)`

= `-hati + hatk`

`vec(AC) = vecC - vecA`

= `(hati + hatk) - (hati + hatj)`

= `-hatj + hatk`

The normal vector `vecn` to the plane containing points A, B, C is:

`vecn = vec(AB) xx vec(AC)`

= `|(hati, hatj, hatk),(-1, 0, 1),(0, -1, 1)|`

= `hati(0 xx 1 - 1 xx (-1)) - hatj((-1) xx 1 - 1 xx 0) + hatk((-1) xx (-1) - 0 xx 0)`

= `hati + hatj + hatk`

The vector equation of a line passing through a point with position vector `vec(r_0)` and having direction vector `vecn` is:

`vecr = vec(r_0) + λvecn`

Given the point (1,2,2), its position vector is:

`vec(r_0) = hati + 2hatj + 2hatk`

Substituting:

`vecr = (hati + 2hatj + 2hatk) + λ(hati + hatj + hatk)`