Question

Find the vector equation of the plane passing through points A(1, 1, 2), B(0, 2, 3) and C(4, 5, 6).

Answer 1

The vector equation of the plane passing through three non-collinear points `A(bara), B(barb)` and `C(barc)` is `barr.(bar(AB) xx bar(AC)) = bara.(bar(AB) xx bar(AC))`  ...(1)

Here, `bara = hati + hatj + 2hatk, barb = 2hatj + 3hatk, barc = 4hati + 5hatj + 6hatk`

∴ `bar(AB) = barb - bara`

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

= `-hati + hatj + hatk`

∴ `bar(AC) = barc - bara`

= `(4hati + 5hatj + 6hatk) - (hati + hatj + 2hatk)`

= `3hati + 4hatj + 4hatk`

∴ `bar(AB) xx bar(AC) = |(hati,hatj,hatk), (-1,1,1),(3,4,4)|`

= `(4 - 4)hati - (-4 - 3)hatj + (-4 - 3)hatk`

= `0hati - 7hatj + 7hatk`

Now, `bara.(bar(AB) xx bar(AC))`

= `(hati + hatj + 2hatk).(0hati - 7hatj + 7hatk)`

= (1)(0) − (1)(7) + (2)(7)

= 0 − 7 + 14

∴ From (1), the vector equation of the required plane is `barr.(0hati - 7hatj + 14hatk)` = −7.