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(bar a), B(bar b)` and `C(bar c)` is `bar r * (bar(AB) xx bar(AC)) = bar a * (bar(AB) xx bar(AC))`     ...(1)

Here, `bar a = hat i + hat j + 2 hat k, bar b = 2 hat j + 3 hat k, bar c = 4 hat i + 5 hat j + 6 hat k`

∴ `bar(AB) = bar b - bar a`

= `(2 hat j + 3 hat k) - (hat i + hat j + 2 hat k)`

= `-hat i + hat j + hat k`

∴ `bar(AC) = bar c - bar a`

= `(4 hat i + 5 hat j + 6 hat k) - (hat i + hat j + 2 hat k)`

= `3 hat i + 4 hat j + 4 hat k`

∴ `bar(AB) xx bar(AC) = |(hat i, hat j, hat k), (-1, 1, 1),(3, 4, 4)|`

= `(4 - 4)hat i - (-4 - 3)hat j + (-4 - 3)hat k`

= `0 hat i + 7 hat j - 7 hat k`   ...(i)

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

= `(hat i + hat j + 2 hat k) * (0 hat i - 7 hat j + 7 hat k)`

= (1)(0) + (1)(−7) + (2)(7)

= 0 − 7 + 14

= 7

∴ From (i), the vector equation of the required plane is `r(0 hat i + 7 hat j - 7 hat k)` = 7.