Question

Find a unit vector perpendicular to the plane containing the point (a, 0, 0), (0, b, 0) and (0, 0, c). What is the area of the triangle with these vertices?

Answer 1

The position vectors `bar"p", bar"q", bar"r"` of the points A(a, 0, 0), B(0, b, 0), C(0, 0, c) are

`bar"p" = "a"hat"i", bar"q" = "b"hat"j", bar"r" = "c"hat"k"`

`bar"AB" = bar"q" - bar"p" = "b"hat"j" - "a"hat"i" = - "a"hat"j" + "b"hat"j"`

`bar"BC" = bar"r" - bar"q" = "c"hat"k" - "b"hat"j" = - "b"hat"j" + "c"hat"k"`

`bar"AB" xx bar"BC" = |(hat"i",hat"j",hat"k"),(-"a","b",0),(0,-"b","c")|`

`= ("bc" - 0)hat"i" - (- "ac" - 0)hat"j" + ("ab" - 0)hat"k"`

`= "bc"hat"i" + "ac"hat"j" + "ab"hat"k"`

`|bar"AB" xx bar"BC"| = sqrt(("bc")^2 + ("ac")^2 + ("ab")^2)`

`= sqrt("b"^2"c"^2 + "a"^2"c"^2 + "a"^2"b"^2)`

`bar"AB" xx bar"BC"` is perpendicular to the plane containing A, B, C.

∴ the required unit vector

`= (bar"AB" xx bar"BC")/(|bar"AB" xx bar"BC"|) = ("bc"hat"i" + "ca"hat"j" + "ab"hat"k")/sqrt("b"^2"c"^2 + "c"^2"a"^2 + "a"^2"b"^2)`

Area of Δ ABC = `1/2 |bar"AB" xx bar"BC"|`

`= 1/2 sqrt("b"^2"c"^2 + "a"^2"c"^2 + "a"^2"b"^2)` sq.units.