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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?
Solution
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.
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