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Question
Solve the following :
A plane makes non zero intercepts a, b, c on the coordinate axes. Show that the vector equation of the plane is `bar"r".(bchat"i" + cahat"j" + abhat"k")` = abc.
Solution
A plane makes intercepts a, b, c on the coordinate axes.
Let, the plane intercept the coordinate axis at points A, B and C respectively.
A (a, 0, 0)
B (0, b, 0)
C (0, 0, C)
Let, `bar"a", bar"b", bar"c"` be the positions vectors of points A, B, and C respectively.
`bar"a" = "a"hat"i", bar"b" = "b"hat"j", bar"c" = "c"hat"k"`
⇒ `bar"AB" = bar"b" - bar"a" = "b"hat"j" - "a"hat"i" = -"a"hat"i" + "b"hat"j"`
and `bar"AC" = bar"c" - bar"a" = "c"hat"k" - "a"hat"i" = - "a"hat"i" + "c"hat"k"`
∴ `bar"AB" xx bar"AC" = |(hat"i",hat"j",hat"k"),(-"a","b",0),(-"a",0,"c")|`
= `"bc"hat"i" + "ac"hat"j" + "ab"hat"k"`
Vector equations of the plane are
`bar"r".(bar"AB" xx bar"AC") = bar"a".(bar"AB" xx bar"AC")`
`bar"r".("bc"hat"i" + "ac"hat"j" + "ab"hat"k") = "a"hat"i".("bc"hat"i" + "ac"hat"j" + "ab"hat"k")`
`bar"r".("bc"hat"i" + "ca"hat"j" + "ab"hat"k")` = abc
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