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प्रश्न
Find the Cartesian and vector equation of the plane which makes intercepts 1, 1, 1 on the coordinate axes
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उत्तर
The plane makes intercepts 1, 1, 1 on the co-ordinate axes.
Let A(1, 0, 0), B(0, 1, 0), C(0, 0, 1) be the intersecting point of a plane with co-ordinate axes.
Vector equation of a plane passing through non-collinear points
`"A"(bar"a"),"B"(bar"b")` and `"C"(bar"c")` is `(bar"r" - bar"a")*(bar"b" - bar"a") xx (bar"c" - bar"a")` = 0
∴ `(bar"b" - bar"a") xx (bar"c" - bar"a") = |(hat"i", hat"j", hat"k"),(-1, 1, 0),(-1, 0, 1)|`
= `hat"i" + hat"j" + hat"k"`
∴ `(bar"r" - hat"i")*(hat"i" + hat"j" + hat"k")` = 0
∴ `bar"r"(hat"i" + ht"j" + hat"k") - hat"i"*(hat"i" + hat"j" + hat"k")` = 0
∴ `bar"r"*(hat"i" + hat"j" + hat"k") - 1` = 0 ......`[∵ hat"i"*hat"i" = 1, hat"i"*hat"j" = 0, hat"i"*hat"k" = 0]`
∴`bar"r"*(hat"i" + hat"j" + hat"k")` = 1 ......(i)
Putting `bar"r" = (xhat"i" + yhat"j" + zhat"k")` in (i), we get
`(xhat"i" + yhat"j" + zhat"k")*(hat"i" + hat"j" + hat"k")` = 1
∴ x + y + z = 1, which is the required Cartesian equation
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