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Question
Find the vector equation of the plane which makes intercepts 1, 1, 1 on the co-ordinates axes.
Solution
The vector equation of the plane passing through `"A"(bara), "B"(barb), "C"(barc)`, where A, B, C are non-collinear is `bar"r".(bar"AB" xx bar"AC") = bar"a".(bar"AB" xx bar"AC")` ...(1)
The required plane makes intercepts 1, 1, 1 on the coordinate axes.
∴ It passes through the three non-collinear points
A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)
∴ `bar"a" = hat"i", bar"b" = hat"j", bar"c" = hat"k"`
`bar"AB" = bar"b" - bar"a" = hat"j" - hat"i" = -hat"i" + hat"j"`
∴ `bar"AC" = bar"c" - bar"a" = hat"k" - hat"i" = -hat"i" + hat"k"`
∴ `bar"AB" xx bar"AC" = |(hat"i", hat"j", hat"k"),(-1, 1, 0),(-1, 0, 1)|`
= `(1 - 0)hat"i" - (- 1 + 0)hat"j" + (0 + 1)hat"k"`
= `hat"i" + hat"j" + hat"k"`
Also, `bar"a".(bar"AB" xx bar"AC")` = `hat"i".(hat"i" + hat"j" + hat"k")`
= 1 × 1 + 0 × 1 + 0 × 1
= 1
∴ From (1), the vector equation of the required plane is `bar"r".(hat"i" + hat"j" + hat"k")` = 1.
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