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Question
Find the vector equation of the plane passing through the points A(1, -2, 1), B(2, -1, -3) and C(0, 1, 5).
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Solution
The vector equation of the plane passing through three non-collinear points `"A"(bara), "B"(barb) and "C"(barc) "is" bar"r".(bar"AB" xx bar"AC") = bar"a".(bar"AB" xx bar"AC")` ...(1)
Here, `bar"a" = hat"i" - 2hat"j" + hat"k", bar"b" = 2hat"i" - hat"j" - 3hat"k", bar"c" = hat"j" + 5hat"k"`
∴ `bar"AB" = bar"b" - bar"a" = (2hat"i" - hat"j" - 3hat"k") - (hat"i" - 2hat"j" + hat"k")`
= `hat"i" + hat"j" - 4hat"k"`
`bar"AC" = bar"c" - bar"a" = (hat"j" + 5hat"k") - (hat"i" - 2hat"j" + hat"k")`
= `hat"i" + 3hat"j" + 4hat"k"`
∴ `bar"AB" xx bar"AC" = |(hati hatj hatk), (1 1-4), (-1 3 4 )|`
= `(4 + 12)hat"i" - (4 - 4)hat"j" + (3 + 1)hat"k"`
= `16hat"i" + 4hat"k"`
Now, `bar"a".(bar"AB" xx bar"AC") = (hat"i" - 2hat"j" + hat"k").(16hat"i" + 4hat"k")`
= (1)(16) + (– 2)(0) + (1)(4) = 20
∴ from(1), the vector equation of the required plane is `bar"r".(16hat"i" + 4hat"k")` = 20.
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