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प्रश्न
Find the cartesian equation of the plane `bar"r" = (5hat"i" - 2hat"j" - 3hat"k") + lambda(hat"i" + hat"j" + hat"k") + mu(hat"i" - 2hat"j" + 3hat"k")`.
उत्तर
The equation `bar"r" = bar"a" + lambdabar"b" + mubar"c"` represents a plane passing through a point having position vector `bar"a"` and parallel to vectors `bar"b" and bar"c"`.
Here,
`bar"a" = 5hat"i" - 2hat"j" - 3hat"k",`
`bar"b" = hat"i" + hat"j" + hat"k"`,
`bar"c" = hat"i" - 2hat"j" + 3hat"k"`
∴ `bar"b" xx bar"c" = |(hat"i", hat"j", hat"k"),(1, 1, 1),(1, -2, 3)|`
= `(3 + 2)hat"i" - (3 - 1)hat"j" + (-2 - 1)hat"k"`
= `5hat"i" - 2hat"j" - 3hat"k"`
= `bar"a"`
Also,
`bar"a".(bar"b" xx bar"c") = bar"a".bar"a" = |bar"a"|^2`
= (5)2 + (– 2)2 + (3)2
= 38
The vector equation of the plane passing through A`(bara)` and parallel to `bar"b" and bar"c"` is
`bar"r".(bar"b" xx bar"c") = bar"a".(bar"b" xx bar"c")`
∴ The vector equation of the given plane is
`bar"r".(5hat"i" - 2hat"j" - 3hat"k")` = 38
If `bar"r" = xhat"i" + yhat"j" + zhat"k"`, then this equation becomes
`(xhat"i" + yhat"j" + zhat"k").(5hat"i" - 2hat"j" - 3hat"k")` = 38
∴ 5x – 2y – 3z = 38.
This is the cartesian equation of the required plane.
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