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Question
Solve the following :
Find the vector equation of the plane passing through the origin and containing the line `bar"r" = (hat"i" + 4hat"j" + hat"k") + lambda(hat"i" + 2hat"j" + hat"k")`.
Solution
The vector equation of the plane passing through `"A"(bara)` and perpendicular to the vector `bar"n"` is `bar"r".bar"n" = bar"a".bar"n"` ...(1)
We can take `bar"a" = bar"0"` since the plane passes through the origin.
The point M with position vector `bar"m" = hat"i" + 4hat"j" + hat"k"` lies on the line and hence it lies on the plane.
∴ `bar"OM" = bar"m" = hat"i" + 4hat"j" + hat"k"` lies on the plane.
The plane contains the given line which is parallel to `bar"b" = hat"i" + 2hat"j" + hat"k"`.
Let `bar"n"` be normal to the plane. Then `bar"n"` is perpendicular to `bar"OM"` as well as `bar"b"`
∴ `bar"n" = bar"OM" xx bar"b" = |(hat"i" ,,),(1, 4, 1),(1, 2, 1)|`
= `(4 - 2)hat"i" - (1 - 1)hat"j" + (2 - 4)hat"k"`
= `2hat"i" - 2hat"k"`
∴ from (1), the vector equation of the required plane is
`bar"r".(2hat"i" - 2hat"k") = bar"0".bar"n"` = 0
i.e. `bar"r".(hat"i" - hat"k")` = 0.
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