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Question
Find the values of c so that for all real x, the vectors `"xc"hat"i" - 6hat"j" + 3hat"k"` and `"x"hat"i" + 2hat"j" + 2"cx"hat"k"` make an obtuse angle.
Solution
Let `bar"a" = "xc"hat"i" - 6hat"j" + 3hat"k"` and `bar"b" = "x"hat"i" + 2hat"j" + 2"cx"hat"k"`
Consider `bar"a".bar"b" = ("xc"hat"i" - 6hat"j" + 3hat"k").("x"hat"i" + 2hat"j" + 2"cx"hat"k")`
`= ("xc")("x") + (-6)(2) + (3)(2"cx")`
`= "cx"^2 - 12 + 6"cx"`
`= "cx"^2 + 6"cx" - 12`
If the angle between `bar"a"` and `bar"b"` is obtuse, `bar"a".bar"b" < 0`
∴ cx2 + 6cx - 12 < 0
∴ cx2 + 6cx < 12
∴ c(x2 + 6x) < 12
∴ c < `12/("x"^2 + 6"x")`
∴ c < `12/(("x"^2 + "6x" + 9) - 9) = 12/(("x + 3")^2 - 9)`
∴ c < min `{12/(("x + 3")^2 - 9)}`
Now, `12/(("x + 3")^2 - 9)` is minimum if (x + 3)2 - 9 is maximum
i.e. (x + 3)2 - 9 = ∞ - 9 = ∞
∴ c < min `{12/∞} = 0`
∴ c < 0.
Hence, the angle between `bar"a"` and `bar"b"` is obtuse if c < 0.
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