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.

Answer 1

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.