# Find the values of c so that for all real x, the vectors xcijkxci^-6j^+3k^ and xijcxkxi^+2j^+2cxk^ make an obtuse angle. - Mathematics and Statistics

Sum

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.

Concept: Vector Product of Vectors (Cross)
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