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Question
Find the value of ‘a’ so that the volume of parallelopiped formed by hat"i" + "a"hat"j" + hat"k", hat"j" + "a"hat"k" and "a"hat"i" + hat"k" becomes minimum.
Solution
Let `bar"p" = hat"i" + "a"hat"j" + hat"k", bar"q" = hat"j" + "a"hat"k", bar"r" = "a"hat"i" + hat"k"`
Let V be the volume of the parallelopiped formed by `bar"p",bar"q",bar"r"`.
Then V = `[bar"p"bar"q"bar"r"]`
`= |(1,"a",1),(0,1,"a"),("a",0,1)|`
`= 1(1 - 0) - "a"(0 - "a"^2) + 1(0 - "a")`
`= 1 + "a"^3 - "a"`
`∴ "dV"/"da" = "d"/"da"(1 + "a"^3 - "a")`
`= 0 + 3"a"^2 - 1 = 3"a"^2 - 1`
and `("d"^2"V")/("da"^2) = "d"/"da"(3"a"^2 - 1)`
= 3 × 2a - 0 = 6a
For maximum and minimum V, `"dV"/"da"` = 0
∴ 3a2 - 1 = 0
∴ `"a"^2 = 1/3`
∴ a = `+- 1/sqrt3`
Now, `(("d"^2"V")/"da"^2)_("at a" = 1/sqrt3) = 6(1/sqrt3) = 2sqrt3 > 0`
∴ V is minimum when a = `1/sqrt3`
Also, `(("d"^2"V")/"da"^2)_("at a" = - 1/sqrt3) = 6(- 1/sqrt3) = - 2sqrt3 < 0`
∴ V is minimum when a = `- 1/sqrt3`
Hence, a = `1/sqrt3`
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