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If abca¯,b¯,c¯ are unit vectors such that abca¯+b¯+c¯=0¯, then find the value of abbccaa¯.b¯+b¯.c¯+c¯.a¯. - Mathematics and Statistics

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Question

If `bar"a", bar"b", bar"c"` are unit vectors such that `bar"a" + bar"b" + bar"c" = bar0,` then find the value of `bar"a".bar"b" + bar"b".bar"c" + bar"c".bar"a".`

Sum

Solution

`bar"a", bar"b", bar"c"` are unit vectors

∴ `|bar"a"| = |bar"b"| = |bar"c"| = 1.`

Also, `bar"a".bar"a" = bar"b".bar"b" = bar"c".bar"c" = 1`

Now, `bar"a" + bar"b" + bar"c" = bar0`    ...(1)

Taking scalar product of both sides with `bar"a", we get

`bar"a".(bar"a" + bar"b" + bar"c") = bar"a".bar0`

∴ `bar"a".bar"a" + bar"a".bar"b" + bar"a".bar"c" = 0`

∴ `bar"a".bar"b" + bar"a".bar"c" = - bar"a".bar"a" = - 1`      ....(2)

Similarly taking scalar product of both sides of (1) with `bar"b" and bar"c",` we get,

`bar"b".bar"a" + bar"b".bar"c" = - 1`     ....(3)

`bar"c".bar"a" + bar"c".bar"b" = - 1`       .....(4)

Adding (2), (3), (4) and using the fact that scalar product is commutative, we get

`2(bar"a".bar"b" + bar"b".bar"c" + bar"c".bar"a") = - 3`

∴ `bar"a".bar"b" + bar"b".bar"c" + bar"c".bar"a" = - 3/2`

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Chapter 5: Vectors - Miscellaneous exercise 5 [Page 191]

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