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Question
Express the vector `bar"a" = 5hat"i" - 2hat"j" + 5hat"k"` as a sum of two vectors such that one is parallel to the vector `bar"b" = 3hat"i" + hat"k"` and other is perpendicular to `bar"b"`.
Solution
Let `bar"a" = bar"c" + bar"d"`, where `bar"c"` is parallel to `bar"b" and bar"d"` is perpendicular to `bar"b"`.
Since, `bar"c"` is parallel to `bar"b", bar"c" = "m"bar"b"`, where m is a scalar.
∴ `bar"c" = "m"(3hat"i" + hat"k")`
i.e. `bar"c" = 3"m"hat"i" + "m"hat"k"`
Let `bar"d" = "x"hat"i" + "y"hat"j"+ zhat"k"`
Since, `bar"d"` is perpendicular to `bar"b" = 3hat"i" + hat"k", bar"d".bar"b" = 0`
∴ `("x"hat"i" + "y"hat"j" + "z"hat"k").(3hat"i" + hat"k") = 0`
∴ 3x + z = 0
∴ z = - 3x
∴ `bar"d" = "x"hat"i" + "y"hat"k" - 3"x"hat"k"`
Now, `bar"a" = bar"c" + bar"d"` gives
∴ `5hat"i" - 2hat"j" + 5hat"k" = (3"m"hat"i" + "m"hat"k") + ("x"hat"i" + "y"hat"j" - 3"x"hat"k")`
`= (3"m" + "x")hat"i" + "y"hat"j" + ("m" - 3"x")hat"k"`
By equality of vectors
3m + x = 5 ....(1)
y = - 2
and m - 3x = 5 ......(2)
From (1) and (2)
3m + x = m - 3x
∴ 2m = - 4x
∴ m = - 2x
Substituting m = - 2x in (1), we get
∴ - 6x + x = 5
∴ - 5x = 5
∴ x = - 1
∴ m = - 2x = 2
∴ `bar"c" = 6hat"i" + 2hat"k"` is parallel to `bar"b" and bar"d" = - hat"i" - 2hat"j" + 3hat"k"` is perpendicular to `bar"b"`
Hence, `bar"a" = bar"c" + bar"d", "where" bar"c" = 6hat"i" + 2hat"k" and bar"d" = - hat"i" - 2hat"j" + 3hat"k"`
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