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"`