# Write the vector u =< 4, -2, -5 > as .vector u = vector a + vector b where vector a is parallel to vector d =< 1,2,3 > and vector b is perpendicular to vector d

*print*Print*list*Cite

### 1 Answer

You need to write the vectors `bar a` and `bar b` , such that:

`bar a = a_x bar i + a_y bar j + a_z bar k`

`bar b = b_x bar i + b_y bar j + b_z bar k`

The problem provides the information that `bar u = bar a + bar b` , such that:

`4bar i - 2 bar j - 5 bar k = (a_x + b_x) bar i + (a_y + b_y) bar k + (a_z + b_z) bar k`

`{(a_x + b_x = 4),(a_y + b_y = -2),(a_z + b_z = -5):}`

The problem provides the information that `bar a` is parallel to `bar d` and `bar b` is perpendicular to `bar b` , such that:

`bar a = k*<1,2,3> => a_x = k`

`a_y = 2k => a_y = 2a_x`

`a_z = 3k => a_z = 3a_x`

`bar b*bar d = 0 => b_x + 2b_y + 3b_z = 0 => b_x = -2b_y - 3b_z`

Since the vector `bar a` is parallel to `bar d` and `bar b` is perpendicular to `bar d` , hence, `bar a` is perpendicular to `bar b` , such that:

`bar a*bar b = 0 => a_x*b_x + 2a_x*b_y + 3a_x*b_z = 0 => a_x(b_x + 2b_y + 3b_z) = 0`

`a_x + b_x = 4 => a_x = 4 - b_x`

`a_y = -2 - b_y => 2a_x = -2 - b_y => 8 - 2b_x = -2 - b_y => -2b_x = -10 - b_y => b_x = 5 + (b_y)/2`

`a_z + b_z = -5 => a_z = -5 - b_z`

`3a_x = -5 - b_z => 12 - 3b_x = -5 - b_z => 3b_x = 17 + b_z => b_x = 17/3 + (b_z)/3`

Since `b_x = 5 + (b_y)/2` and `b_x = 17/3 + (b_z)/3` , hence `5 + (b_y)/2 = 17/3 + (b_z)/3.`

`(b_y)/2 - (b_z)/3 = 17/3 - 5`

`(b_y)/2 - (b_z)/3 = 2/3 => 3b_y - 2b_z = 4`

**Hence, considering `3b_y - 2b_z = 4` , you may write `bar u = bar a + bar b` , such that: **`4bar i - 2 bar j - 5 bar k = (a_x + b_x) bar i + (2a_x + b_y) bar k + (3a_x + (3b_y-4)/2) bar k.`