# Consider three vectors vector u, vector v, vector w.Simplify (2(vector u) -3(vectorv) ) dot product (4(vector u) + 2(vector v)).

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You need to evaluate the following dot product, such that:

`(2(bar u) -3(bar v) )*(4(bar u) + 2(bar v)) = 8(bar u*bar u) + 4(bar u*bar v) - 12(bar u*bar v) - 6(bar v*bar v)`

`(2(bar u) -3(bar v) )*(4(bar u) + 2(bar v)) = 8(|bar u|)^2 - 8(bar u*bar v) - 6(|bar v|)^2`

Considering `bar u = u_x bar i + u_y bar j` and `bar v = v_x bar i + v_y bar j` yields:

`(|bar u|)^2 = (u_x)^2 + (u_y)^2`

`(|bar v|)^2 = (v_x)^2 + (v_y)^2 `

`bar u*bar v = |bar u|*|bar v|* cos alpha`

If `bar u` is perpendicular to `bar v` , hence `cos alpha = cos(pi/2) = 1` and `bar u*bar v = 0` .

`(2(bar u) -3(bar v) )*(4(bar u) + 2(bar v)) = 8(u_x)^2 + 8(u_y)^2 - 8(sqrt(((u_x)^2 + (u_y)^2)((v_x)^2 + (v_y)^2))) cos alpha - 6 (v_x)^2 - 6(v_y)^2`

**Hence, evaluating the given dot product, under the given conditions, yields **`(2(bar u) -3(bar v) )*(4(bar u) + 2(bar v)) = 8(u_x)^2 + 8(u_y)^2 - 8(sqrt(((u_x)^2 + (u_y)^2)((v_x)^2 + (v_y)^2))) cos alpha - 6 (v_x)^2 - 6(v_y)^2.`