# Assume that `vecu*(vecv xx vecw) = 20`  then compute `(2vecu + vecw)*(vecv xx(vecu+vec v - vecw))`

ishpiro | College Teacher | (Level 1) Educator

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First, simplify

`vecv xx (vecu + vecv - vecw)=vecv xx vecu + vecv xx vecv - vecv xx vecw`

Since `vecv xx vecv = 0` (parallel vectors have the cross-product of 0), then

`vecv xx (vecu + vecv - vecw)=vecv xx vecu- vecv xx vecw` .

Thus,

`(2vecu + vecw)*(vecv xx (vecu+vecv-vecw))=(2vecu+vecw)*(vecv xx vecu - vecv xx vecw)=2vecu*(vecv xx vecu) + vecw*(vecv xx vecu) - 2vecu*(vecv xx vecw) - vecw*(vecv xx vecw)`

Since the cross-product of two vectors is perpendicular to both vectors, and the dot product of two perpendicular vectors is 0, the following products above are 0:

`2vecu*(vecv xx vecu) = 0`

`vecw*(vecv xx vecw) = 0`

Since exchanging the order of two vectors in the triple product results in changing its sign, `vecw*(vecv xx vecu)` is equal to `-vecw*(vecu xx vecv)=vecu*(vecw xx vecv) = - vecu*(vecv xx vecw)` .

The original expression therefore equals

`(2vecu + vecw)*(vecv xx (vecu+vecv-vecw) = - vecu*(vecv xx vecw) - 2vecu*(vecv xx vecw)=-3vecu*(vecv xx vecw)`

Plugging in the given value, 20, of this triple product, 20, we get

`(2vecu + vecw)*(vecv xx (vecu+vecv-vecw)=-3*20 = -60`

The value of given expression is -60.