# Show that if a>2 the angle between vectors r=2vec i-a vec j, w= vec i + vec j is not acurte or right?

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### 1 Answer

You need to consider that the angle is obtuse, since the problem provides the information that the angle must not to be right, not acute.

You need to evaluate the dot product of the vectors such that:

`(bar r)*(bar w) = |bar r|*|bar w|*cos (hat(bar r, bar w))`

`(2 bar i - a bar j)(bar i + bar j) = sqrt((4+a^2)(1 + 1))*cos (hat(bar r, bar w))`

`2 - a = sqrt((4+a^2)(1 + 1))*cos (hat(bar r, bar w))`

`cos (hat(bar r, bar w)) = (2 - a)/sqrt(2(4+a^2))`

If the angle is right, hence `cos (hat(bar r, bar w)) = 0` an if the angle is acute, hence `cos (hat(bar r, bar w)) gt0` .

Since the angle is not right, nor acute, hence `cos (hat(bar r, bar w)) lt 0` .

Since the denominator is positive, `sqrt(2(4+a^2)) gt 0` , hence, the numerator must be negative such that:

`2 - a lt 0 =gt 2 lt a`

**Hence, if `agt2` , the angle between vectors is not right, nor acute, but obtuse.**

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