Indicate in each case what bases are orthogonal and/or orthonormal.
- $$\vec{u}=(2,1)$$, $$\ \vec{v}=(1,-2)$$
- $$\vec{u}=(2,1)$$, $$\ \vec{v}=(1,-1)$$
- $$\vec{u}=(1,0)$$, $$\ \vec{v}=(0,1)$$
Development:
In every case we first look to see if the vectors of the basis are perpendicular using the scalar product, and if it is necessary then we calculate the module of the vectors.
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Calculating the scalar product we obtain: $$$ \vec{u}\cdot\vec{v}=2\cdot1+1\cdot(-2)=0$$$ So we can say that the vectors form an angle of $$90^\circ$$, or in other words, they are perpendicular. Also, we might calculate the angle that the vectors form using the formula: $$$ \text{ang}(\vec{u},\vec{v})=\arccos\Big(\dfrac{u_1 v_1+u_2 v_2}{\sqrt{u_1^2+u_2^2}\cdot\sqrt{v_1^2+v_2^2}}\Big) = \arccos \Big( \dfrac{2\cdot1+1\cdot(-2)}{\sqrt{5}\sqrt{5}}\Big) = \arccos(0)=90^\circ$$$ These two vectors form an orthogonal basis. Now we still have to see if they form an orthonormal basis by calculating the module of the vectors. In order to be orthonormal it is necessary that the module of the two vectors is $$1$$. In other words, that the vectors are unitary. $$$|\vec{u}|=\sqrt{2^2+1^2}=\sqrt{5}\quad |\vec{v}|=\sqrt{1^2+(-2)^2}=\sqrt{5}$$$ We can see that these vectors are not unitary, so they do not form an orthonormal basis.
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Calculating the scalar product we obtain: $$$ \vec{u}\cdot\vec{v}=2\cdot1+1\cdot(-1)=1\neq 0$$$ Ao we can affirm that the vectors are not perpendiculars since the product climbing is not equal to zero. Just as in the previous case we might also calculate the angle that they form and we would see that it is different to $$90^\circ$$. So, these vectors cannot form an orthogonal basis and nor, therefore, can they form an ortonormal one either.
- Calculating the scalar product we obtain: $$$ \vec{u}\cdot\vec{v}=1\cdot0+0\cdot1=0$$$ So we can say that the vectors form an angle of $$90^\circ$$, éor in other words, they are perpendicular. Also we might calculate the angle by using the formula: $$$ \text{ang}(\vec{u},\vec{v}) = \arccos \Big( \dfrac{1\cdot0+0\cdot1}{\sqrt{1}\sqrt{1}}\Big) = \arccos(0)=90^\circ$$$ So these two vectors form an orthogonal basis. ANow we still have to see if they form an orthonormal basis by computing the module of the respective vectors. In order to have an orthonormal basis it is necessary that the module of both vectors is $$1$$. In other words, that the vectors are unitary. $$$|\vec{u}|=\sqrt{1^2+0^2}=\sqrt{1}=1\quad |\vec{v}|=\sqrt{0^2+1^2}=\sqrt{1}=1$$$ We can see that the two vectors have module $$1$$, or in other words, they are unitary and therefore they also form an orthonormal basis.
Solution:
- They form an orthogonal basis.
- They do not form an orthogonal basis (and neither do they form a orthonormal one, of course).
- They form an orthonormal basis and, therefore, it is also orthogonal.