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Mathematics 10 Online
OpenStudy (anonymous):

Let U=2i-j v=3i+j w=i+2j Find the specified Scalar (4u).v

OpenStudy (anonymous):

Do you know what the dot product of two vectors is? Where u=(u1, u2) and v=(v1, v2), u.v=u1v1+u2v2. Work out what 4u is, then dot it with v for your answer.

OpenStudy (anonymous):

\[ 4\mathbf u = 4(3\mathbf i+\mathbf j) = 12\mathbf i+4\mathbf j \]

OpenStudy (dan815):

4 *(u.v)

OpenStudy (anonymous):

That is how scalar multiplication of vectors works.

OpenStudy (dan815):

4*(6-2)=16

OpenStudy (dan815):

u=<u1,u2,u3> v=<v1,v2,v3> u.v = u1v1+u2v2+u3v3

OpenStudy (jhannybean):

the arithmetic version of scalar multiplication

OpenStudy (dan815):

oh i looked at w vector for a sec

OpenStudy (dan815):

its4*(6-1)

OpenStudy (anonymous):

@dan815, these vectors are in R2, not R3 so you don't need u3 and v3 in your scalar product.

OpenStudy (anonymous):

The way dot product works is: \[\begin{split} \mathbf u \cdot \mathbf v &= (u_1\mathbf i+u_2\mathbf j)(v_1\mathbf i+v_2\mathbf j) \\ \\&= u_1v_1\mathbf i\cdot \mathbf i+u_2v_1\mathbf j\cdot \mathbf i+u_1 v_2\mathbf i\cdot \mathbf j+ u_2v_2\mathbf j\cdot \mathbf j \end{split}\]The thing is: \(\mathbf i\cdot \mathbf i =1,\ \mathbf i\cdot \mathbf j = 0,\ \mathbf j\cdot \mathbf i=0,\ \mathbf j\cdot \mathbf j =1\) So we end up with: \[ u_1v_1(1)+u_2v_1(0)+u_1 v_2(0)+ u_2v_2(1) = u_1v_1+u_2v_2 \]

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