alinazubairx
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the vectors 2a + kb and 5a + 3b are parallel. find the value of k.
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RDKGames
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(Original post by alinazubairx)
the vectors 2a + kb and 5a + 3b are parallel. find the value of k.
What must you multiply 5 by to get 2?

So multiply 3 by the same thing to get k.

All this is coming from simply comparing the coefficients of vectors a and b.
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alinazubairx
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(Original post by RDKGames)
What must you multiply 5 by to get 2?

So multiply 3 by the same thing to get k.

All this is coming from simply comparing the coefficients of vectors a and b.
so u just multiply 5a + 3b by 2/5?
could you explain it to me pls?
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RDKGames
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(Original post by alinazubairx)
so u just multiply 5a + 3b by 2/5?
could you explain it to me pls?
If the vector 2\mathbf{a} + k \mathbf{b} is parallel to 5\mathbf{a} + 3\mathbf{b}, then this second vector is a scalar multiple of the first one.

So, we have that 2\mathbf{a} + k \mathbf{b} = \lambda(5\mathbf{a} + 3\mathbf{b}) where \lambda is the scalar.

We can rearrange this into (2- 5\lambda)\mathbf{a} + (k - 3\lambda)\mathbf{b} = \mathbf{0}, and because the two vectors \mathbf{a},\mathbf{b} are not scalar multiples of another, we must have that the coefficients of these vectors are zero in order to have the zero vector on the RHS.

This implies that

\begin{cases} 2 - 5\lambda = 0 \\ k - 3\lambda = 0 \end{cases}

and from the first equation we find that this scalar multiple is precisely \lambda = \dfrac{2}{5} which we can substitute into the second equation and that that k = 3 \cdot \dfrac{2}{5}.



That's the expalanation, and the quick way is to find the scalar \lambda by simply observing what you need to multiple the coefficient of \mathbf{a} by in the second vector in order to get the coefficient of \mathbf{a} in the first vector. Then use it on the \mathbf{b} vectors.
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alinazubairx
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(Original post by RDKGames)
If the vector 2\mathbf{a} + k \mathbf{b} is parallel to 5\mathbf{a} + 3\mathbf{b}, then this second vector is a scalar multiple of the first one.

So, we have that 2\mathbf{a} + k \mathbf{b} = \lambda(5\mathbf{a} + 3\mathbf{b}) where \lambda is the scalar.

We can rearrange this into (2- 5\lambda)\mathbf{a} + (k - 3\lambda)\mathbf{b} = \mathbf{0}, and because the two vectors \mathbf{a},\mathbf{b} are not scalar multiples of another, we must have that the coefficients of these vectors are zero in order to have the zero vector on the RHS.

This implies that

\begin{cases} 2 - 5\lambda = 0 \\ k - 3\lambda = 0 \end{cases}

and from the first equation we find that this scalar multiple is precisely \lambda = \dfrac{2}{5} which we can substitute into the second equation and that that k = 3 \cdot \dfrac{2}{5}.



That's the expalanation, and the quick way is to find the scalar \lambda by simply observing what you need to multiple the coefficient of \mathbf{a} by in the second vector in order to get the coefficient of \mathbf{a} in the first vector. Then use it on the \mathbf{b} vectors.
oh i see. thank you very much!
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