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The vector $\vec{A}$ is $3 \hat{i} + \hat{j} + 2 \hat{k}$.

Any vector that is parallel to $\vec{A}$ has to point in the same direction.

Now, here's the trick; for the vector $\vec{B}$ to point in the same direction, its components have to be the same.

However, they don't have to be the same vector; $\vec{B}$ can be any scaled version of $\vec{A}$ and still point in the same direction.

So, knowing this, we see that $\vec{B}$ has first two components twice as much as $\vec{A}$. For it to maintain the property of a scaled version of $\vec{A}$, it must have the 3rd component to be twice as much as the 3rd component of $\vec{A}$.

Therefore $\boxed{S = 4}$

Addendum:

The main point this intuitive solution makes is that for a vector to be parallel to another vector, it must point in the same direction as the second vector, and for that to happen numerically, it has to have any scaled version of vector 2's components.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestOk, I'm going to give a more intuitive answer:

The vector $\vec{A}$ is $3 \hat{i} + \hat{j} + 2 \hat{k}$.

Any vector that is parallel to $\vec{A}$ has to point in the same direction.

Now, here's the trick; for the vector $\vec{B}$ to point in the same direction, its components have to be the same.

However, they don't have to be the same vector; $\vec{B}$ can be any scaled version of $\vec{A}$ and still point in the same direction.

So, knowing this, we see that $\vec{B}$ has first two components twice as much as $\vec{A}$. For it to maintain the property of a scaled version of $\vec{A}$, it must have the 3rd component to be twice as much as the 3rd component of $\vec{A}$.

Therefore $\boxed{S = 4}$

Addendum:The main point this intuitive solution makes is that for a vector to be parallel to another vector, it must point in the same direction as the second vector, and for that to happen numerically, it has to have any scaled version of vector 2's components.

@shruthi srinivasan

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($2$) - $4$ because:

$2A = B$

So:

$2(3i + j + 2k) = B$

$6i + 2j + 4k = B$

$S = \fbox 4$

or:

$A =$$\frac{1}{2}$$B$

So:

$\frac{S}{2}$$= 2$

$S = \fbox 4$

@shruthi srinivasan

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thank you.

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You're welcome!

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That's not technically right. Just because they are parallel, that doesn't mean that $2A = B$. Why not $A = \tfrac12 B$?

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That's also true.

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I'll add that in.

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