# vector question

Hi, can anyone explain the answer? thanks

Note by Shruthi Srinivasan
2 months, 4 weeks ago

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Ok, 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 $\boxed{S = 4}$

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.

- 2 months, 2 weeks ago

($2$) - $4$ 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$

$S = \fbox 4$

- 2 months, 4 weeks ago

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

- 2 months, 3 weeks ago

- 2 months, 3 weeks ago

That's also true.

- 2 months, 3 weeks ago

You're missing the point. Why must one of the vector be double of the other vector?

- 2 months, 3 weeks ago

thank you.

- 2 months, 4 weeks ago

You're welcome!

- 2 months, 4 weeks ago