vector question

Hi, can anyone explain the answer? thanks

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Comments

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 $\boxed{S = 4}$

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

Krishna Karthik - 10 months, 1 week 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:

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

$S = \fbox 4$

@shruthi srinivasan

Yajat Shamji - 10 months, 3 weeks ago

thank you.

shruthi srinivasan - 10 months, 3 weeks ago

You're welcome!

Yajat Shamji - 10 months, 2 weeks ago

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

Pi Han Goh - 10 months, 2 weeks ago

That's also true.

Yajat Shamji - 10 months, 2 weeks ago

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

Pi Han Goh - 10 months, 2 weeks ago

I'll add that in.

Yajat Shamji - 10 months, 2 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$