# Vectors [Solved!]

### My question

In the figure (also at http://prnt.sc/cqth0m), vector OB is three times longer than OA, BC is parallel to OA and twice its length. CD is parallel to BA.

Angle AOB = 60 degrees. If A and B have position vectors a and b respectively relative to O; and CD = kBA, find:

a) vector expression for BA and OD
b) find k if OD is perpendicular to AB

Vectors

### What I've done so far

(a) BA = BO + OA = -b + a = a-b
Now how to find OD?

Since OB = 3OA, does this mean b=3a? And what is use of this information?

BC=2OA
CD=pBA

OD= OB+BC+CD = b+2a+p(a-b),which is not correct I know

For (b),AB would be =b-a, and OD would be found from (a), but what happens if it is perpendicular?

X

<img src="/forum/uploads/imf-3239-ques.png" width="409" height="288" alt="vectors" />

In the figure (also at <a href="http://prnt.sc/cqth0m">http://prnt.sc/cqth0m</a>), vector OB is three times longer than OA, BC is parallel to OA and twice its length. CD is parallel to BA.

Angle AOB = 60 degrees. If A and B have position vectors a and b respectively relative to O; and CD = kBA, find:

a) vector expression for BA and OD
b) find k if OD is perpendicular to AB
Relevant page

<a href="http://www.intmath.com/forum/vectors-39/vectors:108">Vectors</a>

What I've done so far

(a) BA = BO + OA = -b + a = a-b
Now how to find OD?

Since OB = 3OA, does this mean b=3a? And what is use of this information?

BC=2OA
CD=pBA

OD= OB+BC+CD = b+2a+p(a-b),which is not correct I know

For (b),AB would be =b-a, and OD would be found from (a), but what happens if it is perpendicular?

## Re: Vectors

As some general hints for this kind of question, it's good to separate the vector diagram (where you use vector addition and subtraction, as you've done in the first part of your answer), and the related scalar diagram, where you just deal with the lengths involved.

(Just a small point - your question has k as the unknown variable, and then you have used p in your answer. Let's stick to k.)

It's not so easy to see what's going on in the given diagram (that's probably deliberate on the part of the question writers).

Here's the vector diagram, this time drawn to scale (I've used GeoGebra to draw the diagram):

You wrote:

(a) BA = BO + OA = -b + a = a-b

Yes, OK.

Now how to find OD?

Since OB = 3 OA, does this mean b = 3a?

Actually, no. This is what I meant about separating out the vector and scalar diagrams. Vector b is 3 times longer than vector a, but they are pointing in different directions, so we can't say b = 3a.

And what is use of this information?

We'll use it in the scalar diagram.

BC = 2OA
CD = kBA

OD = OB + BC + CD = b + 2a + k(a-b), which is not correct I know

Actually, I think it is correct (with the p changed back to k)!

For (b), AB? would be =b-a, and OD would be found from (a), but what happens if it is perpendicular?

Before going on to this, can you draw a scalar diagram now, showing the known lengths? It will help a lot, I think.

X

Hi Asad

As some general hints for this kind of question, it's good to separate the <b>vector</b> diagram (where you use vector addition and subtraction, as you've done in the first part of your answer), and the related <b>scalar</b> diagram, where you just deal with the lengths involved.

(Just a small point - your question has k as the unknown variable, and then you have used p in your answer. Let's stick to k.)

It's not so easy to see what's going on in the given diagram (that's probably deliberate on the part of the question writers).

Here's the vector diagram, this time drawn to scale (I've used GeoGebra to draw the diagram):

<img src="/forum/uploads/imf-2519-vector-diagram.png" width="360" height="307" alt="Vectors" />

You wrote:

<blockquote>(a) <b>BA = BO + OA = -b + a = a-b</b></blockquote>

Yes, OK.

<blockquote>Now how to find <b>OD</b>?

Since <b>OB</b> = 3 <b>OA</b>, does this mean <b>b = 3a</b>?</blockquote>

Actually, no. This is what I meant about separating out the vector and scalar diagrams. Vector <b>b</b> is 3 times longer than vector <b>a</b>, but they are pointing in different directions, so we can't say <b>b = 3a</b>.

<blockquote>And what is use of this information?</blockquote>

We'll use it in the scalar diagram.

<blockquote><b>BC</b> = 2<b>OA</b>
<b>CD</b> = k<b>BA</b>

<b>OD = OB + BC + CD = b + </b>2<b>a</b> + k<b>(a-b)</b>, which is not correct I know</blockquote>

Actually, I think it is correct (with the p changed back to k)!

<blockquote>For (b), <b>AB</b>? would be <b>=b-a</b>, and <b>OD</b> would be found from (a), but what happens if it is perpendicular?</blockquote>

Before going on to this, can you draw a scalar diagram now, showing the known lengths? It will help a lot, I think.

## Re: Vectors

It seems Asad has disappeared. Anyone else like to have a go at finishing it?

X

It seems Asad has disappeared. Anyone else like to have a go at finishing it?

## Re: Vectors

OK, I'll have a go.

I used GeoGebra (great tool for these problems) for a scalar diagram, as Murray suggested.

There may be an easier way, but this is how I did it.

I've put all the known values on the graph. I extended OA to intersect CD at G.

We can find length AB using cosine rule:

c^2 = a^2 + b^2 - 2ac cosC

v^2 = (3p)^2 + p^2 - 2(3p)(p)cos60°

v^2 = 9p^2 + p^2 - 6p^2(0.5)

v^2 = 7p^2

v = p sqrt(7)

Using that, we can find the angle OAB and the sine rule

a/(sin A) = c/(sin c)

(3p)/(sin A) = (psqrt7)/(sin 60°)

Cancelling the p's and turning it all upside down:

(sin A)/3 = (sin 60°)/sqrt(7)

sin A = 0.98198050606

A is more than 90 deg, so

A = 180 - 79.1066054 = 100.8933946°

Since AB and CD are parallel, by alternate angles we can say angle OGD is 79.1066054°.

Using that, we can find length GD using the fact OGD is a right triangle.

cos 79.1066054° = (GD)/(3p)

So GD = 3p (0.18898223566) = 0.56694670698p

So CD = CG + GD = p sqrt(7)+0.56694670698p = 3.21269801805p

Since AB = p sqrt(7),

k = (3.21269801805p)/(psqrt(7)) = 1.21428571333

According to the measuring tool in GeoGebra, this is a correct answer.

Does it look right?

Yours
Stephen

X

OK, I'll have a go.

I used GeoGebra (great tool for these problems) for a scalar diagram, as Murray suggested.

There may be an easier way, but this is how I did it.

<img src="/forum/uploads/imf-4812-vector-diagram-scalar.png" width="339" height="297" alt="Vectors" />

I've put all the known values on the graph. I extended OA to intersect CD at G.

We can find length AB using cosine rule:

c^2 = a^2 + b^2 - 2ac cosC

v^2 = (3p)^2 + p^2 - 2(3p)(p)cos60°

v^2 = 9p^2 + p^2 - 6p^2(0.5)

v^2 = 7p^2

v = p sqrt(7)

Using that, we can find the angle OAB and the sine rule

a/(sin A) = c/(sin c)

(3p)/(sin A) = (psqrt7)/(sin 60°)

Cancelling the p's and turning it all upside down:

(sin A)/3 = (sin 60°)/sqrt(7)

sin A = 0.98198050606

A is more than 90 deg, so

A = 180 - 79.1066054 = 100.8933946°

Since AB and CD are parallel, by alternate angles we can say angle OGD is 79.1066054°.

Using that, we can find length GD using the fact OGD is a right triangle.

cos 79.1066054° = (GD)/(3p)

So GD = 3p (0.18898223566) = 0.56694670698p

So CD = CG + GD = p sqrt(7)+0.56694670698p = 3.21269801805p

Since AB = p sqrt(7),

k = (3.21269801805p)/(psqrt(7)) = 1.21428571333

According to the measuring tool in GeoGebra, this is a correct answer.

Does it look right?

Yours
Stephen

## Re: Vectors

Stephen

This looks good to me!

Regards
Murray

X

Stephen

This looks good to me!

Regards
Murray

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