# Show that P, Q and D are collinear in the vector problem

• chwala
In summary, for part (b), the approach for finding the vector PQ is to first find the vectors PB, AD, and QD, and then use the relationships between these vectors to find the value of PQ. This can be done by setting up equations using the vector addition and scalar multiplication properties, and then solving for PQ.

#### chwala

Gold Member
Homework Statement
See attached
Relevant Equations
Vectors

My interest is only on part (b),

For part (i), My approach is as follows,
##PB=PO+OB##
##PB=-\dfrac{2}{5}a +b##

For part (ii), we shall have;

##QD=QB+BD##

##QD=\dfrac{2}{7}AB+\dfrac{3}{2}b##

##QD=-\dfrac{2}{7}a+\dfrac{2}{7}b+\dfrac{3}{2}b##

##QD=\dfrac{25}{14}b-\dfrac{2}{7}a##

also,
##PQ=PA+AQ##

##PQ=\dfrac{3}{5}a-\dfrac{5}{7}a+\dfrac{5}{7}b##

##PQ=\dfrac{5}{7}b-\dfrac{4}{35}a##

Therefore,

##PQ=\dfrac{1}{7}\left[5b-\dfrac{4}{5}a\right]##
&
##QD=\dfrac{1}{7}\left[\dfrac{25}{2}b-2a\right]##

It follows that,

##PQ=\dfrac{5}{7}\left[25b-4a\right]##

##QD=\dfrac{2}{7}\left[25b-4a\right]##

##OQ=\dfrac{2}{5} PQ## thus shown as there is a scalar connecting the two...seeking any alternative approach and i hope my working is correct as i do not have solutions to these questions...

Last edited:
Delta2
for (b) (i) I find it correct
for (b) (ii) there seem to be some typos after the phrase "It follows that" please correct them. For example I think it should $$\vec{PQ}=\frac{1}{5\cdot 7}[25\vec{b}-4\vec{a}]$$ and the final equation should involve PQ and QD not OQ(?)

Last edited:
Delta2 said:
for (b) (i) I find it correct
for (b) (ii) there seem to be some typos after the phrase "It follows that" please correct them. For example I think it should $$\vec{PQ}=\frac{1}{5\cdot 7}[25b-4a]$$ and the final equation should involve PQ and QD not OQ(?)
Ok...I will check on that ...

Delta2
The correct final equation should be ##\vec{PQ}=\frac{2}{5}\vec{QD}##, at least that's what i think.

chwala
Delta2 said:
The correct final equation should be ##\vec{PQ}=\frac{2}{5}\vec{QD}##, at least that's what i think.
That should be correct...my working is fine, but the division part at the end was wrong...

Delta2
chwala said:
Homework Statement:: See attached
Relevant Equations:: Vectors

View attachment 301993

My interest is only on part (b),

For part (i), My approach is as follows,
##PB=PO+OB##
##PB=-\dfrac{2}{5}a +b##

For part (ii), we shall have;

##QD=QB+BD##

##QD=\dfrac{2}{7}AB+\dfrac{3}{2}b##

##QD=-\dfrac{2}{7}a+\dfrac{2}{7}b+\dfrac{3}{2}b##

##QD=\dfrac{25}{14}b-\dfrac{2}{7}a##

also,
##PQ=PA+AQ##

##PQ=\dfrac{3}{5}a-\dfrac{5}{7}a+\dfrac{5}{7}b##

##PQ=\dfrac{5}{7}b-\dfrac{4}{35}a##

Therefore,

##PQ=\dfrac{1}{7}\left[5b-\dfrac{4}{5}a\right]##
&
##QD=\dfrac{1}{7}\left[\dfrac{25}{2}b-2a\right]##

It follows that,

##PQ=\dfrac{5}{7}\left[25b-4a\right]##

##QD=\dfrac{2}{7}\left[25b-4a\right]##

##QD=\dfrac{2}{5} PQ## thus shown as there is a scalar connecting the two...seeking any alternative approach and i hope my working is correct as i do not have solutions to these questions...
Amended to ##QD##...I will amend division part later, using my android to type...

I find the expressions of PQ and QD wrong after the phrase "It follows that" regarding the multiplicative constant of each, please review and try to correct them.

Delta2 said:
I find the expressions of PQ and QD wrong after the phrase "It follows that" regarding the multiplicative constant of each, please review and try to correct them.
Yes I will...thanks for your input...

Delta2
chwala said:
Homework Statement:: See attached
Relevant Equations:: Vectors

View attachment 301993

My interest is only on part (b),

For part (i), My approach is as follows,
##PB=PO+OB##
##PB=-\dfrac{2}{5}a +b##

For part (ii), we shall have;

##QD=QB+BD##

##QD=\dfrac{2}{7}AB+\dfrac{3}{2}b##

##QD=-\dfrac{2}{7}a+\dfrac{2}{7}b+\dfrac{3}{2}b##

##QD=\dfrac{25}{14}b-\dfrac{2}{7}a##

also,
##PQ=PA+AQ##

##PQ=\dfrac{3}{5}a-\dfrac{5}{7}a+\dfrac{5}{7}b##

##PQ=\dfrac{5}{7}b-\dfrac{4}{35}a##

Therefore,

##PQ=\dfrac{1}{7}\left[5b-\dfrac{4}{5}a\right]##
&
##QD=\dfrac{1}{7}\left[\dfrac{25}{2}b-2a\right]##

It follows that,

##PQ=\dfrac{5}{7}\left[25b-4a\right]##

##QD=\dfrac{2}{7}\left[25b-4a\right]##

##OQ=\dfrac{2}{5} PQ## thus shown as there is a scalar connecting the two...seeking any alternative approach and i hope my working is correct as i do not have solutions to these questions...
We ought to have;

It follows that,

##PQ=\dfrac{1}{35}\left[25b-4a\right]##

##QD=\dfrac{1}{14}\left[25b-4a\right]##

##35 PQ=(25b-4a)##
##14QD=(25b-4a)##

##35PQ=14QD##
##PQ=\dfrac{14}{35} QD##
##PQ=\dfrac{2}{5} QD##

Delta2

## 1. How do you show that three vectors are collinear?

To show that three vectors P, Q, and D are collinear, you can use the scalar triple product. If the scalar triple product of the three vectors is equal to zero, then they are collinear. This means that the vectors lie on the same line.

## 2. What is the scalar triple product?

The scalar triple product is a mathematical operation that takes in three vectors and produces a scalar value. It is calculated by taking the dot product of one vector with the cross product of the other two vectors.

## 3. Can you explain the concept of collinearity in vector problems?

In vector problems, collinearity refers to the property of three or more vectors lying on the same line. This means that the vectors have the same direction or are parallel to each other.

## 4. How does the scalar triple product help in determining collinearity?

The scalar triple product helps in determining collinearity by providing a mathematical test. If the scalar triple product of three vectors is equal to zero, then the vectors are collinear. If the scalar triple product is non-zero, then the vectors are not collinear.

## 5. Are there any other methods to prove collinearity in vector problems?

Yes, there are other methods to prove collinearity in vector problems. These include using the concept of linear dependence, where one vector can be written as a linear combination of the other two vectors. Also, if the three vectors have the same direction or are parallel, then they are collinear.