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

[chwala]

Homework Statement

See attached

Relevant Equations

Vectors

My interest is only on part (b),

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

$AD=AO+OD$
$AD=-a+\dfrac{5}{2}b$

Therefore, $PB=\dfrac{2}{5}AD$

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...

 

[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(?)

 

[chwala]

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]

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...

 

[chwala]

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,
$PB=PO+OB$
$PB=-\dfrac{2}{5}a +b$

$AD=AO+OD$
$AD=-a+\dfrac{5}{2}b$

Therefore, $PB=\dfrac{2}{5}AD$

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...

 

[Delta2]

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.

 

[chwala]

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...

 

[chwala]

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,
$PB=PO+OB$
$PB=-\dfrac{2}{5}a +b$

$AD=AO+OD$
$AD=-a+\dfrac{5}{2}b$

Therefore, $PB=\dfrac{2}{5}AD$

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,

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

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

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

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