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# Question on Mechanics - Vectors

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OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AP and MQ meet at P.

If O to A = a, O to C = c

express O to P in terms of a and c.

Somone help me plz...?

physics
Dec 11, 2014

#9
+270
+10

$$\vec{AC}=-\mathbf{a}+\mathbf{c}$$

$$\vec{MQ}=\frac{1}{2}\mathbf{a}+\mathbf{c}-\frac{1}{4}\mathbf{a}$$

$$=\frac{1}{4}\mathbf{a}+\mathbf{c}$$

$$\vec{OP}=\vec{OM}+\vec{MP}$$

$$=\vec{OA}+\vec{AP}$$

$$\vec{MP}=\lambda(\frac{1}{4}\mathbf{a}+\mathbf{c})$$

$$=\frac{\lambda}{4}\mathbf{a}+\lambda \mathbf{c}$$

$$\vec{AP}=\mu(-\mathbf{a}+\mathbf{c})$$

$$=-\mu\mathbf{a}+\mu\mathbf{c}$$

As $$\vec{AP}$$ and $$\vec{MP}$$ intersect, if I let them equal, I can find the $$P$$.

$$\frac{\lambda}{4}\mathbf{a}+\lambda \mathbf{c}$$ $$=-\mu\mathbf{a}+\mu\mathbf{c}$$

Therefore, $$\frac{\lambda}{4}=-\mu$$ and $$\lambda = \mu$$

Then if I subtitute $$\mu$$ to $$\frac{\lambda}{4}=-\mu$$,

$$\frac{\mu}{4}=-\mu$$

$$\mu=-4$$

Sub -4 to $$\vec{MP}=\lambda(\frac{1}{4}\mathbf{a}+\mathbf{c})$$    (As  $$\mu = \lambda$$)

$$\vec{MP}=-4(\frac{1}{4}\mathbf{a}+\mathbf{c})$$

$$=-\mathbf{a}-4\mathbf{c}$$

Finally,

$$\vec{OP} = \vec{OM}+\vec{MP}$$

$$=\frac{1}{2}\mathbf{a}+(-\mathbf{a}-4\mathbf{c})$$

$$=\frac{1}{2}\mathbf{a}-\mathbf{a}-4\mathbf{c}$$

$$=-\frac{1}{2}\mathbf{a}-4\mathbf{c}$$

.
Dec 11, 2014

#1
+112011
+5

AP and MQ meet at P

I assume that you mean

AC and MQ meet at P  ??

Dec 11, 2014
#2
+112011
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It is a square - so

OA=OC

so why have 2 different pronumerals?

Dec 11, 2014
#3
+270
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Here you are.

Dec 11, 2014
#4
+270
+8

It has different pronumerals, because it's vector.

vector is not only about how long or how big, it's also about the direction of it.

Dec 11, 2014
#7
+270
0

I don't know the answer yet, I'm gonna ask it to my teacher today.

But I think you are on right line.

Anyway, I will post the full answer here today.

Dec 11, 2014
#8
+25654
+5

OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AC and MQ meet at P.  If O to A = a, O to C = c .  Express O to P in terms of a and c.

1.  $$\small{\text{ If  \vec{Q} = \vec{c} + \frac{3}{4}\vec{a}  and  \vec{M} = \frac{1}{2}\vec{a}  so line  \overline{MQ}  is \frac{1}{2}\vec{a} + \lambda ( \vec{Q}-\vec{M}) = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a})  }}$$

2.  $$\small{\text{ Line  \overline{AC}  is \vec{a} + \mu ( \vec{c}-\vec{a})  }}$$

3.

$$\\\small{\text{ The intersection of the line \overline{MQ} with the line \overline{AC} is }} \\\small{\text{ the point \vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) .}} \\\small{\text{ We equate:  \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) = \vec{a} + \mu ( \vec{c}-\vec{a})  }} \\\small{\text{ so  \lambda (\vec{c}+\frac{1}{4}\vec{a}) - \mu ( \vec{c}-\vec{a}) = \frac{1}{2}\vec{a}  }} \\\small{\text{ The perpendicular vector of (\vec{c}-\vec{a}) is (\vec{c}+\vec{a}), because we have a square and }} \\\small{\text{ (\vec{c}-\vec{a}) * ( \vec{c}+\vec{a} ) = 0, because the dot-product of vectors }} \\\small{\text{ perpendicular to each other is a Zero. }} \\\small{\text{ We multiply our equation with ( \vec{c}+\vec{a} ) and see what passed: }}$$

$$\\\small{\text{  \lambda (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) - \mu \underbrace{( \vec{c}-\vec{a}) (\vec{c}+\vec{a})}_{dot-product = 0} = \frac{1}{2}\vec{a} (\vec{c}+\vec{a})  }} \\\small{\text{ Now we can dissolve after \lambda and receive for \lambda: }} \\\small{\text{  \lambda = \dfrac{ \frac{1}{2}\vec{a} (\vec{c}+\vec{a}) } {(\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) }  }} \\ \boxed{ \small{\text{ \vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) }} = \lambda \vec{c} + (\frac{1}{2}+\frac{\lambda}{4})\vec{a}  }}}$$

Dec 11, 2014
#9
+270
+10

$$\vec{AC}=-\mathbf{a}+\mathbf{c}$$

$$\vec{MQ}=\frac{1}{2}\mathbf{a}+\mathbf{c}-\frac{1}{4}\mathbf{a}$$

$$=\frac{1}{4}\mathbf{a}+\mathbf{c}$$

$$\vec{OP}=\vec{OM}+\vec{MP}$$

$$=\vec{OA}+\vec{AP}$$

$$\vec{MP}=\lambda(\frac{1}{4}\mathbf{a}+\mathbf{c})$$

$$=\frac{\lambda}{4}\mathbf{a}+\lambda \mathbf{c}$$

$$\vec{AP}=\mu(-\mathbf{a}+\mathbf{c})$$

$$=-\mu\mathbf{a}+\mu\mathbf{c}$$

As $$\vec{AP}$$ and $$\vec{MP}$$ intersect, if I let them equal, I can find the $$P$$.

$$\frac{\lambda}{4}\mathbf{a}+\lambda \mathbf{c}$$ $$=-\mu\mathbf{a}+\mu\mathbf{c}$$

Therefore, $$\frac{\lambda}{4}=-\mu$$ and $$\lambda = \mu$$

Then if I subtitute $$\mu$$ to $$\frac{\lambda}{4}=-\mu$$,

$$\frac{\mu}{4}=-\mu$$

$$\mu=-4$$

Sub -4 to $$\vec{MP}=\lambda(\frac{1}{4}\mathbf{a}+\mathbf{c})$$    (As  $$\mu = \lambda$$)

$$\vec{MP}=-4(\frac{1}{4}\mathbf{a}+\mathbf{c})$$

$$=-\mathbf{a}-4\mathbf{c}$$

Finally,

$$\vec{OP} = \vec{OM}+\vec{MP}$$

$$=\frac{1}{2}\mathbf{a}+(-\mathbf{a}-4\mathbf{c})$$

$$=\frac{1}{2}\mathbf{a}-\mathbf{a}-4\mathbf{c}$$

$$=-\frac{1}{2}\mathbf{a}-4\mathbf{c}$$

flflvm97 Dec 11, 2014
#10
+25654
+8

$$P.S. \qquad \lambda= \frac{ \frac{1}{2}\vec{a}(\vec{c}+\vec{a} ) } { (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) } = \frac{ \frac{1}{2}\vec{a}\vec{c} + \frac{1}{2}a^2 }{ c^2 + \vec{c}\vec{a} + \frac{1}{4}\vec{a}\vec{c} +\frac{1}{4}a^2 } =\frac{0 + \frac{1}{2}a^2 }{ a^2+0+0+ \frac{1}{4}a^2 } = \frac{ \frac{1}{2}a^2 }{\frac{5}{4}a^2 } = \frac{2}{5}$$

$$\small{\text{  \vec{a}*\vec{c} =0 \quad \vec{a}  and  \vec{c}  are perpendicular and  a^2=c^2  is a square. }}$$

$$\lambda= \frac{2}{5}$$

$$\vec{OP}=\frac{2}{5}\vec{c}+\frac{3}{5}\vec{a}$$

Sorry flflvm97 you have a missing term in your equation:

$$\vec{MA}+\vec{AP}-\vec{MP}=0 \\ \vec{MP}=\vec{MA}+\vec{AP} \\ \lambda(\frac{1}{4}\vec{a}+\vec{c})={\frac{1}{2}\vec{a}} +\mu(-\vec{a}+\vec{c})$$

.
Dec 14, 2014