Find the value of "m" given two vector.

[ratezz]

So I have been struggling with this question for a whole afternoon, I tried many methods such as dot product and by multiplying cos 60 which is 1/2 into the equation and solve for "m," but the answer keeps showing a false number that does not correspond to the textbook answer. Can anyone please help me and provide me with steps throughout? I would really appreciate! Thank you!

 

[Deleted member 4993]

So I have been struggling with this question for a whole afternoon, I tried many methods such as dot product and by multiplying cos 60 which is 1/2 into the equation and solve for "m," but the answer keeps showing a false number that does not correspond to the textbook answer. Can anyone please help me and provide me with steps throughout? I would really appreciate! Thank you!

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[pka]

We are given that \(\vec{a}\cdot\vec{a}=\vec{b}\cdot\vec{b}=1\) also \(\vec{a}\cdot\vec{b}=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
Can you post reasons for those? Please do.
Of course we must expand \((\vec{a}-3\vec{b})\cdot(m\vec{a}+\vec{b})=~?\) Please show us.

 

[skeeter]

you could use components, aligning vector a with the positive x-axis …

[MATH]\vec{a} = \hat{i} + 0 \hat{j}[/MATH][MATH]\vec{b} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}[/MATH]
you can then express [MATH]\vec{a} - 3\vec{b}[/MATH] and [MATH]m\vec{a} + \vec{b}[/MATH] in component form, then set the scalar product of the two resultant vectors equal to zero.

 

[Deleted member 4993]

you could use components, aligning vector a with the positive x-axis …

[MATH]\vec{a} = \hat{i} + 0 \hat{j}[/MATH][MATH]\vec{b} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}[/MATH]
you can then express [MATH]\vec{a} - 3\vec{b}[/MATH] and [MATH]m\vec{a} + \vec{b}[/MATH] in component form, then set the scalar product of the two resultant vectors equal to zero.

Or:

(a − 3b)⋅(ma + b) = 0 (since those are perpendicular to each other)

Distribute & expand left-hand-side of the equation above and use the results provided in response #3

 

[pka]

you could use components, aligning vector a with the positive x-axis …
[MATH]\vec{a} = \hat{i} + 0 \hat{j}[/MATH][MATH]\vec{b} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}[/MATH]you can then express [MATH]\vec{a} - 3\vec{b}[/MATH] and [MATH]m\vec{a} + \vec{b}[/MATH]

We do not know that these are three dimensional vectors in the form \((\vec{i},\vec{j},\vec{k})\)
As unit vectors we do know \(\|\vec{a}\|=\|\vec{b}\|=1\)
We know that \(\vec{a}\cdot\vec{a}=\vec{b}\cdot\vec{b}=1\) also \(\vec{a}\cdot\vec{b}=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
Moreover expansion \((\vec{a}-3\vec{b})\cdot(m\vec{a}+\vec{b})=m+\vec{a}\cdot\vec{b}-3m\left(\vec{a}\cdot\vec{b}\right)-3=0\)

 

[skeeter]

We do not know that these are three dimensional vectors in the form \((\vec{i},\vec{j},\vec{k})\)
As unit vectors we do know \(\|\vec{a}\|=\|\vec{b}\|=1\)
We know that \(\vec{a}\cdot\vec{a}=\vec{b}\cdot\vec{b}=1\) also \(\vec{a}\cdot\vec{b}=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
Moreover expansion \((\vec{a}-3\vec{b})\cdot(m\vec{a}+\vec{b})=m+\vec{a}\cdot\vec{b}-3m\left(\vec{a}\cdot\vec{b}\right)-3=0\)

So, you're saying vectors a and b do not lie in the same plane or can be translated to lie in the same plane?