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[SOLVED] If a and b are unit vectors...

Raerin

Member
Oct 7, 2013
46
If a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?

Is the answer -1 by any chance? If not...

I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: If a nd b are unit vecotrs...

If a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?

Is the answer -1 by any chance? If not...

I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.
Note that

\[\begin{aligned} (2\mathbf{a}-\mathbf{b}) \cdot (\mathbf{a}+3\mathbf{b}) &= 2\mathbf{a}\cdot\mathbf{a} +6\mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} -3\mathbf{b}\cdot\mathbf{b} \\ &= 2\|\mathbf{a}\|^2 +5\mathbf{a}\cdot\mathbf{b} - 3\|\mathbf{b}\|^2\\ &= 5\mathbf{a}\cdot\mathbf{b} - 1\quad\text{since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors}\end{aligned}\]

Since $\|\mathbf{a}+\mathbf{b}\| = \sqrt{2}$, squaring both sides and expanding via dot product leaves you with
\[\|\mathbf{a}\|^2+ 2\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 = 2 \implies 2\mathbf{a}\cdot\mathbf{b} = 0\implies \mathbf{a}\cdot\mathbf{b} = 0\]

Therefore, we now have that

\[(2\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}+3\mathbf{b}) = 5\mathbf{a}\cdot\mathbf{b} - 1 = -1\]

So yes, your answer is correct.
 

Raerin

Member
Oct 7, 2013
46
Re: If a nd b are unit vecotrs...

Note that

\[\begin{aligned} (2\mathbf{a}-\mathbf{b}) \cdot (\mathbf{a}+3\mathbf{b}) &= 2\mathbf{a}\cdot\mathbf{a} +6\mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} -3\mathbf{b}\cdot\mathbf{b} \\ &= 2\|\mathbf{a}\|^2 +5\mathbf{a}\cdot\mathbf{b} - 3\|\mathbf{b}\|^2\\ &= 5\mathbf{a}\cdot\mathbf{b} - 1\quad\text{since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors}\end{aligned}\]

Since $\|\mathbf{a}+\mathbf{b}\| = \sqrt{2}$, squaring both sides and expanding via dot product leaves you with
\[\|\mathbf{a}\|^2+ 2\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 = 2 \implies 2\mathbf{a}\cdot\mathbf{b} = 0\implies \mathbf{a}\cdot\mathbf{b} = 0\]

Therefore, we now have that

\[(2\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}+3\mathbf{b}) = 5\mathbf{a}\cdot\mathbf{b} - 1 = -1\]

So yes, your answer is correct.
I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0?

Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: If a nd b are unit vecotrs...

I don't understand how 2a . b = 0 becomes a . b = 0.

Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?
Since $\mathbf{a}\cdot\mathbf{b}$ is a scalar, then by the zero product property $2\mathbf{a}\cdot \mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\mathbf{a}\cdot\mathbf{b}=0$ (which is the correct choice). With that result, you can now substitute zero in for $\mathbf{a}\cdot\mathbf{b}$ in the simplified form of $(2\mathbf{a}-\mathbf{b})\cdot(a+3\mathbf{b})$ to get $5\mathbf{a}\cdot\mathbf{b} - 1 = 5(0) - 1 = -1$.

I hope this clarifies things! (Bigsmile)