Dot product of vectors #2

[elleocin]

Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b if |a| = 2|b|

When I subbed the a value into each vector point I ended up with [4b+2b] • [-b]. If I apply the dot product, is it just (4b)(-b) + (2b)? Or do I have to multiply 2b with -b also? :?

This is what I've written:
[2(2b) + b] • [(2b)-3b] = 0
=[4b+2b] • [-b] <------ here is where I'm asking about. Do you have to multiply the 2b by anything if there is no second value in the point?
=(4b)(-b) + ?

 

[Deleted member 4993]

Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b if |a| = 2|b|

When I subbed the a value into each vector point I ended up with [4b+2b] • [-b]. If I apply the dot product, is it just (4b)(-b) + (2b)? Or do I have to multiply 2b with -b also? :?

Here a and b are each vectors

\(\displaystyle (2\vec{a} + \vec{b}) \bullet (\vec{a} - 3\vec{b}) = 0\)

\(\displaystyle ( 2(|a|)^2 - 6 \vec{a}\bullet\vec{b} + \vec{a}\bullet\vec{b} - 3(|b|)^2 = 0\)

\(\displaystyle ( 5(|b|)^2 - 5 \vec{a}\bullet\vec{b} = 0\)

Now continue....

 

[elleocin]

Yeah, simplifying would've really been a smart idea first..

If I'm not mistaken, did you add 2|a|² and 3|b|² to get 5|b|²?

Shouldn't the right equation be: 2(|a|)² - 5a ∙ b - 3|b|² =0

Also, when do you know if you're supposed to distribute? In some questions we multiply the vector points by their corresponding order.

 

[Deleted member 4993]

Yeah, simplifying would've really been a smart idea first..

If I'm not mistaken, did you add 2|a|² and 3|b|² to get 5|b|²? No 2|a|² - 3|b|² = 2(2b)² - 3b² = 8b² - 3b² = 5b²

Shouldn't the right equation be: 2(|a|)² - 5a ∙ b - 3|b|² =0

Also, when do you know if you're supposed to distribute? In some questions we multiply the vector points by their corresponding order.

.

 

[elleocin]

I meant to say, where did you get 5|b|² in your first post?