Beginner Vector Geometry #3

Geometry Level 2

Let θ \theta be the angle formed by two vectors a = ( 3 , 4 ) \overrightarrow{a}=(3,~-4) and b = ( 15 , 8 ) \overrightarrow{b}=(15,~8) .

Then cos θ = q p \cos\theta=\dfrac{q}{p} .

Given that 0 < θ < π 2 0<\theta<\dfrac{\pi}{2} , and p , q p,~q are coprime integers, find the value of p + q p+q .


This problem is a part of <Beginner Vector Geometry> series .


The answer is 98.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Boi (보이)
Jul 2, 2017

Note that a b = a b cos θ \overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta .

From the question, we see that a b = 45 32 = 13 \overrightarrow{a}\cdot\overrightarrow{b}=45-32=13 .

Also, a = 3 2 + ( 4 ) 2 = 5 |\overrightarrow{a}|=\sqrt{3^2+(-4)^2}=5 and b = 1 5 2 + 8 2 = 17 |\overrightarrow{b}|=\sqrt{15^2+8^2}=17 .


Therefore,

cos θ = a b cos θ a b = a b a b = 13 85 \begin{aligned} \cos\theta&=\dfrac{|\overrightarrow{a}||\overrightarrow{b}|\cos\theta}{|\overrightarrow{a}||\overrightarrow{b}|} \\ &=\dfrac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|} \\ &=\boxed{\dfrac{13}{85}} \end{aligned}

p + q = 85 + 13 = 98 \therefore~p+q=85+13=\boxed{98} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...