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Given the interval 0<θ<π/2. Find the angle θ which is formed by the line y = -2x+4 and y = 3x-3

Show your work as well, thank you!​

User Rsteg
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2 Answers

1 vote

Consider two vector-valued functions,


\vec r(t) = \left\langle t, -2t+4\right\rangle \text{ and } \vec s(t) = \left\langle t, 3t-3\right\rangle

Differentiate both to get the corresponding tangent/direction vectors:


(\mathrm d\vec r(t))/(\mathrm dt) = \left\langle1,-2\right\rangle \text{ and } (\mathrm d\vec s(t))/(\mathrm dt) = \left\langle1,3\right\rangle

Recall the dot product identity: for two vectors
\vec a and
\vec b, we have


\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)

where
\theta is the angle between them.

We have


\langle1,-2\rangle \cdot \langle1,3\rangle = \|\langle1,-2\rangle\| \|\langle1,3\rangle\| \cos(\theta) \\\\ 1*1 + (-2)*3 = √(1^2 + (-2)^2) * √(1^2+3^2) \cos(\theta) \\\\ \cos(\theta) = (-5)/(\sqrt5*√(10)) = -\frac1{\sqrt2} \\\\ \implies \theta = \cos^(-1)\left(-\frac1{\sqrt2}\right) = \frac{3\pi}4

Then the acute angle between the lines is π/4.

User Kyle KIM
by
8.7k points
4 votes

Answer:


\rm\displaystyle \theta = (\pi)/(4)

Explanation:

we want to find the acute angle θ (as θ is between (0,π/2)) formed by the line y=-2x+4 and y=3x-3 to do so we can consider the following formula:


\displaystyle\tan( \theta) = \bigg| ( m_(2) - m_(1) )/(1 + m_(1) m_(2) ) \bigg |


\rm \displaystyle \implies\theta = \arctan \left( \bigg | ( m_(2) - m_(1) )/(1 + m_(1) m_(2) ) \bigg | \right)

From the first equation we obtain that
m_1 is -2 and from the second that
m_2 is 3 therefore substitute:


\rm\displaystyle \theta = \arctan \left( \bigg | ( 3 - ( - 2) )/(1 + ( - 2) (3)) \bigg | \right)

simplify multiplication:


\rm\displaystyle \theta = \arctan \left( \bigg | ( 3 - ( - 2) )/(1 + ( - 6)) \bigg | \right)

simplify Parentheses:


\rm\displaystyle \theta = \arctan \left( \bigg | ( 3 + 2 )/(1 + ( - 6)) \bigg | \right)

simplify addition:


\rm\displaystyle \theta = \arctan \left( \bigg | ( 5 )/( - 5) \bigg | \right)

simplify division:


\rm\displaystyle \theta = \arctan \left( | - 1| \right)

calculate the absolute of -1:


\rm\displaystyle \theta = \arctan \left( 1\right)

calculate the inverse function:


\rm\displaystyle \theta = (\pi)/(4)

hence,

the angle θ which is formed by the line y = -2x+4 and y = 3x-3 is π/4

(for more info about the formula refer the attachment thank you!)

User Fraser Dale
by
8.9k points

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