Question

Find the angle between v = 2i - j and w = 3i + 4jRound nearest tenth of a degree

Answer 1

Answer:

$\theta=79.7^{\circ}$

Step-by-step explanation:

Given that,

v = 2i - j and w = 3i + 4j

We need to find the angle between v and w.

Magnitude of |v|, $|v|=\sqrt{2^2+(-1)^2} =\sqrt5$

Magnitude of |w|, $|w|=\sqrt{3^2+4^2} =5$

The dot product of v and w,

$u{\cdot}w=2(3)+(-1)4\\\\=2$

The formula for the dot product is given by :

$u{\cdot}w=|u||w|\cos\theta\\\\\cos\theta=\dfrac{u{\cdot}w}{|u||w|}\\\\=\dfrac{2}{\sqrt5\times 5}\\\\\theta=\cos^{-1}(\dfrac{2}{\sqrt5\times 5})\\\\\theta=79.69^{\circ}\\\\=79.7^{\circ}$

So, the angle between u and v is $79.7^{\circ}$.

Answer 2

Answer:

The angle between two vectors

 $\alpha = cos^{-1} (\frac{2}{5\sqrt{5} } )$

  ∝ = 79.700°

Step-by-step explanation:

Explanation

Given V =  2i - j and w = 3 i + 4 j

Let '∝' be the angle between the two vectors

           $cos \alpha = \frac{v^{-} .w^{-} }{|v||w|}$

         $cos \alpha = \frac{(2i-j).(3i+4j) }{\sqrt{2^{2}+1^{2} )\sqrt{3^{2} +4^{2} } } }$

        $cos \alpha = \frac{(2(3)-4(1) }{\sqrt{5 )\sqrt{25 } } } = \frac{2}{\sqrt{5})5 } = \frac{2}{5\sqrt{5} }$

          $cos\alpha = \frac{2}{5\sqrt{5} } \\\alpha = cos^{-1} (\frac{2}{5\sqrt{5} } )$

 The angle between two vectors

        ∝ = 79.77°