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

Question

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

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Thiên Hương 6 months 2021-08-13T16:53:44+00:00 2 Answers 1 views 0

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    0
    2021-08-13T16:54:44+00:00

    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}.

    0
    2021-08-13T16:55:10+00:00

    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°

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )