Math  /  Geometry

QuestionSuppose that v=22\|\vec{v}\|=22 and w=6\|\vec{w}\|=6. Suppose also that, when drawn starting at the same point, v\vec{v} and w\vec{w} make an angle of 5π6\frac{5 \pi}{6} radians. (A.) Find w+v\|\vec{w}+\vec{v}\| and round to two decimal places. w+v=\|\vec{w}+\vec{v}\|=\square (B.) Find wv\|\vec{w}-\vec{v}\| and round to two decimal places. wv=\|\vec{w}-\vec{v}\|=\square

Studdy Solution

STEP 1

1. The magnitudes of vectors v\vec{v} and w\vec{w} are given as v=22\|\vec{v}\| = 22 and w=6\|\vec{w}\| = 6.
2. The angle between v\vec{v} and w\vec{w} is 5π6\frac{5\pi}{6} radians.
3. We will use the law of cosines for vectors to find the magnitudes of w+v\|\vec{w}+\vec{v}\| and wv\|\vec{w}-\vec{v}\|.

STEP 2

1. Calculate w+v\|\vec{w}+\vec{v}\| using the law of cosines.
2. Calculate wv\|\vec{w}-\vec{v}\| using the law of cosines.

STEP 3

To find w+v\|\vec{w}+\vec{v}\|, use the formula for the magnitude of the sum of two vectors:
w+v=w2+v2+2wvcos(θ)\|\vec{w}+\vec{v}\| = \sqrt{\|\vec{w}\|^2 + \|\vec{v}\|^2 + 2\|\vec{w}\|\|\vec{v}\|\cos(\theta)}
where θ=5π6\theta = \frac{5\pi}{6}.

STEP 4

Substitute the known values into the formula:
w+v=62+222+2622cos(5π6)\|\vec{w}+\vec{v}\| = \sqrt{6^2 + 22^2 + 2 \cdot 6 \cdot 22 \cdot \cos\left(\frac{5\pi}{6}\right)}

STEP 5

Calculate the cosine of 5π6\frac{5\pi}{6}:
cos(5π6)=32\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}

STEP 6

Substitute cos(5π6)=32\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} into the equation:
w+v=62+222+2622(32)\|\vec{w}+\vec{v}\| = \sqrt{6^2 + 22^2 + 2 \cdot 6 \cdot 22 \cdot \left(-\frac{\sqrt{3}}{2}\right)}

STEP 7

Simplify the expression:
w+v=36+4841323\|\vec{w}+\vec{v}\| = \sqrt{36 + 484 - 132\sqrt{3}}

STEP 8

Calculate the numerical value and round to two decimal places:
w+v520132318.68\|\vec{w}+\vec{v}\| \approx \sqrt{520 - 132\sqrt{3}} \approx 18.68

STEP 9

To find wv\|\vec{w}-\vec{v}\|, use the formula for the magnitude of the difference of two vectors:
wv=w2+v22wvcos(θ)\|\vec{w}-\vec{v}\| = \sqrt{\|\vec{w}\|^2 + \|\vec{v}\|^2 - 2\|\vec{w}\|\|\vec{v}\|\cos(\theta)}
where θ=5π6\theta = \frac{5\pi}{6}.

STEP 10

Substitute the known values into the formula:
wv=62+2222622cos(5π6)\|\vec{w}-\vec{v}\| = \sqrt{6^2 + 22^2 - 2 \cdot 6 \cdot 22 \cdot \cos\left(\frac{5\pi}{6}\right)}

STEP 11

Substitute cos(5π6)=32\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} into the equation:
wv=62+222+262232\|\vec{w}-\vec{v}\| = \sqrt{6^2 + 22^2 + 2 \cdot 6 \cdot 22 \cdot \frac{\sqrt{3}}{2}}

STEP 12

Simplify the expression:
wv=36+484+1323\|\vec{w}-\vec{v}\| = \sqrt{36 + 484 + 132\sqrt{3}}

STEP 13

Calculate the numerical value and round to two decimal places:
wv520+132328.68\|\vec{w}-\vec{v}\| \approx \sqrt{520 + 132\sqrt{3}} \approx 28.68
The solutions are: (A.) w+v18.68\|\vec{w}+\vec{v}\| \approx 18.68 (B.) wv28.68\|\vec{w}-\vec{v}\| \approx 28.68

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