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Problem 315

# Find the distance between the vectors $$\mathrm{u}$$ and $$\mathrm{v}$$ where i) $$\mathrm{u}=(1,7), \mathrm{v}=(6,-5) ;$$ ii) $\mathrm{u}=(3,-5,4), \mathrm{v}=(6,2,-1)$ iii) $$\mathrm{u}=(5,3,-2,-4,1) \cdot \mathrm{v}=(2,-1,0,-7,2)$$

Expert verified
i) $$d(\mathrm{u}, \mathrm{v}) = 13$$ ii) $$d(\mathrm{u}, \mathrm{v}) = \sqrt{83}$$ iii) $$d(\mathrm{u}, \mathrm{v}) = \sqrt{39}$$
See the step by step solution

## Step 1: Subtract Vectors

Subtract the vectors component-wise: $$\mathrm{w} = \mathrm{u} - \mathrm{v} = (1 - 6, 7 - (-5)) = (-5, 12)$$

## Step 2: Calculate the Magnitude of the Resulting Vector

Find the magnitude of the vector $$\mathrm{w}$$ using the formula: $$\|\mathrm{w}\| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$ So, the distance between the vectors $$\mathrm{u}$$ and $$\mathrm{v}$$ is: $$d(\mathrm{u}, \mathrm{v}) = 13$$ ii) For given vectors $$\mathrm{u} = (3, -5, 4)$$ and $$\mathrm{v} = (6, 2, -1)$$:

## Step 1: Subtract Vectors

Subtract the vectors component-wise: $$\mathrm{w} = \mathrm{u} - \mathrm{v} = (3-6, -5-2, 4-(-1))=( -3,-7, 5)$$

## Step 2: Calculate the Magnitude of the Resulting Vector

Find the magnitude of the vector $$\mathrm{w}$$ using the formula: $$\|\mathrm{w}\| = \sqrt{(-3)^2 + (-7)^2 + 5^2} = \sqrt{9 + 49 + 25} = \sqrt{83}$$ So, the distance between the vectors $$\mathrm{u}$$ and $$\mathrm{v}$$ is: $$d(\mathrm{u}, \mathrm{v}) = \sqrt{83}$$ iii) For given vectors $$\mathrm{u} = (5, 3, -2, -4, 1)$$ and $$\mathrm{v} = (2, -1, 0, -7, 2)$$:

## Step 1: Subtract Vectors

Subtract the vectors component-wise: $$\mathrm{w} = \mathrm{u} - \mathrm{v} = (5-2,3-(-1),-2-0,-4-(-7),1-2)=(3,4,-2,3,-1)$$

## Step 2: Calculate the Magnitude of the Resulting Vector

Find the magnitude of the vector $$\mathrm{w}$$ using the formula: $$\|\mathrm{w}\| = \sqrt{3^2 + 4^2 + (-2)^2 + 3^2 + (-1)^2} = \sqrt{9 + 16 + 4 + 9 + 1} = \sqrt{39}$$ So, the distance between the vectors $$\mathrm{u}$$ and $$\mathrm{v}$$ is: $$d(\mathrm{u}, \mathrm{v}) = \sqrt{39}$$

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