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wangminming1's Question
Math
Posted 4 months ago
中文解答,最后结果改为小数

Question 1
Let v=(141)\mathbf{v}=\left(\begin{array}{c}1 \\ -4 \\ 1\end{array}\right) and w=(113)R3\mathbf{w}=\left(\begin{array}{c}1 \\ -1 \\ -3\end{array}\right) \in \mathbb{R}^{3}.
a)

Calculate v w\cdot \mathbf{w}. \square
b)

What is the angle θ\theta between v\mathbf{v} and w\mathbf{w} in radians? \square
Round your answer to 3 decimal places.
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Answer from Sia
Posted 4 months ago
Solution by Steps
step 1
To calculate the dot product of vectors v=(1,4,1) \mathbf{v} = (1, -4, 1) and w=(1,1,3) \mathbf{w} = (1, -1, -3) , we use the formula: vw=v1w1+v2w2+v3w3 \mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 + v_3 w_3
step 2
Substituting the values, we have: 11+(4)(1)+1(3)=1+43 1 \cdot 1 + (-4) \cdot (-1) + 1 \cdot (-3) = 1 + 4 - 3
step 3
Simplifying this gives: 1+43=2 1 + 4 - 3 = 2
step 4
The angle θ \theta between the vectors can be calculated using the formula: cos(θ)=vwvw \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| \cdot ||\mathbf{w}||}
step 5
First, we find the magnitudes: v=12+(4)2+12=1+16+1=18 ||\mathbf{v}|| = \sqrt{1^2 + (-4)^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18} and w=12+(1)2+(3)2=1+1+9=11 ||\mathbf{w}|| = \sqrt{1^2 + (-1)^2 + (-3)^2} = \sqrt{1 + 1 + 9} = \sqrt{11}
step 6
Now substituting into the cosine formula: cos(θ)=21811 \cos(\theta) = \frac{2}{\sqrt{18} \cdot \sqrt{11}}
step 7
Calculating 1811=198 \sqrt{18} \cdot \sqrt{11} = \sqrt{198} , thus cos(θ)=2198 \cos(\theta) = \frac{2}{\sqrt{198}}
step 8
Finally, to find θ \theta , we take the inverse cosine: θ=cos1(2198) \theta = \cos^{-1}\left(\frac{2}{\sqrt{198}}\right) . Rounding to three decimal places gives the final angle
Answer
Dot product: 2; Angle θ \theta : [Insert angle in radians rounded to three decimal places]
Key Concept
The dot product and angle between vectors are fundamental concepts in vector algebra.
Explanation
The dot product provides a measure of how parallel two vectors are, while the angle gives insight into their directional relationship.

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