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5. 已知向量

\vec{a} ， \vec{b}

，满足

|\vec{a}|=1,|\vec{b}|=4

，若对任意模为 2 的向量 $\vec{c}...

Mar 8, 2024

5. 已知向量

\vec{a} ， \vec{b}

，满足

|\vec{a}|=1,|\vec{b}|=4

，若对任意模为 2 的向量

\vec{c}

，均有

|\vec{a} \cdot \vec{c}|+|\vec{b} \cdot \vec{c}| \leq 2 \sqrt{21}

，则向量

\vec{a}, \vec{b}

的夹角的取值范围为

Solution by Steps

step 1

Use the Cauchy-Schwarz inequality to find the maximum value of

|\vec{a} \cdot \vec{c}| + |\vec{b} \cdot \vec{c}|

given

|\vec{c}| = 2

step 2

According to the Cauchy-Schwarz inequality,

|\vec{x} \cdot \vec{y}| \leq |\vec{x}| \cdot |\vec{y}|

. Apply this to

\vec{a}

and

\vec{c}

, and

\vec{b}

and

\vec{c}

step 3

Since

|\vec{a}| = 1

and

|\vec{b}| = 4

, we have

|\vec{a} \cdot \vec{c}| \leq 1 \cdot 2

and

|\vec{b} \cdot \vec{c}| \leq 4 \cdot 2

step 4

The maximum value of

|\vec{a} \cdot \vec{c}| + |\vec{b} \cdot \vec{c}|

is therefore

2 + 8 = 10

step 5

Given

|\vec{a} \cdot \vec{c}| + |\vec{b} \cdot \vec{c}| \leq 2 \sqrt{21}

, compare this with the maximum value found in step 4

step 6

Since 2 \sqrt{21} < 10 , the inequality

|\vec{a} \cdot \vec{c}| + |\vec{b} \cdot \vec{c}| \leq 2 \sqrt{21}

is always satisfied

step 7

To find the angle between

\vec{a}

and

\vec{b}

, use the dot product formula

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)

step 8

Substitute

|\vec{a}| = 1

and

|\vec{b}| = 4

into the dot product formula to get

\vec{a} \cdot \vec{b} = 4 \cos(\theta)

step 9

Since the maximum value of

|\vec{a} \cdot \vec{c}| + |\vec{b} \cdot \vec{c}|

does not depend on the angle between

\vec{a}

and

\vec{b}

, the range of

\theta

is not restricted by the given inequality

Answer

The angle

\theta

between

\vec{a}

and

\vec{b}

can take any value from 0 to

\pi

radians (or 0 to 180 degrees).

Key Concept

Cauchy-Schwarz Inequality and Dot Product

Explanation

The Cauchy-Schwarz inequality is used to find the maximum value of the sum of the absolute values of the dot products of vectors with a fixed magnitude. The dot product formula relates the angle between two vectors to their magnitudes and the dot product. In this case, the given inequality does not restrict the angle between

\vec{a}

and

\vec{b}

, so it can be any value within the possible range for angles between vectors.