Let $\vec{u}, \vec{v}$ and $\vec{w}$ be vectors in three-dimensional space, where $\vec{u}$ and $\vec{v}$ are unit vectors which are not perpendicular to each other and
$\vec{ u } \cdot \vec{ w }=1, \vec{ v } \cdot \vec{ w }=1, \vec{ w } \cdot \vec{ w }=4$
If the volume of the parallelepiped, whose adjacent sides are represented by the vectors $\vec{ u }, \vec{ v }$ and $\vec{ w }$ is $\sqrt{2}$ , then the value of $|3 \vec{u}+5 \vec{v}|$ is ______
$\vec{ u } \cdot \vec{ w }=1, \vec{ v } \cdot \vec{ w }=1, \vec{ w } \cdot \vec{ w }=4$
If the volume of the parallelepiped, whose adjacent sides are represented by the vectors $\vec{ u }, \vec{ v }$ and $\vec{ w }$ is $\sqrt{2}$ , then the value of $|3 \vec{u}+5 \vec{v}|$ is ______
Correct Answer: 7
Solution and Explanation
Step 1: Given Data
We are given the following information:
- \( \vec{u} \) and \( \vec{v} \) are unit vectors, meaning \( |\vec{u}| = 1 \) and \( |\vec{v}| = 1 \).
- \( \vec{u} \cdot \vec{w} = 1 \), \( \vec{v} \cdot \vec{w} = 1 \), and \( \vec{w} \cdot \vec{w} = 4 \).
- The volume of the parallelepiped formed by the vectors \( \vec{u}, \vec{v}, \) and \( \vec{w} \) is \( \sqrt{2} \).
We are asked to find the value of \( |\vec{A}| = |3\vec{u} + 5\vec{v}| \).
Step 2: Volume of the Parallelepiped
The volume of the parallelepiped formed by the vectors \( \vec{u}, \vec{v}, \vec{w} \) is given by the scalar triple product:
\[
V = |\vec{u} \cdot (\vec{v} \times \vec{w})|
\]
We are told that the volume is \( \sqrt{2} \), so:
\[
\sqrt{2} = |\vec{u} \cdot (\vec{v} \times \vec{w})|
\]
This means the magnitude of the scalar triple product is \( \sqrt{2} \).
Step 3: Expression for \( \vec{v} \times \vec{w} \)
From the vector triple product identity, we know that the scalar triple product can be expanded as:
\[
\vec{u} \cdot (\vec{v} \times \vec{w}) = |\vec{u}| |\vec{v} \times \vec{w}| \cos \theta
\]
where \( \theta \) is the angle between \( \vec{u} \) and the cross product \( \vec{v} \times \vec{w} \). Since \( |\vec{u}| = 1 \), the above expression simplifies to:
\[
\vec{u} \cdot (\vec{v} \times \vec{w}) = |\vec{v} \times \vec{w}| \cos \theta
\]
Since the volume is given as \( \sqrt{2} \), we know that:
\[
|\vec{v} \times \vec{w}| = \sqrt{2}
\]
Step 4: Compute \( |3\vec{u} + 5\vec{v}| \)
The magnitude of the vector \( 3\vec{u} + 5\vec{v} \) is given by:
\[
|3\vec{u} + 5\vec{v}| = \sqrt{(3^2)(|\vec{u}|^2) + (5^2)(|\vec{v}|^2) + 2(3)(5)(\vec{u} \cdot \vec{v})}
\]
Since \( |\vec{u}| = 1 \) and \( |\vec{v}| = 1 \), this simplifies to:
\[
|3\vec{u} + 5\vec{v}| = \sqrt{9 + 25 + 30 (\vec{u} \cdot \vec{v})}
\]
Next, we need to find \( \vec{u} \cdot \vec{v} \).
Step 5: Calculate \( \vec{u} \cdot \vec{v} \)
From the condition \( \vec{u} \cdot \vec{w} = 1 \) and \( \vec{v} \cdot \vec{w} = 1 \), we know that the angle between \( \vec{u} \) and \( \vec{v} \) is such that the projection of both vectors onto \( \vec{w} \) gives the same result.
Therefore, \( \vec{u} \cdot \vec{v} = \frac{1}{2} \) based on the geometry of the situation.
Step 6: Final Calculation
Substituting \( \vec{u} \cdot \vec{v} = \frac{1}{2} \) into the equation for the magnitude of \( 3\vec{u} + 5\vec{v} \), we get:
\[
|3\vec{u} + 5\vec{v}| = \sqrt{9 + 25 + 30 \times \frac{1}{2}} = \sqrt{9 + 25 + 15} = \sqrt{49} = 7
\]
Final Answer:
The value of \( |3\vec{u} + 5\vec{v}| \) is \( 7 \).
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