Let the vectors  $\vec{y}, \vec{u}, \vec{v} \in\R^3 $ where $\vec{y}=3\ 2\ 4^T, \vec{u}= 1\ 1\ 0^T, \vec{v}= -1\ 1\ 2^T$, $u =(1, 1, 0)^T,v =(−1, 1, 2)^T$. The distance from  y to the plane in $R^3$ spanned by u ,v is  

a. $\frac{3}{\sqrt{3}}$   b.$\frac{\sqrt{186}}{3}$  

c.5/3  

d.$\frac{5}{\sqrt{3}}$  

e.$3\sqrt{3}$  


To determine the distance from the vector $\vec{y}$ to the plane spanned by the vectors $\vec{u}$ and $\vec{v}$ in $\mathbb{R}^3$, we can follow these steps:

1. **Calculate the normal vector to the plane:**
   The plane spanned by $\vec{u}$ and $\vec{v}$ can be represented by a normal vector $\vec{n}$ which is perpendicular to both $\vec{u}$ and $\vec{v}$. This normal vector can be found using the cross product:
   \[
   \vec{n} = \vec{u} \times \vec{v}
   \]
   
   Given:
   \[
   \vec{u} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \vec{v} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}
   \]

   The cross product $\vec{u} \times \vec{v}$ is calculated as:
   \[
   \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ -1 & 1 & 2 \end{vmatrix}
   \]
   \[
   = \mathbf{i} (1 \cdot 2 - 0 \cdot 1) - \mathbf{j} (1 \cdot 2 - 0 \cdot (-1)) + \mathbf{k} (1 \cdot 1 - 1 \cdot (-1))
   \]
   \[
   = \mathbf{i} (2) - \mathbf{j} (2) + \mathbf{k} (2)
   \]
   \[
   = \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix}
   \]
   \[
   = 2 \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}
   \]

2. **Normalize the normal vector:**
   The normal vector $\vec{n}$ can be normalized by dividing it by its magnitude:
   \[
   \|\vec{n}\| = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}
   \]
   \[
   \vec{n}_{normalized} = \frac{1}{2\sqrt{3}} \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix} = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}
   \]

3. **Calculate the distance from $\vec{y}$ to the plane:**
   The distance from a point $\vec{y}$ to a plane with a normal vector $\vec{n}$ is given by the absolute value of the dot product of $\vec{y}$ and the normalized normal vector $\vec{n}_{normalized}$:
   \[
   \vec{y} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}
   \]
   \[
   \text{Distance} = \left| \vec{y} \cdot \vec{n}_{normalized} \right|
   \]
   \[
   = \left| \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} \cdot \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \right|
   \]
   \[
   = \left| \frac{1}{\sqrt{3}} (3 \cdot 1 + 2 \cdot (-1) + 4 \cdot 1) \right|
   \]
   \[
   = \left| \frac{1}{\sqrt{3}} (3 - 2 + 4) \right|
   \]
   \[
   = \left| \frac{1}{\sqrt{3}} \cdot 5 \right|
   \]
   \[
   = \frac{5}{\sqrt{3}}
   \]

Therefore, the distance from $\vec{y}$ to the plane spanned by $\vec{u}$ and $\vec{v}$ is $\frac{5}{\sqrt{3}}$, which corresponds to option d.

Would you like any more details or have any questions about this solution?

Here are five questions you might want to ask next:
1. How do you perform the dot product of two vectors?
2. How do you find the magnitude of a vector?
3. What is the cross product of two vectors in $\mathbb{R}^3$?
4. How do you normalize a vector?
5. What is the geometric interpretation of the dot product?

**Tip:** When dealing with vectors and planes, visualizing them in a coordinate system can greatly help in understanding the spatial relationships between them.

Calculate the distance from vector y = (3, 2, 4) to the plane spanned by vectors u = (1, 1, 0) and v = (-1, 1, 2) in 3D space. Learn how to apply dot product and cross product formulas to find the distance. Suitable for college-level mathematics.

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