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Sagot :
Let's solve this problem step-by-step.
### Step 1: Define the Vectors
Given vectors:
[tex]\[ u = \langle -2, 3, -2 \rangle \][/tex]
[tex]\[ v = \langle -4, 3, 2 \rangle \][/tex]
### Step 2: Compute the Cross Product
To find a vector normal to the plane containing [tex]\( u \)[/tex] and [tex]\( v \)[/tex], we calculate their cross product:
[tex]\[ u \times v \][/tex]
Using the determinant method:
[tex]\[ u \times v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 3 & -2 \\ -4 & 3 & 2 \end{vmatrix} \][/tex]
Expanding this determinant:
[tex]\[ u \times v = \mathbf{i} \left( 3 \cdot 2 - (-2) \cdot 3 \right) - \mathbf{j} \left( -2 \cdot 2 - (-2) \cdot -4 \right) + \mathbf{k} \left( -2 \cdot 3 - (-4) \cdot 3 \right) \][/tex]
Simplify each component:
[tex]\[ u \times v = \mathbf{i} (6 + 6) - \mathbf{j} (-4 - 8) + \mathbf{k} (-6 + 12) \][/tex]
[tex]\[ u \times v = \mathbf{i} (12) - \mathbf{j} (-12) + \mathbf{k} (6) \][/tex]
Thus:
[tex]\[ u \times v = 12 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k} \][/tex]
So the normal vector is:
[tex]\[ \langle 12, 12, 6 \rangle \][/tex]
### Step 3: Calculate the Magnitude of the Normal Vector
To find the magnitude of the normal vector, we use the formula:
[tex]\[ \| \mathbf{n} \| = \sqrt{n_1^2 + n_2^2 + n_3^2} \][/tex]
Substitute the components of the normal vector:
[tex]\[ \| \mathbf{n} \| = \sqrt{12^2 + 12^2 + 6^2} \][/tex]
[tex]\[ \| \mathbf{n} \| = \sqrt{144 + 144 + 36} \][/tex]
[tex]\[ \| \mathbf{n} \| = \sqrt{324} \][/tex]
[tex]\[ \| \mathbf{n} \| = 18 \][/tex]
### Step 4: Find the Unit Vectors
The unit vectors normal to the plane are found by dividing the normal vector by its magnitude.
The unit vector in the direction of the normal vector is:
[tex]\[ \mathbf{u_1} = \frac{1}{18} \langle 12, 12, 6 \rangle \][/tex]
Simplifying each component:
[tex]\[ \mathbf{u_1} = \left\langle \frac{12}{18}, \frac{12}{18}, \frac{6}{18} \right\rangle \][/tex]
[tex]\[ \mathbf{u_1} = \left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right\rangle \][/tex]
And the unit vector in the opposite direction is simply the negative of this vector:
[tex]\[ \mathbf{u_2} = -\mathbf{u_1} \][/tex]
[tex]\[ \mathbf{u_2} = \left\langle -\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right\rangle \][/tex]
### Final Answer:
The two unit vectors normal to the plane containing [tex]\( u \)[/tex] and [tex]\( v \)[/tex] are:
[tex]\[ \mathbf{u_1} = \left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right\rangle \][/tex]
[tex]\[ \mathbf{u_2} = \left\langle -\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right\rangle \][/tex]
### Step 1: Define the Vectors
Given vectors:
[tex]\[ u = \langle -2, 3, -2 \rangle \][/tex]
[tex]\[ v = \langle -4, 3, 2 \rangle \][/tex]
### Step 2: Compute the Cross Product
To find a vector normal to the plane containing [tex]\( u \)[/tex] and [tex]\( v \)[/tex], we calculate their cross product:
[tex]\[ u \times v \][/tex]
Using the determinant method:
[tex]\[ u \times v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 3 & -2 \\ -4 & 3 & 2 \end{vmatrix} \][/tex]
Expanding this determinant:
[tex]\[ u \times v = \mathbf{i} \left( 3 \cdot 2 - (-2) \cdot 3 \right) - \mathbf{j} \left( -2 \cdot 2 - (-2) \cdot -4 \right) + \mathbf{k} \left( -2 \cdot 3 - (-4) \cdot 3 \right) \][/tex]
Simplify each component:
[tex]\[ u \times v = \mathbf{i} (6 + 6) - \mathbf{j} (-4 - 8) + \mathbf{k} (-6 + 12) \][/tex]
[tex]\[ u \times v = \mathbf{i} (12) - \mathbf{j} (-12) + \mathbf{k} (6) \][/tex]
Thus:
[tex]\[ u \times v = 12 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k} \][/tex]
So the normal vector is:
[tex]\[ \langle 12, 12, 6 \rangle \][/tex]
### Step 3: Calculate the Magnitude of the Normal Vector
To find the magnitude of the normal vector, we use the formula:
[tex]\[ \| \mathbf{n} \| = \sqrt{n_1^2 + n_2^2 + n_3^2} \][/tex]
Substitute the components of the normal vector:
[tex]\[ \| \mathbf{n} \| = \sqrt{12^2 + 12^2 + 6^2} \][/tex]
[tex]\[ \| \mathbf{n} \| = \sqrt{144 + 144 + 36} \][/tex]
[tex]\[ \| \mathbf{n} \| = \sqrt{324} \][/tex]
[tex]\[ \| \mathbf{n} \| = 18 \][/tex]
### Step 4: Find the Unit Vectors
The unit vectors normal to the plane are found by dividing the normal vector by its magnitude.
The unit vector in the direction of the normal vector is:
[tex]\[ \mathbf{u_1} = \frac{1}{18} \langle 12, 12, 6 \rangle \][/tex]
Simplifying each component:
[tex]\[ \mathbf{u_1} = \left\langle \frac{12}{18}, \frac{12}{18}, \frac{6}{18} \right\rangle \][/tex]
[tex]\[ \mathbf{u_1} = \left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right\rangle \][/tex]
And the unit vector in the opposite direction is simply the negative of this vector:
[tex]\[ \mathbf{u_2} = -\mathbf{u_1} \][/tex]
[tex]\[ \mathbf{u_2} = \left\langle -\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right\rangle \][/tex]
### Final Answer:
The two unit vectors normal to the plane containing [tex]\( u \)[/tex] and [tex]\( v \)[/tex] are:
[tex]\[ \mathbf{u_1} = \left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right\rangle \][/tex]
[tex]\[ \mathbf{u_2} = \left\langle -\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right\rangle \][/tex]
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