Mathematical Problem with Vectors in French
The image shows a handwritten math problem in French, which includes two parts. The first part is about computing the vectorial product of two vectors, and the second part is about determining the expression of a vector. Let's solve both parts step by step.
1. Compute the vector product (vecteur) of:
\(\overrightarrow{u}(x_u, y_u, z_u) = (x - y, y^2 - x^2, 3z)\)
\(\overrightarrow{v}(x_v, y_v, z_v) = (x, y, z)\)
The cross product of two vectors in three-dimensional space is given by the determinant of the following matrix:
\[\vec{u} \times \vec{v} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
x_u & y_u & z_u \\
x_v & y_v & z_v \\
\end{vmatrix}\]
Substituting the given vector components:
\[
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
x - y & y^2 - x^2 & 3z \\
x & y & z \\
\end{vmatrix}
\]
= \(\hat{i}((y^2 - x^2)z - y(3z))\)
- \(\hat{j}((x - y)z - (3z)x)\)
+ \(\hat{k}((x - y)y - (y^2 - x^2)x)\)
Now let's compute each component individually:
For \(\hat{i}\):
\[ (y^2 - x^2)z - y(3z) = y^2z - x^2z - 3yz \]
For \(\hat{j}\) (note the change of sign because it's the second component):
\[ -((x - y)z - (3z)x) = -xz + yz - 3zx = 2zx - yz \]
For \(\hat{k}\):
\[ (x - y)y - (y^2 - x^2)x = xy - y^2 - y^2x + x^3 \]
Finally, put these components together to get the cross product:
\(\overrightarrow{u} \times \overrightarrow{v} = (y^2z - x^2z - 3yz)\hat{i} + (2zx - yz)\hat{j} + (xy - y^2 - y^2x + x^3)\hat{k}\)
2. Determine the expression of the vector \( \overrightarrow{w} \) as a linear combination of the base vectors \( \hat{i}, \hat{j}, \hat{k} \) knowing that \( \overrightarrow{w} \cdot \overrightarrow{u} = 3 \) and \( \overrightarrow{w} \cdot \overrightarrow{v} = 1 \)
The scalar product (dot product) of two vectors is defined as \( \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \), where \( a_x, a_y, a_z \) and \( b_x, b_y, b_z \) are the components of vectors \( \vec{a} \) and \( \vec{b} \), respectively.
Let's assume \( \overrightarrow{w} = (w_x, w_y, w_z) \).
We have two equations from the problem statement:
\( \overrightarrow{w} \cdot \overrightarrow{u} = w_x(x - y) + w_y(y^2 - x^2) + w_z(3z) = 3 \)
\( \overrightarrow{w} \cdot \overrightarrow{v} = w_xx + w_yy + w_zz = 1 \)
Unfortunately, with only two equations, we cannot uniquely determine three unknowns (\( w_x, w_y, w_z \)). Additional information would be needed to find a unique solution for \( \vec{w} \), but we can write \( \vec{w} \) as the following, leaving the coefficients \( w_x, w_y, w_z \) as variables:
\( \vec{w} = w_x \hat{i} + w_y \hat{j} + w_z \hat{k} \)
Without more information, we can't determine the exact values for \( w_x, w_y, w_z \) that satisfy both given conditions.
