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QuestionTrova il vettore normale ai seguenti piani: a. 4x−y+5=04x - y + 5 = 0, b. 2y−z+1=02y - z + 1 = 0, c. x+2y−z=0x + 2y - z = 0, d. 12x−y+7z+4=012x - y + 7z + 4 = 0.

Studdy Solution

STEP 1

Assumptions1. The equations given are of planes in three-dimensional space. . The normal vector to a plane is given by the coefficients of x, y, and z in the equation of the plane.

STEP 2

For plane a, the equation is 4x−y+5=04x-y+5=0. The coefficients of x, y, and z (in this case z is not present, so its coefficient is0) are the components of the normal vector.
na⃗=[4−10]\vec{n_a} = \begin{bmatrix}4 \\ -1 \\0 \end{bmatrix}

STEP 3

For plane b, the equation is 2y−z+1=02y-z+1=0. Again, the coefficients of x, y, and z are the components of the normal vector. Here, x is not present, so its coefficient is0.
nb⃗=[02−1]\vec{n_b} = \begin{bmatrix}0 \\2 \\ -1 \end{bmatrix}

STEP 4

For plane c, the equation is x+2y−z=0x+2y-z=0. The coefficients of x, y, and z are the components of the normal vector.
nc⃗=[12−1]\vec{n_c} = \begin{bmatrix}1 \\2 \\ -1 \end{bmatrix}

STEP 5

For plane d, the equation is 12x−y+7z+4=012x-y+7z+4=0. The coefficients of x, y, and z are the components of the normal vector.
nd⃗=[12−17]\vec{n_d} = \begin{bmatrix}12 \\ -1 \\7 \end{bmatrix}The normal vectors to the given planes area. na⃗=[4−10]\vec{n_a} = \begin{bmatrix}4 \\ -1 \\0 \end{bmatrix}b. nb⃗=[02−1]\vec{n_b} = \begin{bmatrix}0 \\2 \\ -1 \end{bmatrix}c. nc⃗=[12−1]\vec{n_c} = \begin{bmatrix}1 \\2 \\ -1 \end{bmatrix}d. nd⃗=[12−17]\vec{n_d} = \begin{bmatrix}12 \\ -1 \\7 \end{bmatrix}

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