Math  /  Geometry

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

Studdy Solution
For plane d, the equation is 12xy+7z+4=012x-y+7z+4=0. The coefficients of x, y, and z are the components of the normal vector.
nd=[1217]\vec{n_d} = \begin{bmatrix}12 \\ -1 \\7 \end{bmatrix}The normal vectors to the given planes area. na=[410]\vec{n_a} = \begin{bmatrix}4 \\ -1 \\0 \end{bmatrix}b. nb=[021]\vec{n_b} = \begin{bmatrix}0 \\2 \\ -1 \end{bmatrix}c. nc=[121]\vec{n_c} = \begin{bmatrix}1 \\2 \\ -1 \end{bmatrix}d. nd=[1217]\vec{n_d} = \begin{bmatrix}12 \\ -1 \\7 \end{bmatrix}

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