Question

Quickly find normal vectors for the planes:

1. P₁:−4(x+5)−3(y+3)+5(z−5)=0 n=
2. P₂ :5y+7z+4=0 n=
3. P₃ :−3x−3y−4z−1=5 n=
4. P₄ :7y=4z+3 n=
5. P₅ :3+5y=−6−4x−5z n=

Answer 1

Normal vectors for the planes are directly taken from the coefficients of x, y, and z in their equations, resulting in vectors: (-4, -3, 5), (0, 5, 7), (-3, -3, -4), (0, 7, -4), and (4, 5, 5).

Step-by-step explanation:

To find the normal vectors for the given planes, we use the coefficients of the variables x, y, and z from the equations of the planes. These coefficients directly form the normal vectors to the planes. Here are the normal vectors for each plane:

1. Plane P₁: The equation is −4(x+5) −3(y+3) + 5(z−5)=0. The normal vector n is (−4, −3, 5).
2. Plane P₂: The equation is 5y + 7z + 4 = 0. Since there is no x term, its coefficient is 0. The normal vector n is (0, 5, 7).
3. Plane P₃: The equation is −3x −3y −4z −1 = 5 which simplifies to −3x −3y −4z = 6. The normal vector n is (−3, −3, −4).
4. Plane P₄: The equation is 7y = 4z + 3. We can rewrite this equation as 0x + 7y −4z = 3 to see all components. The normal vector n is (0, 7, −4).
5. Plane P₅: The equation is 3 + 5y = −6 − 4x − 5z. Rearranging gives 4x + 5y + 5z = −9. The normal vector n is (4, 5, 5).