Please give full explanation to the answer. **Question:**Is the vector **v** = ⟨−1, −2, 3⟩ perpendicular to the plane 3x + 6y − 9z + 5 = 0?**Explanation:**In order to determine if the vector **v** is perpendicular to the plane, we need to find out if **v** is normal to the plane. A vector is normal (perpendicular) to a plane if it is parallel to the normal vector of the plane.The equation of the plane is given as 3x + 6y − 9z + 5 = 0. The coefficients of x, y, and z (3, 6, and -9) form the normal vector of the plane.Thus, the normal vector **n** of the plane is ⟨3, 6, −9⟩.To check if **v** is perpendicular to the plane, it should be parallel to **n**. Two vectors are parallel if one is a scalar multiple of the other.By comparing **v** = ⟨−1, −2, 3⟩ with **n** = ⟨3, 6, −9⟩, we can see that there is no scalar k such that k * ⟨3, 6, −9⟩ equals ⟨−1, −2, 3⟩.**Conclusion:**Therefore, the vector **v** = ⟨−1, −2, 3⟩ is not perpendicular to the plane 3x + 6y − 9z + 5 = 0.

Name: Please give full explanation to the answer. **Question:**Is the vector
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: Please give full explanation to the answer. **Question:**Is the vector **v** = ⟨−1, −2, 3⟩ perpendicular to the plane 3x + 6y − 9z + 5 = 0?**Explanation:**In order to determine if the vector **v** is perpendicular to the plane, we need to find out if