Problem 2 - Calculus Early Transcendentals

Question

Find a vector normal to the plane \(-2 x-3 y+4 z=12\)

Step-by-Step Solution

Answer

Answer: The vector normal to the plane is \(\begin{pmatrix} -2 \\ -3 \\ 4\end{pmatrix}\).

Step 1: Identify coefficients

From the given equation of the plane, \(-2 x - 3 y + 4z = 12\), we can identify the coefficients \(a = -2\), \(b = -3\), and \(c = 4\).

Step 2: Form the normal vector

With the identified coefficients, we can form the normal vector to the given plane. The normal vector will have components \((a, b, c)\) directly. Therefore, the normal vector is \(\begin{pmatrix} -2 \\ -3 \\ 4\end{pmatrix}\). So the vector normal to the plane \(-2 x - 3 y + 4z = 12\) is \(\begin{pmatrix} -2 \\ -3 \\ 4\end{pmatrix}\).