Find the area of the part of the plane $5x+4y+z=20$ that lies in the first octant.

• A. $40$
• B. $30$
• C. $24$
• D. $20$

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the problem. The plane 5x + 4y + z = 20 intersects the coordinate axes in the first octant. The first octant is where x, y, and z are all positive. To find the area of the triangular region formed by the plane in the first octant, we need to determine the points where the plane intersects the x-axis, y-axis, and z-axis.
2. Step 2: Find the intercepts of the plane. Set two variables to zero at a time to find the intercepts: - For the x-intercept, set y = 0 and z = 0, solve 5x = 20 to get x = 4. - For the y-intercept, set x = 0 and z = 0, solve 4y = 20 to get y = 5. - For the z-intercept, set x = 0 and y = 0, solve z = 20 to get z = 20.
3. Step 3: Visualize the triangular region. The intercepts (4, 0, 0), (0, 5, 0), and (0, 0, 20) form a triangle in the first octant. The area of this triangle can be calculated using the formula for the area of a triangle in 3D space: \( \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| \), where \( \vec{AB} \) and \( \vec{AC} \) are vectors formed by the vertices of the triangle.
4. Step 4: Compute the vectors \( \vec{AB} \) and \( \vec{AC} \). Using the points (4, 0, 0), (0, 5, 0), and (0, 0, 20): - \( \vec{AB} = (0 - 4, 5 - 0, 0 - 0) = (-4, 5, 0) \). - \( \vec{AC} = (0 - 4, 0 - 0, 20 - 0) = (-4, 0, 20) \).
5. Step 5: Calculate the cross product \( \vec{AB} \times \vec{AC} \). Use the determinant formula for the cross product: \( \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 5 & 0 \\ -4 & 0 & 20 \end{vmatrix} \). Then, find the magnitude of the resulting vector and divide by 2 to get the area of the triangle.