Final answer:
The shortest distance, d, from the point (3, 0, -4) to the plane x + y + z = 4 is 5 / sqrt(3) units.
Step-by-step explanation:
To find the shortest distance, d, from the point (3, 0, -4) to the plane x + y + z = 4, we can use the formula for the distance between a point and a plane. This formula states that the distance d between a point (x1, y1, z1) and a plane Ax + By + Cz + D = 0 is given by: d = |Ax1 + By1 + Cz1 + D| / sqrt(A^2 + B^2 + C^2). In this case, A = B = C = 1, and D = -4. Substituting the values into the formula, we get: d = |1*3 + 1*0 + 1*(-4) + (-4)| / sqrt(1^2 + 1^2 + 1^2). Simplifying, we have: d = |3 - 4 + (-4) - 4| / sqrt(3) = |-5| / sqrt(3) = 5 / sqrt(3). Therefore, the shortest distance d from the point (3, 0, -4) to the plane x + y + z = 4 is 5 / sqrt(3) units.