Find the equation of the plane whose 𝑥-, 𝑦-, and 𝑧-intercepts are negative seven, three, and negative four, respectively.
In this question, we’re asked to find the equation of a plane. And to do this, we’re given that it has an 𝑥-intercept at negative seven, a 𝑦-intercept at three, and its 𝑧-intercept is at negative four. And we can recall we know how to find the equation of a plane from its intercepts by using the intercept form of the equation of a plane. This tells us if the 𝑥-, 𝑦-, and 𝑧-intercepts of a plane are 𝑎, 𝑏, and 𝑐 for nonzero values, then 𝑥 over 𝑎 plus 𝑦 over 𝑏 plus 𝑐 over 𝑧 is equal to one is the equation of this plane. And since the three intercepts we’re given are nonzero, we can just substitute these values into our equation for the plane
We substitute 𝑎 is equal to negative seven, 𝑏 is equal to three, and 𝑐 is equal to negative four into our intercept equation of the plane. We get 𝑥 over negative seven plus 𝑦 over three plus 𝑧 over negative four is equal to one. We can then simplify this slightly. In the first and third term on the left-hand side of our equation, we can bring the factor of negative one into the expression. This gives us the following equation. And it’s worth noting we could write the coefficients of 𝑥, 𝑦, and 𝑧 as integers by multiplying the equation through by 84. However, we can just leave our answer in this form. This gives us the answer of negative 𝑥 over seven plus 𝑦 over three minus 𝑧 over four is equal to one.