Question

find the equation of the plane through the point ( p = (3, 4, 5) ) and parallel to the plane ( - (x + 4y + 4z) = 5 ). ( z = ) input box submit answer next item answers # your answer score 01 - /1 - /1

Explanation:

Step1: Recall the formula for a plane parallel to a given plane

If two planes are parallel, they have the same normal vector. The given plane is \(-(x + 4y + 4z)=5\), which can be rewritten as \(x + 4y + 4z=- 5\). The general equation of a plane is \(ax+by + cz=d\), where \(\vec{n}=(a,b,c)\) is the normal vector. For the given plane, the normal vector \(\vec{n}=(1,4,4)\). So the plane we want to find, being parallel, will have the same normal vector, so its equation is of the form \(x + 4y+4z = k\), where \(k\) is a constant to be determined.

Step2: Substitute the point \(P=(3,4,5)\) into the equation

We know the plane passes through the point \(P=(3,4,5)\). Substitute \(x = 3\), \(y = 4\), and \(z = 5\) into the equation \(x + 4y+4z=k\).
\[

$$\begin{align*} k&=3+4\times4 + 4\times5\\ &=3 + 16+20\\ &=39 \end{align*}$$

\]
So the equation of the plane is \(x + 4y + 4z=39\). Now we need to solve for \(z\):
\[

$$\begin{align*} 4z&=39 - x - 4y\\ z&=\frac{39 - x - 4y}{4}\\ z&=-\frac{1}{4}x - y+\frac{39}{4} \end{align*}$$

\]

Answer:

\(z = -\frac{1}{4}x - y+\frac{39}{4}\)