To find the equation of the plane that is orthogonal to the plane $5x + 4z = 4$ and contains the line of intersection of the planes $2x - 3y + z = 35$ and $x + 2y - 3z = 42$, we can proceed step by step.

Step 1: Equation of the plane orthogonal to $5x + 4z = 4$

The normal vector to the plane $5x + 4z = 4$ is given by $\mathbf{n_1} = \langle 5, 0, 4 \rangle$.

Thus, the required plane should have a normal vector that is parallel to this one, i.e., the form of the plane will be: $5x + 0y + 4z + d = 0$ or simply: $5x + 4z + d = 0$ We need to determine $d$ by using the next steps.

Step 2: Find the line of intersection of the two planes

The line of intersection of the planes $2x - 3y + z = 35$ and $x + 2y - 3z = 42$ can be found by solving these two plane equations simultaneously.

Step 2.1: Eliminate one variable

Let's solve these equations step by step to eliminate one variable and find a parametric form of the line.

1. Multiply the second equation by 2 to align the coefficients of $x$: $2(x + 2y - 3z) = 2 \times 42$ This gives: $2x + 4y - 6z = 84$

2. Subtract the first equation $2x - 3y + z = 35$ from this new equation: $(2x + 4y - 6z) - (2x - 3y + z) = 84 - 35$ Simplifying: $7y - 7z = 49$ Dividing by 7: $y - z = 7$ Hence, $y = z + 7$.

Step 2.2: Substitute $y = z + 7$ into one of the original equations

Substitute $y = z + 7$ into $2x - 3y + z = 35$: $2x - 3(z + 7) + z = 35$ Simplifying: $2x - 3z - 21 + z = 35$ $2x - 2z = 56$ $x - z = 28$ Thus, $x = z + 28$.

Step 2.3: Parametric equations of the line of intersection

Now, we have: $x = z + 28, \quad y = z + 7$ Let $z = t$ be the parameter. Then the parametric equations of the line are: $x = t + 28, \quad y = t + 7, \quad z = t$ These represent the line of intersection.

Step 3: Find $d$ using a point on the line

We now substitute a point on the line into the equation $5x + 4z + d = 0$ to find $d$.

Using $t = 0$ in the parametric equations, we get the point $(28, 7, 0)$.

Substitute $x = 28$ and $z = 0$ into $5x + 4z + d = 0$: $5(28) + 4(0) + d = 0$ $140 + d = 0$ $d = -140$

Step 4: Final equation of the plane

The equation of the required plane is: $5x + 4z - 140 = 0$ or equivalently: $5x + 4z = 140$

This is the equation of the plane that is orthogonal to $5x + 4z = 4$ and contains the line of intersection of the given planes.

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Would you like more details or have any questions on this solution?

5 Related Questions:

1. How do you find the normal vector to a plane given its equation?
2. What is the geometric interpretation of the line of intersection of two planes?
3. How can parametric equations of a line be derived from simultaneous plane equations?
4. How do you solve systems of linear equations involving three variables?
5. How does the orientation of a plane change if its normal vector changes?

Tip:

When finding the line of intersection between two planes, using elimination to solve for one variable simplifies the process significantly.