Question

Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -7) and is perpendicular to the plane 8x 7y 4z

Answer 1

Answer:

$3x - 4y + z = 1$

Step-by-step explanation:

Given

$Point\ 1 = (4,3,1)$

$Point\ 2 = (4,1,-7)$

Perpendicular to $8x + 7y + 4z = 18$

Required

Determine the plane equation

The general equation of a plane is:

$a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

For $n =$

$(x_1,y_1,z_1) = (4,3,1)$

$(x_2,y_2,z_2) = (4,1,-7)$

First, we need to determine parallel vector $V_1$

$V_1 =$

$V_1 =$

$V_1 =$

$V_1$ is parallel to the required plane

From the question, the required plane is perpendicular to $8x + 7y + 4z = 18$

Next, we determine vector $V_2$

$V_2 =$

This implies that the required plane is parallel to $V_2$

Hence: $V_1$ and $V_2$ are parallel.

So, we can calculate the cross product $V_1 * V_2$

$V_1 =$

$V_2 =$

$n = V_1 * V_2$

$V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]$

The product is always of the form + - +

So:

$V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]$  $-j\left[\begin{array}{cc}0&8\\8&4\end{array}\right]$ $+k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]$

Calculate the product

$V_1 * V_2 = i(2*4- 8*7) - j(0*4- 8*8) + k(0*7 - 2 * 8)$

$V_1 * V_2 = i(8- 56) - j(0- 64) + k(0 - 16)$

$V_1 * V_2 = i(-48) - j(- 64) + k(- 16)$

$V_1 * V_2 = -48i +64j - 16k$

So, the resulting vector, n is:

$n =$

Recall that:

$n =$

By comparison:

$a = -48$   $b = 64$   $c = -16$

Substitute these values in $a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

$-48(x-x_1) + 64(y - y_1) -16(z-z_1) =0$

Recall that:$(x_1,y_1,z_1) = (4,3,1)$

So, we have:

$-48(x-4) + 64(y - 3) -16(z-1) =0$

$-48x + 192 + 64y -192 - 16z +16 = 0$

Collect Like Terms

$-48x + 64y - 16z = 0 - 192 + 192 - 16$

$-48x + 64y - 16z = -16$

Divide through by -16

$3x - 4y + z = 1$

Hence, the equation of the plane is$3x - 4y + z = 1$