Krishna
2

Step 1:  Recall the Cartesian equation of the plane

Equation:  Ax + By + Cz + D = 0

Where A,B,C and D are constants and not all A,B,C are zero, can be taken

to be an equation of a plane in space. The coefficients A, B and C are the

components of a normal vector

Step 2: Note down the coordinates of the three points from the given equations

GIVEN \frac{x-4}{5} = \frac{y - 5}{8} = \frac{z - 10}{3}

And (6, 9, 4)

POINTS:     \frac{x-4}{5} = \frac{y - 5}{8} = \frac{z - 10}{3} = t       \because assume\ t

Now \frac{x-4}{5} = t  \Rightarrow  x = 5t + 4

\frac{y - 5}{8} = t   \Rightarrow y = 8t + 5

\frac{z - 10}{3}= t \Rightarrow z = 3t + 10

take t = 0

Then,  x = 5t + 4 \Rightarrow x = 5(0) + 4 \Rightarrow x = 4

y = 8t + 5 \Rightarrow y = 8(0) + 5 \Rightarrow y = 5

z = 3t + 10 \Rightarrow z = 3(0) + 10 \Rightarrow z = 10

So the point (x, y, z) = (4, 5, 10)

Similarly, take  x = 1

then, the other point is (x, y, z) = (9, 13, 13)

Therefore, the points are

A(4, 5, 10), B(9, 13,13) and C(6, 9, 4)

Step 3: Find the normal of the vectors AB and AC

A(4,5,10), B(9, 13, 13)

AB = A - B = (-5, -8, -3)

A(4, 5, 10), C(6, 9, 4)

AC =  A - C = (-2, -4, 6)

normal vector (n) = AB X AC = \left( \begin{array}{ccc} i & j & k \\ -5 & -8 & -3 \\ -2 & -4 & 6 \end{array} \right)

=$i\left[\left(-8\right)\left(6\right)\ -\left(-4)\left(-3\right)\right]\ -\ j\left[\left(-5\right)\left(6\right)-\left(-2\right)\left(-3\right)\right]\ +\ k\left[\left(-5\right)\left(-4\right)\ -\ \left(-8\right)\left(-2\right)\right]\right]$

= i\left(-48\ -12\right)-\ j\left(-30\ -\ 6\right)\ +k\left(20\ -16\right)

= (-60i+36j+4k)

= 4\left(-15i+9j+k\right)

Step 4: Determine the Cartesian equation by using the normal

Ax + By + Cz + D =0

n =  -15i+9j+k

n=(A,B,C)=(-15,\ 9,\ 1)

So, The Cartesian equation

-15x + 9y + z + D = 0

At point (6, 9, 4) find D

-15(6)+9(9)+1(4)+D=0

-90+81+4+D=0

-5+D=0

D=5

The Cartesian equation

-15x+9y+z+5=0

Anonymous
1
Thank you .I appreciate this!