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Math  /  Geometry

Question(A) Find the parametric equations for the line through the point P=(5,−5,3)\mathrm{P}=(5,-5,3) that is perpendicular to the plane 1x+3y+4z=11 x+3 y+4 z=1. x=□y=□z=□\begin{array}{l} x=\square \\ y=\square \\ z=\square \end{array} (B) At what point QQ does this line intersect the yzy z-plane? Q=(□,□)\mathrm{Q}=(\square, \square) □\square

Studdy Solution

STEP 1

1. The line is perpendicular to the given plane.
2. The normal vector of the plane is the direction vector of the line.
3. The line passes through the point P=(5,−5,3) P = (5, -5, 3) .
4. The yz yz -plane is defined by x=0 x = 0 .

STEP 2

1. Determine the direction vector of the line.
2. Write the parametric equations for the line.
3. Find the intersection of the line with the yz yz -plane.

STEP 3

Determine the direction vector of the line:
The normal vector of the plane 1x+3y+4z=1 1x + 3y + 4z = 1 is n=⟨1,3,4⟩ \mathbf{n} = \langle 1, 3, 4 \rangle .
Since the line is perpendicular to the plane, the direction vector of the line is the same as the normal vector:
d=⟨1,3,4⟩ \mathbf{d} = \langle 1, 3, 4 \rangle

STEP 4

Write the parametric equations for the line:
The parametric equations for a line through a point P(x0,y0,z0) P(x_0, y_0, z_0) with direction vector d=⟨a,b,c⟩ \mathbf{d} = \langle a, b, c \rangle are:
x=x0+at x = x_0 + at y=y0+bt y = y_0 + bt z=z0+ct z = z_0 + ct
Substitute P=(5,−5,3) P = (5, -5, 3) and d=⟨1,3,4⟩ \mathbf{d} = \langle 1, 3, 4 \rangle :
x=5+1t x = 5 + 1t y=−5+3t y = -5 + 3t z=3+4t z = 3 + 4t

STEP 5

Find the intersection of the line with the yz yz -plane:
The yz yz -plane is defined by x=0 x = 0 .
Set the parametric equation for x x to zero:
5+1t=0 5 + 1t = 0
Solve for t t :
t=−5 t = -5
Substitute t=−5 t = -5 into the parametric equations for y y and z z :
y=−5+3(−5)=−5−15=−20 y = -5 + 3(-5) = -5 - 15 = -20 z=3+4(−5)=3−20=−17 z = 3 + 4(-5) = 3 - 20 = -17
The intersection point Q Q is:
Q=(0,−20,−17) Q = (0, -20, -17)
The parametric equations for the line are:
x=5+ty=−5+3tz=3+4t \begin{array}{l} x = 5 + t \\ y = -5 + 3t \\ z = 3 + 4t \end{array}
The intersection point Q Q with the yz yz -plane is:
Q=(0,−20,−17) Q = (0, -20, -17)

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