Math  /  Geometry

Question 直红果 x2=y13=z+25 与平面 2x+3y+8 ○。的是昆位置关系 \begin{array}{l} \text { 直红果 } \frac{x}{2}=\frac{y-1}{3}=\frac{z+2}{5} \text { 与平面 } 2 x+3 y+8 \\ \text { ○。的是昆位置关系 } \end{array}

Studdy Solution

STEP 1

What is this asking? We're checking if a line and a plane are parallel, intersecting, or if the line lies within the plane! Watch out! Don't mix up the direction vector of the line and a point on the line.
Also, remember a line can be *in* a plane, not just intersecting it!

STEP 2

1. Rewrite the line equation
2. Find a point on the line
3. Find the direction vector of the line
4. Check if the direction vector is orthogonal to the plane's normal vector
5. Check if the point lies on the plane

STEP 3

Let's rewrite the line equation in a more useful format.
We'll set the given ratios equal to a parameter, let's call it tt.
This gives us x2=t\frac{x}{2} = t, y13=t\frac{y-1}{3} = t, and z+25=t\frac{z+2}{5} = t.

STEP 4

Now, we can express xx, yy, and zz in terms of tt: x=2tx = 2t, y=3t+1y = 3t + 1, and z=5t2z = 5t - 2.
This parametric form is super helpful for visualizing the line!

STEP 5

To find a point on the line, we can pick any value for tt.
Let's choose t=0t = \mathbf{0} because that makes the math easy!

STEP 6

Plugging t=0t = \mathbf{0} into our parametric equations gives us x=20=0x = 2 \cdot \mathbf{0} = \mathbf{0}, y=30+1=1y = 3 \cdot \mathbf{0} + 1 = \mathbf{1}, and z=502=2z = 5 \cdot \mathbf{0} - 2 = \mathbf{-2}.
So, the point (0,1,2)( \mathbf{0}, \mathbf{1}, \mathbf{-2} ) is on our line.
Boom!

STEP 7

The coefficients of tt in our parametric equations give us the direction vector of the line.
That's (2,3,5)( \mathbf{2}, \mathbf{3}, \mathbf{5} ).
This vector tells us which way the line is pointing!

STEP 8

The normal vector to the plane 2x+3y+8z=02x + 3y + 8z = 0 is given by the coefficients of xx, yy, and zz, which is (2,3,8)( \mathbf{2}, \mathbf{3}, \mathbf{8} ).

STEP 9

Two vectors are orthogonal (perpendicular) if their dot product is zero.
Let's calculate the dot product of the direction vector of the line and the normal vector of the plane: (2,3,5)(2,3,8)=(22)+(33)+(58)=4+9+40=53( \mathbf{2}, \mathbf{3}, \mathbf{5} ) \cdot ( \mathbf{2}, \mathbf{3}, \mathbf{8} ) = (\mathbf{2} \cdot \mathbf{2}) + (\mathbf{3} \cdot \mathbf{3}) + (\mathbf{5} \cdot \mathbf{8}) = \mathbf{4} + \mathbf{9} + \mathbf{40} = \mathbf{53}.

STEP 10

Since the dot product is 53\mathbf{53}, which is not zero, the direction vector and the normal vector are *not* orthogonal.
This means the line is *not* parallel to the plane.
Exciting!

STEP 11

We found the point (0,1,2)( \mathbf{0}, \mathbf{1}, \mathbf{-2} ) lies on the line.
Let's see if it's also on the plane by plugging its coordinates into the plane's equation: 2(0)+3(1)+8(2)=0+316=132(\mathbf{0}) + 3(\mathbf{1}) + 8(\mathbf{-2}) = \mathbf{0} + \mathbf{3} - \mathbf{16} = \mathbf{-13}.

STEP 12

Since the result is 13\mathbf{-13}, and not zero, the point is *not* on the plane.
This means the line doesn't lie within the plane.

STEP 13

Since the line is not parallel to the plane and doesn't lie within it, the line must *intersect* the plane!

STEP 14

The line and the plane intersect!

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord