Find the angle between the line x23=y+33=z11 and the plane 2x3y+4z=0 ?

2 Answers
Jun 22, 2018

look carefully, θ is the anlge b/w plane and line.

Explanation:

given the direction cosine of line as b=3ˆi+3ˆj+ˆk and that of plane as n=2ˆi3ˆj+4ˆk
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clearly , using dot product of vectors,
cos(90θ)=ˆb.ˆn
Nowit is very easy to find the numerical value of θ

Jun 22, 2018

θ0.0565 rad (3.238)

Explanation:

We are given the symmetric form of the equation of the line:

x23=y+33=z11

Convert the symmetric form of the line to the parametric form by setting each expression equal to t:

x23=t

y+33=t

z11=t

Multiply both sides by the denominators:

x2=3t

y+3=3t

z1=t

Move the constants to the right:

x=3t+2

y=3t3

z=t+1

Convert to the vector form:

(x,y,z)=(2,3,1)+t(3ˆi+3ˆj+ˆk)

Please understand the u=3ˆi+3ˆj+ˆk is the direction of the line.

Find any two points in the plane, 2x3y+4z=0:

Pick (0,0,?):

2(0)3(0)+4z=0

z=0

The first point is (0,0,0)

Pick (4,4,?)

2(4)3(4)+4z=0

z=1

The second point is (4,4,1)

Make a vector from the first point to the second:

v=(40)ˆi+(40)ˆj+(10)ˆk

v=4ˆi+4ˆj+1ˆk

Please understand that v is in the plane and the angle between u and v is, also, the angle between the line and the plane.

Compute the dot-product of u and v:

uv=3(4)+3(4)+1(1)=25

Compute the magnitudes of u and v:

u=32+32+12=19

v=42+42+12=33

Using the formula uv=uvcos(θ), we can find the angle between the two vectors:

25=1933cos(θ)

θ=cos1(251933)

θ0.0565 rad (3.238)