Acute angle between a plane and y axis

Hey guys. The question goes like this :

Coordinates of A(-1,2,5) and B(2,-2,11). Plane p passes through B and is perpendicular to AB. It is required to find

(a) Equation of plane p

(b) Acute angle between plane p and y axis.

An equation for p is 3x - 4y + 6z = 80

What is the next step to calculate the angle ?

i use the formula cos θ = A . B / |A| x |B| ,

i can substitute the value of A with normal vector of plane p. Then how should i proceed? Is there a direction vector for the y axis ??

Thanks in advance :)

Re: Acute angle between a plane and y axis

Quote:

Originally Posted by

**nks2427** Hey guys. The question goes like this :

Coordinates of A(-1,2,5) and B(2,-2,11). Plane p passes through B and is perpendicular to AB. It is required to find

(a) Equation of plane p

(b) Acute angle between plane p and y axis.

An equation for p is 3x - 4y + 6z = 80

What is the next step to calculate the angle ?

Given a plane $\displaystyle N\cdot(R-P)=0$ and line $\displaystyle Q+tD$ if the line is not parallel to the plane, i.e. $\displaystyle N\cdot D\ne 0$, then the acute angle between the line and the plane is the complement of the acute angle between $\displaystyle D~\&~N$.

In this case $\displaystyle D=<0,1,0>~\&~N=<3,-4,6>$.

Re: Acute angle between a plane and y axis

can D be taken as D = < 0, 2 , 0 > ?

Re: Acute angle between a plane and y axis

Quote:

Originally Posted by

**nks2427** can D be taken as D = < 0, 2 , 0 > ?

Yes. Any vector parallel to the y-axis.

Re: Acute angle between a plane and y axis

Re: Acute angle between a plane and y axis

Hi, why is D = < 0,1,0> ?

Re: Acute angle between a plane and y axis

Quote:

Originally Posted by

**Jasmine17** Hi, why is D = < 0,1,0> ?

That is parallel to the $y$-axis.