- #1

chwala

Gold Member

- 2,725

- 381

## Homework Statement

a) Find the acute angle between the line ## r = i+3j+5k+t (2i+4j+k) ## and the plane x-y+z=0

b)Two planes are defined by the equations ## x+2y+z=4## and ## 2x-3y=6 ##

i. find the acute angle between them.

ii. find a vector equation of their line of intersection.

## Homework Equations

## The Attempt at a Solution

a) [/B]I considered the directional vector ##p= 2i+4j+k## and the normal to plane ## q= i-j+k##

##p.q=|p||q| cos θ## using this i got ##θ=97.2^0 ##degrees. The textbook answer is given as ## θ-0.5π= 7.2 ##degrees

**b)**

i.I let ##m=i+2j+k## and ## k=2i-3j## using ##m.k=|m||k| cos Φ## i got ## θ=116.9 ## degrees. The textbook answer is ##63.1## degrees, now my question is can we conclude that to get an acute angle α between a line and a plane we can use ##α=θ-0.5π## and to get an acute angle α between two planes then we use ## α=180-θ## is that correct?

i.