Question

Hey, how would i find the cosine of the angle between the plane z−3x+2y = 11 and the y-axis? Thanks for helps =)

Answer 1

@AliceMath99
Write the equation of the plane in standard from -3x+2y+z=11.

The easiest way to go about this problem is to do the following:

First, find a vector that parallels the y-axis. The vector u = (0,1,0) should work nicely.

Next, the find a vector running through the plane -3x+2y+z=11. To do this, pick two points (x1,y1,z1) and (x2,y2,z2) that satisfy the plane equation. I chose (0,0,11) and (0,5,1). Then, a vector v that lies in the plane would be given as

v = (0,0,11) - (0,5,1)

= (0,-5,10)

Then, remember that an alternative way to calculuate the dot product is given as

u*v = ||u||*||v||*cosθ

where θ is the angle between the two vectors.

So, we have

u*v = (0,1,0)*(0,-5,10)

= 0*0+1*(-5)+0*10

= -5

We also have that

||u|| = √(02+12+02)

= 1

and

||v|| = √(02+(-5)2+102)

= √125

= 5√5

Hence, by the alternative formula for the dot product, we have

-5 = 1*5√5 * cosθ

and so

cosθ = -5/(5√5)

= -1/√5

or

cosθ = -√(5)/5

≈-0.4472