Question

Find the cosine of the angle between the plane 4x-5y+z=4 and the plane 5x-4y+3z=-2

Answer 1

Given the planes
`4x-5y+z=4` and
`5x-4y+3z=-2`, find the angle between them.

Using the formula,

`\theta=cos^(-1) (\frac{A_(x)B_(x)+A_(y)B_(y)+A_(z)B_(z) }{\sqrt{A^2_(x)+A^2_(y)+A^2_(z)}\sqrt{B^2_(x)+B^2_(y)+B^2_(z)} } )`

Where vector,
`A= < 4,-5,1 >` and vector
`B= < 5,-4,3 >`.

Plug these values into the formula above...

`\Longrightarrow \theta=cos^(-1) (\frac{A_(x)B_(x)+A_(y)B_(y)+A_(z)B_(z) }{\sqrt{A^2_(x)+A^2_(y)+A^2_(z)}\sqrt{B^2_(x)+B^2_(y)+B^2_(z)} } )`

`\Longrightarrow \theta=cos^(-1) (((4)(5)+(-5)(-4)+(1)(3) )/(√((4)^2+(-5)^2+(1)^2)√((5)^2+(-4)^2+(3)^2) ) )`

`\Longrightarrow \theta=cos^(-1) ((43 )/(√(42)√(50) ) )`

`\Longrightarrow \theta=cos^(-1) ((43 )/(10√(21) ) )`

`\Longrightarrow \theta \approx 20.2\textdegree`