Question

Find the cosine of the angle between the planes −1x+3y+1z=0 and the plane 5x+5y+4z=−4

Answer 1

To find the cosine of the angle between two planes, calculate the dot product of their normal vectors.

Step-by-step explanation:

To find the cosine of the angle between two planes, we need to determine the normal vectors of the planes and then calculate the dot product of the two normal vectors. The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them.

Given the planes -x+3y+z=0 and 5x+5y+4z=-4, the normal vectors are (-1,3,1) and (5,5,4) respectively.

Calculating the dot product of the two normal vectors, we get: (-1)(5) + (3)(5) + (1)(4) = 0. Therefore, the cosine of the angle between the planes is 0.

Answer 2

The he cosine of the angle between the planes is
`(14)/(11√(6))`.

Step-by-step explanation:

Using the definition of the dot product:

`\cos\theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}`

The given planes are

`-1x+3y+1z=0`

`5x+5y+4z=-4`

The angle between two normal vectors of the planes is the same as one of

the angles between the planes. We can find a normal vector to each of the

planes by looking at the coefficients of x, y, z.

`\overrightarrow{n_1}=<-1,3,1>`

`\overrightarrow{n_2}=<5,5,4>`

`\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(-1)(5)+(3)(5)+(1)(4)=14`

`|n_1|=√((-1)^2+(3)^2+(1)^2)=√(11)`

`|n_2|=√((5)^2+(5)^2+(4)^2)=√(66)`

The cosine of the angle between the planes

`\cos\theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}`

`\cos\theta =(14)/(√(11)√(66))`

`\cos\theta =(14)/(11√(6))`

Therefore the cosine of the angle between the planes is
`(14)/(11√(6))`.