In the rectangular prism ABCDEFGH,with sides EF=5cm,FG=8cm and BF=6cm,calculate the size of the angle between the space diagonal BH and the plane BFGC.

Answer 1

To find the size of the angle between the space diagonal BH and the plane BFGC, we first need to find the length of the space diagonal BH.

Using the Pythagorean theorem, we can find the length of BH:
BH^2 = BF^2 + FH^2
BH = sqrt(6^2 + 5^2)
BH = sqrt(36 + 25)
BH = sqrt(61)

Next, we need to find the angle between BH and the plane BFGC. The cosine of the angle between a diagonal of a rectangular prism and one of its faces is the dot product of the two vectors divided by the product of their magnitudes.

Let's consider BH as vector a and the normal vector of the plane BFGC as vector b.
Vector a = <6,5,√61>
Vector b = <8,0,0>

Now, we'll calculate the dot product of the two vectors:
a • b = 6*8 + 5*0 + √61*0
= 48

Next, we'll calculate the magnitudes of the two vectors:
|a| = sqrt(6^2 + 5^2 + (√61)^2)
= sqrt(36 + 25 + 61)
= sqrt(122)
≈ 11.05

|b| = sqrt(8^2 + 0 + 0)
= 8

Now we can find the cosine of the angle between BH and the plane BFGC:
cos(angle) = (a • b) / (|a| * |b|)
cos(angle) = 48 / (11.05 * 8)
cos(angle) = 48 / 88.4
angle = arccos(0.5421)
angle ≈ 57.26 degrees

Therefore, the size of the angle between the space diagonal BH and the plane BFGC is approximately 57.26 degrees.