Answer:
Explanation:
To find the angle between the line BH and plane AEHD, we first need to identify these two elements in the given cube.
From the description of the cube, we can see that the line BH lies on the front plane of the cube, which is formed by the lines AB, BC, CD, and AD.
Also, the plane AEHD is the base plane of the cube, which is formed by the lines AE, EH, HD, and AD.
Therefore, to find the angle between line BH and plane AEHD, we need to determine the direction vectors of both BH and the plane AEHD.
Let's start with BH:
The line BH passes through points B and H. To find the direction vector of BH, we can subtract the position vectors of B and H:
BH = vector(H) - vector(B)
To do this, we need to find the position vectors of B and H with respect to some chosen origin. We can use the vector AD as one of the basis vectors for our coordinate system, and choose A as the origin.
The position vector of B is:
vector(B) = vector(A) + vector(AB)
= (0, 0, 0) + (7, 0, 0)
= (7, 0, 0)
The position vector of H is:
vector(H) = vector(A) + vector(AD) + vector(AE) + vector(AH)
= (0, 0, 0) + (0, 0, 21) + (7, 0, 0) + (0, 23, 0)
= (7, 23, 21)
Therefore, the direction vector of BH is:
BH = vector(H) - vector(B)
= (7, 23, 21) - (7, 0, 0)
= (0, 23, 21)
Next, let's find the normal vector of the plane AEHD:
The plane AEHD is the base plane of the cube, which means it is parallel to the X-Y plane. Therefore, its normal vector is in the Z direction.
To find the normal vector of the plane AEHD, we can take the cross product of two vectors lying in the plane, such as AE and AD:
normal vector = AE x AD
= (7, -23, 0) x (0, 0, 21)
= (-483, -147, 0)
(Note: We could also take the cross product of other pairs of vectors lying in the plane, such as EH and HD, to get the same normal vector.)
Now that we have the direction vector of BH and the normal vector of the plane AEHD, we can find the angle between them using the dot product:
cos(angle) = (BH . normal vector) / (|BH| |normal vector|)
where BH . normal vector is the dot product of BH and the normal vector, and |BH| and |normal vector| are the magnitudes of BH and the normal vector, respectively.
BH . normal vector = (0, 23, 21) . (-483, -147, 0)
= 0 - 3393 + 0
= -3393
|BH| = sqrt(0^2 + 23^2 + 21^2) = sqrt(1102)
|normal vector| = sqrt((-483)^2 + (-147)^2 + 0^2) = sqrt(234018)
Therefore,
cos(angle) = -3393 / (sqrt(1102) * sqrt(234018))
= -0.0662
Finally, we can find the angle between BH and the plane AEHD by taking the inverse cosine. cos(angle) = -0.0662
To find the angle, we take the inverse cosine of both sides:
angle = arccos(-0.0662)
Using a calculator, we get:
angle ≈ 1.649 radians ≈ 94.53 degrees
Therefore, the angle between the line BH and plane AEHD is approximately 94.53 degrees.