The base of the roof, 𝑂𝐴𝐵𝐶, is rectangular and horizontal with 𝑂𝐴 = 𝐶𝐵 = 14 𝑚 and 𝑂𝐶 = 𝐴𝐵 = 8 𝑚. The top of the roof 𝐷𝐸 is 5 𝑚 above the base and DE = 6m. The sloping edges 𝑂𝐷,𝐶𝐷,𝐴𝐸 and 𝐵𝐸 are all equal in length.

Question

The base of the roof, 𝑂𝐴𝐵𝐶, is rectangular and horizontal with 𝑂𝐴 = 𝐶𝐵 = 14 𝑚 and 𝑂𝐶 = 𝐴𝐵 = 8 𝑚. The top of the roof 𝐷𝐸 is 5 𝑚 above the base and DE = 6m. The sloping edges 𝑂𝐷,𝐶𝐷,𝐴𝐸 and 𝐵𝐸 are all equal in length.

Unit vector i and j are parallel to OA and OC respectively and the unit vectors k is vertically upwards.
(i) Express the vector 𝑂𝐷⃗⃗⃗⃗⃗⃗ in terms of i , j and k, and find its magnitude (4 marks)
(ii) Use a scalar product to find angle 𝐷𝑂𝐵. (5 marks)

Answer 1

D is 4m in from AC, so

OD = 4i+4j+4k

OB = 14i+8j
If θ = ∡DOB, then
OB•OD = |OB| * |OD| cosθ
I assume you know how to find the dot product and magnitudes

Answer 2

(i) To find the vector OD, we can express it as the sum of vectors OA and AD. Let's start with vector OA:

OA = 14i

Now, let's find vector AD. Since DE is 5m above the base and DE = 6m, vector AD can be represented as:

AD = 5k + 6i

Adding these two vectors together gives us vector OD:

OD = OA + AD = 14i + (5k + 6i) = 20i + 5k

The magnitude of vector OD can be found using the formula:

|OD| = √(OD · OD)

where · denotes the dot product. Let's calculate the magnitude:

|OD| = √((20i + 5k) · (20i + 5k))
= √(400i · i + 200i · k + 200i · k + 25k · k)
= √(400 + 400 + 25)
= √(825)
= √33 m

So, the magnitude of vector OD is √33 m.

(ii) To find the angle DOB, we can use the scalar product formula:

OD · OB = |OD| * |OB| * cos(∠DOB)

Since vector OD = 20i + 5k and vector OB = 8i, we can calculate the scalar product:

OD · OB = (20i + 5k) · (8i)
= 160i · i + 40i · k
= 160 + 0
= 160

The magnitude of vector OB can be found using the formula:

|OB| = √(OB · OB)
= √(8i · 8i)
= √(64)
= 8

Plugging these values into the scalar product formula:

160 = √33 * 8 * cos(∠DOB)

Simplifying the equation:

cos(∠DOB) = 160 / (8 * √33)
= 5 / √33

To find the angle ∠DOB, we can take the inverse cosine (or arccosine) of both sides:

∠DOB = arccos(5 / √33)

Thus, the angle ∠DOB is given by arccos(5 / √33).

Answer 3

(i) To express the vector OD→ in terms of i, j, and k, we need to break it down into its components along the x, y, and z directions.

First, let's calculate the components along the x and y directions. Since OD→ is parallel to OA→, its x-component is the same as OA→, which is 14 m. Similarly, OD→ is parallel to OC→, so its y-component is the same as OC→, which is 8 m.

Now, let's calculate the component along the z-direction. Since DE→ is 5 m above the base, and OD→ is parallel to DE→, its z-component is also 5 m.

Therefore, the vector OD→ can be expressed as:
OD→ = 14i + 8j + 5k

To find the magnitude of OD→, we use the formula for the magnitude of a vector:
|OD→| = √(ODx² + ODy² + ODz²)
|OD→| = √((14)^2 + (8)^2 + (5)^2)
|OD→| = √(196 + 64 + 25)
|OD→| = √285
|OD→| ≈ 16.88 m

(ii) To find the angle θ between DO→ and OB→, we can use the scalar product (dot product):
DO→ · OB→ = |DO→| |OB→| cos(θ)

The scalar product of two vectors A→ and B→ is calculated as:
A→ · B→ = AxBx + AyBy + AzBz

In this case, DO→ and OB→ are perpendicular to each other because the roof is rectangular. Therefore, cos(θ) = 0, and the scalar product DO→ · OB→ is zero.

DO→ · OB→ = (14)(14) + (8)(0) + (5)(-14) (Substituting the components)
0 = 196 + 0 - 70
0 = 126

Since the scalar product is zero, we can conclude that the angle θ is 90 degrees, or a right angle.

Therefore, angle DOB = θ = 90 degrees.