was there something wrong with the solution I gave you earlier?
www.jiskha.com/questions/1884051/the-base-of-the-roof-is-rectangular-and-horizontal-with
The base of the roof, ππ΄π΅πΆ, is rectangular and horizontal with ππ΄ = πΆπ΅ = 14 π and ππΆ = π΄π΅ = 8 π. The top of the roof π·πΈ is 5 π above the base and DE = 6m. The sloping edges ππ·,πΆπ·,π΄πΈ and π΅πΈ 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 ππ·ββββββ in terms of i , j and k, and find its magnitude (4 marks)
(ii) Use a scalar product to find angle π·ππ΅. (5 marks)
3 answers
sorry about that. I forgot you had answered the question.
Could you go a bit more in-depth with (ii) because I haven't done vectors before.
Could you go a bit more in-depth with (ii) because I haven't done vectors before.
OD = 4i+4j+4k
OB = 14i+8j
If ΞΈ = β‘DOB, then
OBβ’OD = |OB| * |OD| cosΞΈ
OBβ’OD = 4*14 + 4*8 + 4*0 = 88
|OD| = β(4^2+4^2+4^2) = 4β3
|OB| = β(14^2+8^2) = 2β65
so,
cosΞΈ = 88/(4β3 * 2β65) = 0.7877
ΞΈ = 38Β°
OB = 14i+8j
If ΞΈ = β‘DOB, then
OBβ’OD = |OB| * |OD| cosΞΈ
OBβ’OD = 4*14 + 4*8 + 4*0 = 88
|OD| = β(4^2+4^2+4^2) = 4β3
|OB| = β(14^2+8^2) = 2β65
so,
cosΞΈ = 88/(4β3 * 2β65) = 0.7877
ΞΈ = 38Β°