Given a cuboid PQRS TUVW of length PR = 9cm, width PS = 5cm and height PT = 12cm. Calculate the angle between the line: PU and the base PQS R (ii) SU and the plane WTUV (iii) AV and the plane R QU, where point A is the midpoint of PQ

Haya General, twende kazi! Hili ni swali la jiometri ya 3D. Tutatumia mfumo wa kuratibu (coordinates) kutatua.

Kwanza, hebu tueleze kuratibu za pembe za mchemraba (cuboid) PQRSTUVW. Kutokana na mchoro na vipimo vilivyotolewa (PQ = 9 cm, PS = 5 cm, PT = 12 cm), tutachukulia P kama asili (origin) $(0,0,0)$.
P = $(0,0,0)$
Q = $(9,0,0)$ (urefu wa PQ)
S = $(0,5,0)$ (upana wa PS)
T = $(0,0,12)$ (urefu wa PT)
R = $(9,5,0)$ (Q + S - P)
U = $(9,0,12)$ (Q + T - P)
V = $(0,5,12)$ (S + T - P)
W = $(9,5,12)$ (R + T - P)

Kumbuka: Maelezo ya swali "length PR = 9m" yanapingana na mchoro unaoonyesha PQ = 9cm kama urefu wa upande. Tutatumia vipimo vya mchoro (PQ=9, PS=5, PT=12) kwa sababu ndivyo vinavyoendana na umbo la mchemraba.

i) Kokotoa pembe kati ya mstari PU na msingi PQRS.

Mstari ni PU. Msingi ni ndege PQRS (ndege $z=0$).
Makadirio ya PU kwenye ndege PQRS ni mstari PQ.
Hivyo, pembe kati ya mstari PU na ndege PQRS ni pembe $\angle UPQ$.
Tuna pembetatu mraba $\triangle UPQ$.
U = $(9,0,12)$
P = $(0,0,0)$
Q = $(9,0,0)$

Urefu wa PQ = 9 cm.
Urefu wa UQ = PT = 12 cm.

Katika $\triangle UPQ$:

$$\tan(\angle UPQ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{UQ}{PQ} = \frac{12}{9} = \frac{4}{3}$$ $$\angle UPQ = \arctan\left(\frac{4}{3}\right)$$ $$\angle UPQ \approx 53.1301^\circ$$

Ikiwa imezungushwa hadi sehemu moja ya desimali:

$$\boxed{53.1^\circ}$$

ii) Kokotoa pembe kati ya mstari SU na ndege WTUV.

Mstari ni SU. Ndege ni WTUV (ndege $z=12$, uso wa juu).
Makadirio ya S kwenye ndege WTUV ni V (kwa sababu SV iko wima kwa ndege WTUV).
Makadirio ya U kwenye ndege WTUV ni U yenyewe (kwa sababu U iko kwenye ndege).
Hivyo, makadirio ya mstari SU kwenye ndege WTUV ni mstari VU.
Pembe kati ya mstari SU na ndege WTUV ni pembe $\angle SUV$.
Tuna pembetatu mraba $\triangle SVU$.
S = $(0,5,0)$
U = $(9,0,12)$
V = $(0,5,12)$

Urefu wa SV = PT = 12 cm.
Urefu wa VU = PQ = 9 cm.

Katika $\triangle SVU$:

$$\tan(\angle SUV) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{SV}{VU} = \frac{12}{9} = \frac{4}{3}$$ $$\angle SUV = \arctan\left(\frac{4}{3}\right)$$ $$\angle SUV \approx 53.1301^\circ$$

Ikiwa imezungushwa hadi sehemu moja ya desimali:

$$\boxed{53.1^\circ}$$

iii) Kokotoa pembe kati ya mstari AV na ndege RQUV, ambapo A ni sehemu ya kati ya PQ.

Kwanza, pata kuratibu za A.
P = $(0,0,0)$, Q = $(9,0,0)$.
A ni sehemu ya kati ya PQ:

$$A = \left(\frac{0+9}{2}, \frac{0+0}{2}, \frac{0+0}{2}\right) = (4.5, 0, 0)$$

Mstari ni AV.
V = $(0,5,12)$.
Vekta ya mwelekeo wa mstari AV ni:

$$\vec{l} = \vec{AV} = V - A = (0 - 4.5, 5 - 0, 12 - 0) = (-4.5, 5, 12)$$

Ndege ni RQUV.
R = $(9,5,0)$
Q = $(9,0,0)$
U = $(9,0,12)$
V = $(0,5,12)$

Kumbuka: Pointi Q, R, U, V haziko kwenye ndege moja. Hii inafanya swali kuwa na utata. Tutachukulia "ndege RQUV" inamaanisha ndege inayopita kwenye pointi Q, R, na U, ambayo ni ndege ya uso wa kulia wa mchemraba (plane $x=9$).

Ikiwa ndege ni $x=9$:
Vekta ya kawaida (normal vector) ya ndege $x=9$ ni $\vec{n} = (1, 0, 0)$.

Pembe $\theta$ kati ya mstari na ndege inatolewa na formula:

$$\sin\theta = \frac{|\vec{l} \cdot \vec{n}|}{||\vec{l}|| \cdot ||\vec{n}||}$$

Kokotoa dot product $\vec{l} \cdot \vec{n}$:

$$\vec{l} \cdot \vec{n} = (-4.5)(1) + (5)(0) + (12)(0) = -4.5$$

Kokotoa ukubwa wa vekta $\vec{l}$:

$$||\vec{l}|| = \sqrt{(-4.5)^2 + 5^2 + 12^2} = \sqrt{20.25 + 25 + 144} = \sqrt{189.25}$$

Kokotoa ukubwa wa vekta $\vec{n}$:

$$||\vec{n}|| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1$$

Sasa tumia formula:

$$\sin\theta = \frac{|-4.5|}{\sqrt{189.25} \times 1} = \frac{4.5}{\sqrt{189.25}}$$ $$\sin\theta = \frac{4.5}{13.756816...} \approx 0.32712$$ $$\theta = \arcsin(0.32712) \approx 19.09^\circ$$

Ikiwa imezungushwa hadi sehemu moja ya desimali:

$$\boxed{19.1^\circ}$$