1. **Problem Statement:** We have a cuboid PQRSTUVW with dimensions: - PQ = 19 cm (length) - QR = 16 cm (height) - RW = 4 cm (depth) Points M and N are midpoints of UT and VW respectively. We need to find: (a) Length of line RM. (b)(i) Angle between line RM and plane PQRS. (b)(ii) Angle between lines RM and MQ. (b)(iii) Obtuse angle between planes PMNQ and MNWT. --- 2. **Set up coordinate system:** Place P at origin $(0,0,0)$. - $P=(0,0,0)$ - $Q=(19,0,0)$ (along x-axis) - $R=(19,16,0)$ (height along y-axis) - $S=(0,16,0)$ - $W=(19,16,4)$ (depth along z-axis) - $T=(0,16,4)$ - $U=(0,0,4)$ - $V=(19,0,4)$ Midpoints: - $M$ midpoint of $UT$: $U=(0,0,4)$ and $T=(0,16,4)$ $$M=\left(0,\frac{0+16}{2},4\right)=(0,8,4)$$ - $N$ midpoint of $VW$: $V=(19,0,4)$ and $W=(19,16,4)$ $$N=\left(19,\frac{0+16}{2},4\right)=(19,8,4)$$ --- 3. **(a) Calculate length of RM:** Coordinates: - $R=(19,16,0)$ - $M=(0,8,4)$ Distance formula: $$RM=\sqrt{(0-19)^2+(8-16)^2+(4-0)^2}$$ $$=\sqrt{(-19)^2+(-8)^2+4^2}$$ $$=\sqrt{361+64+16}=\sqrt{441}=21$$ **Answer (a):** $RM=21$ cm --- 4. **(b)(i) Angle between line RM and plane PQRS:** Plane PQRS lies in $z=0$ plane. The angle between line and plane is complementary to angle between line and plane's normal vector. Normal vector to plane PQRS is along $z$-axis: $\vec{n}=(0,0,1)$ Vector $\vec{RM} = M - R = (0-19,8-16,4-0) = (-19,-8,4)$ Calculate angle $\theta$ between $\vec{RM}$ and $\vec{n}$: $$\cos\theta = \frac{|\vec{RM} \cdot \vec{n}|}{|\vec{RM}||\vec{n}|} = \frac{|4|}{21 \times 1} = \frac{4}{21}$$ Angle between line and plane: $$\phi = 90^\circ - \theta = 90^\circ - \cos^{-1}\left(\frac{4}{21}\right)$$ Calculate $\cos^{-1}(4/21)$: $$\cos^{-1}(0.1905) \approx 79.01^\circ$$ So, $$\phi = 90^\circ - 79.01^\circ = 10.99^\circ$$ Rounded to 2 decimal places: $$\boxed{10.99^\circ}$$ --- 5. **(b)(ii) Angle between lines RM and MQ:** Coordinates: - $M=(0,8,4)$ - $R=(19,16,0)$ - $Q=(19,0,0)$ Vectors: $$\vec{RM} = M - R = (-19,-8,4)$$ $$\vec{MQ} = Q - M = (19-0,0-8,0-4) = (19,-8,-4)$$ Dot product: $$\vec{RM} \cdot \vec{MQ} = (-19)(19) + (-8)(-8) + 4(-4) = -361 + 64 -16 = -313$$ Magnitudes: $$|\vec{RM}|=21$$ (from part a) $$|\vec{MQ}|=\sqrt{19^2 + (-8)^2 + (-4)^2} = \sqrt{361 + 64 + 16} = \sqrt{441} = 21$$ Cosine of angle $\alpha$: $$\cos\alpha = \frac{-313}{21 \times 21} = \frac{-313}{441} \approx -0.7098$$ Angle: $$\alpha = \cos^{-1}(-0.7098) \approx 134.99^\circ$$ Rounded to 2 decimal places: $$\boxed{134.99^\circ}$$ --- 6. **(b)(iii) Obtuse angle between planes PMNQ and MNWT:** Planes defined by points: - Plane 1: PMNQ - Plane 2: MNWT Find normal vectors of each plane. **Plane PMNQ:** Points: - $P=(0,0,0)$ - $M=(0,8,4)$ - $N=(19,8,4)$ - $Q=(19,0,0)$ Use vectors: $$\vec{PM} = M - P = (0,8,4)$$ $$\vec{PQ} = Q - P = (19,0,0)$$ Normal vector $\vec{n_1} = \vec{PM} \times \vec{PQ}$ Calculate cross product: $$\vec{n_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 8 & 4 \\ 19 & 0 & 0 \end{vmatrix} = (8 \times 0 - 4 \times 0)\mathbf{i} - (0 \times 0 - 4 \times 19)\mathbf{j} + (0 \times 0 - 8 \times 19)\mathbf{k}$$ $$= 0\mathbf{i} - (-76)\mathbf{j} - 152\mathbf{k} = (0,76,-152)$$ **Plane MNWT:** Points: - $M=(0,8,4)$ - $N=(19,8,4)$ - $W=(19,16,4)$ - $T=(0,16,4)$ Vectors: $$\vec{MN} = N - M = (19,0,0)$$ $$\vec{MT} = T - M = (0,8,0)$$ Normal vector $\vec{n_2} = \vec{MN} \times \vec{MT}$ Calculate cross product: $$\vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 19 & 0 & 0 \\ 0 & 8 & 0 \end{vmatrix} = (0 \times 0 - 0 \times 8)\mathbf{i} - (19 \times 0 - 0 \times 0)\mathbf{j} + (19 \times 8 - 0 \times 0)\mathbf{k}$$ $$= 0\mathbf{i} - 0\mathbf{j} + 152\mathbf{k} = (0,0,152)$$ --- Calculate angle $\beta$ between $\vec{n_1}$ and $\vec{n_2}$: Dot product: $$\vec{n_1} \cdot \vec{n_2} = 0 \times 0 + 76 \times 0 + (-152) \times 152 = -23104$$ Magnitudes: $$|\vec{n_1}| = \sqrt{0^2 + 76^2 + (-152)^2} = \sqrt{5776 + 23104} = \sqrt{28880} \approx 170.01$$ $$|\vec{n_2}| = \sqrt{0^2 + 0^2 + 152^2} = 152$$ Cosine of angle: $$\cos\beta = \frac{-23104}{170.01 \times 152} = \frac{-23104}{25841.52} \approx -0.894$$ Angle: $$\beta = \cos^{-1}(-0.894) \approx 153.58^\circ$$ This is obtuse angle between planes. Rounded to 2 decimal places: $$\boxed{153.58^\circ}$$ --- **Final answers:** - (a) $RM=21$ cm - (b)(i) Angle between RM and plane PQRS $=10.99^\circ$ - (b)(ii) Angle between RM and MQ $=134.99^\circ$ - (b)(iii) Obtuse angle between planes PMNQ and MNWT $=153.58^\circ$