Aussie Maths & Science Teachers: Save your time with SmarterEd
`ABCDEFGH` are the vertices of a rectangular prism.
i. | `A(2, text{−2}, 0),` | `G(text{−2}, 2, 2)` |
`D(2, 2, 0),` | `F (text{−2}, text{−2}, 2)` |
`text(Midpoint)\ AG = ((1/2 (2 – 2)),(1/2 (text{−2} + 2)),(1/2 (0 + 2))) = ((0), (0), (1))`
`text(Midpoint)\ DF = ((1/2 (2 – 2)),(1/2 (2 – 2)),(1/2 (0 + 2))) = ((0), (0), (1))`
`text(S) text(ince midpoints are the same), AG and DF\ text(intersect.)`
ii. | `vec(AG) = ((text{−2}), (2), (2)) – ((2), (text{−2}), (0)) = ((text{−4}), (4), (2))` |
`vec(DF) = ((text{−2}), (text{−2}), (2)) – ((2), (2), (0)) = ((text{−4}), (text{−4}), (2))` |
`vec (AG) ⋅ vec (DF) = |\ vec (AG)\ | ⋅ |\ vec(DF)\ |\ cos theta`
`((text{−4}), (4), (2)) ⋅ ((text{−4}), (text{−4}), (2)) = sqrt 36 sqrt 36 cos theta`
`16 – 16 + 4` | `= 36 cos theta` |
`cos theta` | `= 1/9` |
`theta` | `= 83.62…` |
`= 83^@37′` |
A cube with side length 3 units is pictured below.
i. `A(3, 0 , 0), \ \ G(0, 3, 3)`
`vec(AG)` | `= ((0), (3), (3)) – ((3), (0), (0)) = ((text{−3}), (3), (3))` | |
`|\ vec(AG)\ |` | `= sqrt (9 + 9 + 9)` | |
`= 3 sqrt 3\ text(units)` |
ii. | `H (3, 3, 3)` |
`vec(BH) = ((3), (3), (3))` |
`vec(AG) ⋅ vec(BH)` | `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta` |
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` | `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta` |
`-9 + 9 + 9` | `= 27 cos theta` |
`cos theta` | `= 1/3` |
`theta` | `= 70.52…` |
`= 70^@32′` |