smc-1210-10-Cube/Rect Prism

Vectors, EXT2 V1 EQ-Bank 13

In a rectangular prism, `M` is the midpoint of `AD`.

Use vector methods to find

`angle HBD` to 1 decimal place (2 marks)

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`angle HBM` to 1 decimal place (2 marks)

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i. `text{Consider the position vectors of points using}\ B\ text{as origin:}`

`vec(BH)=((0),(9),(6)), \ abs(vec(BH))=sqrt(9^2+6^2)=sqrt117`

`vec(BD)=((4),(9),(0)), \ abs(vec(BD))=sqrt(4^2+9^2)=sqrt97`

`cos angleHBD`	`=(vec(BH)vec(BD))/(abs(vec(BH))abs(vec(BD)))`
	`=(0xx4+9xx9+6xx0)/(sqrt117 xx sqrt97)`
`angleHBD`	`=cos^(-1)(81/(sqrt117 xx sqrt97))`
	`=40.5°`

ii. `vec(BM)=((4),(9/2),(0)), \ abs(vec(BM))=sqrt(4^2+(9/2)^2)=sqrt(145/4)`

`cos angleHBM`	`=(vec(BH) * vec(BM))/(abs(vec(BH))*abs(vec(BM)))`
	`=(0xx4+9xx9/2+6xx0)/(sqrt117 xx sqrt(145/4))`
`angleHBM`	`=cos^(-1)(40.5/(sqrt117 xx sqrt(145/4)))`
	`=51.5°\ \text{(1 d.p.)}`

Four cubes are placed in a line as shown on the diagram.

Which of the following vectors is equal to `overset->{AB} + overset->{CQ}`

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`B`

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`overset->{AB} \ + \ overset->{CQ}`	`= overset->{CD} + overset->{DP}`
	`= overset->{CP}`

`=>\ B`

A cube with side length 3 units is pictured below.

Calculate the magnitude of vector `vec(AG)`. (1 mark)

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Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`. (3 marks)

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i. `A(3, 0 , 0), \ \ G(0, 3, 3)`

ii.	`H (3, 3, 3)`
	`vec(BH) = ((3), (3), (3))`

`ABCDEFGH` are the vertices of a rectangular prism.

Show that the internal diagonals of the prism, `AG` and `DF`, intersect. (2 marks)

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Calculate the acute angle, `theta`, between the diagonals, to the nearest minute. (2 marks)

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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`