smc-1196-70-2D vectors

Vectors, EXT2 V1 2024 HSC 15a

Consider the three vectors \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\) and \(\underset{\sim}{c}=\overrightarrow{O C}\), where \(O\) is the origin and the points \(A, B\) and \(C\) are all different from each other and the origin.

The point \(M\) is the point such that \(\dfrac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})=\overrightarrow{O M}\).

Show that \(M\) lies on the line passing through \(A\) and \(B\). (1 mark)

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The point \(G\) is the point such that \(\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})=\overrightarrow{O G}\).
Show that \(G\) lies on the line passing through \(M\) and \(C\), and lies between \(M\) and \(C\). (2 marks)

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The complex numbers \(x, w\) and \(z\) are all different and all have modulus 1.
Using part (ii), or otherwise, show that \(\dfrac{1}{3}(x+w+z)\) is never a cube root of \(x w z\). (2 marks)

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i. \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

	\(\overrightarrow{O M}\)	\(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
		\(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
		\(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
		\(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).

ii. \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

	\(\overrightarrow{O G}\)	\(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
		\(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
		\(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
		\(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C} \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).

iii. \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of \(xwz\).}\)

Show Worked Solution

i. \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

	\(\overrightarrow{O M}\)	\(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
		\(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
		\(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
		\(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).

ii. \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

	\(\overrightarrow{O G}\)	\(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
		\(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
		\(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
		\(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C} \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).

♦ Mean mark (ii) 43%.

iii. \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)

♦♦♦ Mean mark (iii) 10%.

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of \(xwz\).}\)

Vectors, EXT2 V1 2022 HSC 11e

Let `ℓ_(1)` be the line with equation `([x],[y])=([-1],[7])+lambda([3],[2]),lambda inRR`.

The line `ℓ_(2)` passes through the point `A(-6,5)` and is parallel to `ℓ_(1)`.

Find the equation of the line `ℓ_(2)` in the form `y=mx+c`. (2 marks)

Show Answers Only

`y=2/3x+9`

Show Worked Solution

`m_(ℓ_(1))=2/3`

`text{Equation of}\ ℓ_(2)\ text{has}\ m=2/3\ text{and passes through}\ (-6,5):`

`y-5`	`=2/3(x+6)`
`y`	`=2/3x+9`

Vectors, EXT2 V1 SM-Bank 6

What vector line equation, `underset~r`, corresponds to the Cartesian equation
`qquad (x + 2)/5 = (y - 5)/4` (1 mark)

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Express `underset~v` in Cartesian form where,
`qquad underset~v = ((1),(−4)) + lambda((3),(1))` (1 mark)

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`underset~r = ((−2),(5)) + lambda((5),(4))`
`y = (x – 13)/3`

Show Worked Solution

i. `underset~r = ((−2),(5)) + lambda((5),(4))`

COMMENT: Ensure you know the format for conversion from Cartesian to vector form (part i)!

ii. `((x),(y)) = ((1),(−4)) + lambda((3),(1))`

`x = 1 + 3lambda \ \ => \ lambda = (x – 1)/3`

`y = −4 + lambda\ \ => \ lambda = y + 4`

`y + 4 = (x – 1)/3`

`:. y = (x – 13)/3`

Vectors, EXT2 V1 SM-Bank 2

Find values of `a`, `b`, `c` and `d` such that `underset~v = ((a),(b)) + 2((c),(d))` is a vector equation of a line that passes through `((3),(1))` and `((−3),(−3))`. (2 marks)

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Determine whether `underset~u = ((4),(6)) + lambda((−2),(3))` is perpendicular to `underset~v`. (1 mark)

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Express `underset~u` in Cartessian form. (1 mark)

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`a = 3, b = 1, c = −3, d = −2`, or

`a = −3, b = −3, c = 3, d = 2`
`text(See worked solutions)`
`y = −3/2x + 12`

Show Worked Solution

i. `text(Method 1)`

`overset(->)(OA) = underset~a = ((3),(1)),\ \ overset(->)(OB) = underset~b = ((−3),(−3))`

`overset(->)(AB)`	`= overset(->)(OB) – overset(->)(OA)`
	`= ((−3),(−3)) – ((3),(1))`
	`= ((−6),(−4))`

`underset~v`	`= underset~a + lambdaunderset~b`
	`= ((3),(1)) + lambda((−6),(−4))`
	`= ((3),(1)) + 2((−3),(−2))`

`:. a = 3, b = 1, c = −3, d = −2`

`text(Method 2)`

`overset(->)(BA)`	`= overset(->)(OA) – overset(->)(OB)`
	`= ((3),(1)) – ((−3),(−3))`
	`= ((6),(4))`

`underset~v = ((−3),(−3)) + 2((3),(2))`

`:. a = −3, b = −3, c = 3, d = 2`

ii. `underset~u = ((4),(6)) + lambda((−2),(3))`

`underset~v = ((3),(1)) + 2((−3),(−2))`

`((−2),(3)) · ((−3),(−2)) = 6 – 6 = 0`

`:. underset~u ⊥ underset~v`

iii. `((x),(y))= ((4),(6)) + lambda((−2),(3))`

`x = 4 – 2lambda\ \ \ …\ (1)`

`y = 6 + 3lambda\ \ \ …\ (2)`

`text(Substitute)\ \ lambda = (4 – x)/2\ \ text{from (1) into (2):}`

`y`	`= 6 + 3((4 – x)/2)`
`y`	`= 6 + 6 – (3x)/2`
`y`	`= −3/2x + 12`

Vectors, EXT2 V1 SM-Bank 8

Use the vector form of the linear equations

`3x - 2y = 4` and `3y + 2x - 6 = 0`

to show they are perpendicular. (3 marks)

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`text(Proof)\ text{(See Worked Solutions)}`

Show Worked Solution

`3x-2y`	`= 4`
`3x`	`= 2y + 4`
`3/2 x`	`=y+2`
`x/(2/3)`	`= y + 2`

`underset ~(v_1) = ((0), (-2)) + lambda ((2/3), (1))`

`3y + 2x-6`	`= 0`
`2x`	`= -3y + 6`
`-2/3 x`	`= y-2`
`x/(-3/2)`	`= y-2`

`underset ~(v_2) = ((0), (2)) + lambda ((-3/2), (1))`

`((2/3), (1)) ((-3/2), (1)) = -1 + 1 = 0`

`:. underset ~(v_1) _|_ underset ~(v_2)`