smc-7425-25-Point lies on line

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 2024 HSC 12e

The line \(\ell\) passes through the points \(A(3,5,-4)\) and \(B(7,0,2)\).

Find a vector equation of the line \(\ell\). (1 mark)

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Determine, giving reasons, whether the point \(C(10,5,-2)\) lies on the line \(\ell\). (2 marks)

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i. \(\text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )\)

ii. \(\text{If \(C\) lies on the line}, \ \exists \lambda \in \mathbb{R}\ \ \text{such that:}\)

\(\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

\(\text{Find \(\lambda\) for \(x\)-component: }\)

\(10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}\)

\(\text{Check \(\lambda=\dfrac{7}{4}\) for \(y\)-component:}\)

\(5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}\)

\(\therefore C\ \text{does not lie on the line.}\)

Show Worked Solution

i. \(A(3,5,-4), \quad B(7,0,2)\)

\(\overrightarrow{A B}=\left(\begin{array}{l}7 \\ 0 \\ 2\end{array}\right)-\left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)=\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

\(\therefore \text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )\)

ii. \(\text{If \(C\) lies on the line}, \ \exists \lambda \in \mathbb{R}\ \ \text{such that:}\)

\(\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

\(\text{Find \(\lambda\) for \(x\)-component: }\)

\(10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}\)

\(\text{Check \(\lambda=\dfrac{7}{4}\) for \(y\)-component:}\)

\(5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}\)

\(\therefore C\ \text{does not lie on the line.}\)

Vectors, EXT2 V1 2023 HSC 5 MC

Which of the following is a true statement about the lines \(\ell_1={\displaystyle\left(\begin{array}{cc}-1 \\ 2 \\ 5\end{array}\right)+\lambda\left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right)}\) and \(\ell_2=\left(\begin{array}{c}3 \\ -10 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right) ?\)

\(\ell_1\) and \(\ell_2\) are the same line.
\(\ell_1\) and \(\ell_2\) are not parallel and they intersect.
\(\ell_1\) and \(\ell_2\) are parallel and they do not intersect.
\(\ell_1\) and \(\ell_2\) are not parallel and they do not intersect.

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\(A\)

Show Worked Solution

\(\text{Since}\ \ \left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right) = -1 \left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right), \ \ell_1\ \text{is parallel to}\ \ell_2 \)

\(\text{Test if point}\ (3,-10,1)\ \text{lies on}\ \ell_1: \)

\(\text{i.e.}\ \ \exists \lambda\ \ \text{such that} \)

♦ Mean mark 49%.

\( \left(\begin{array}{cc}3 \\ -10 \\ 1\end{array}\right) = \left(\begin{array}{cc}-1 \\ 2 \\ 5\end{array}\right) + \lambda \left(\begin{array}{cc}-1 \\ 3 \\ 1\end{array}\right)\)

\(\lambda = -4\ \ \text{satisfies equation} \)

\(\therefore\ \ell_1\ \text{and}\ \ell_2\ \text{are the same line.}\)

\(\Rightarrow A\)

Vectors, EXT2 V1 EQ-Bank 16

Show that the points `A(2, 1, text{−1}), \ B(4, 2, text{−3})` and `C(text{−4}, text{−2}, 5)` are collinear. (2 marks)

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

Show Worked Solution

`vec(AB) = ((4), (2), (text{−3})) – ((2), (1), (text{−1})) = ((2), (1), (text{−2}))`

`vec(AC) = ((text{−4}), (text{−2}), (5)) – ((2), (1), (text{−1})) = ((text{−6}), (text{−3}), (6)) = -3((2), (1), (text{−2}))`

`text(S) text(ince)\ vec(AB)\ text(||)\ vec(AC) and A\ text(lies on both lines,)`

`A, B, C\ \ text(are collinear).`

Vectors, EXT2 V1 EQ-Bank 24

Determine the point of intersection of `underset ~a` and `underset ~b` given.

`qquad underset ~a = ((3), (5), (1)) + lambda ((1), (3), (text{−2})),` and

`qquad underset ~b = ((text{−2}), (2), (text{−1})) + mu ((1), (text{−1}), (2))` (2 marks)

Determine if the point `(2, text{−2}, 5)` lies on `underset ~b`. (1 mark)

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a. `((1), (text{−1}), (5))`

b. `text(See Worked Solutions)`

Show Worked Solution

a. `text(At point of intersection:)`

`3 + lambda`	`= -2 + mu\ \ text{… (1)}`
`5 + 3 lambda`	`= 2-mu\ \ text{… (2)}`
`1-2 lambda`	`= -1 + 2 mu\ \ text{… (3)}`

`(1) + (2)`

`8 + 4 lambda`	`= 0`
`lambda`	`= -2,\ \ mu = 3`

`text{Intersection (using}\ lambda = –2 text{)}:`

`((x), (y), (z)) = ((3-2 xx 1), (5-2 xx 3), (1-2 xx text{−2})) = ((1), (text{−1}), (5))`

b. `text(If)\ (2, text{−2}, text{−10})\ text(lies on)\ underset ~b, ∃ mu\ \ text(that satisfies:)`

`-2 + mu`	`= 2\ \ text{… (1)}\ => \ mu = 4`
`2-mu`	`= text{−2}\ \ text{… (2)}\ => \ mu = 4`
`-1 + 2 mu`	`= 5\ \ text{… (3)}\ => \ mu = 3`

`=>\ text(No solution)`

`:. (2, text{−2}, 5)\ \ text(does not lie on)\ underset ~b.`

Vectors, EXT2 V1 EQ-Bank 11

Find the equation of line vector `underset ~r`, given it passes through `(1, 3, –2)` and `(2, –1, 2)`. (2 marks)

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Determine if `underset ~r` passes through `(4, –9, 10)`. (1 mark)

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a. `underset ~r = ((1), (3), (-2)) + lambda ((1), (-4), (4)) or`

`underset ~r = ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

b. text(See Worked Solutions)`

Show Worked Solution

a. `text(Method 1)`

`text(Let)\ \ A(1, 3, –2) and B(2, –1, 2)`

`vec (AB)`	`= ((2), (-1), (2))-((1), (3), (-2)) = ((1), (-4), (4))`
`underset ~r`	`= ((1), (3), (-2)) + lambda ((1), (-4), (4))`

`text (Method 2)`

`vec (BA)`	`= ((1), (3), (-2))-((2), (-1), (2)) = ((-1), (4), (-4))`
`underset ~r`	`= ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

b. `text(If)\ \ (4, –9, 10)\ \ text(lies on the vector line,)`

`∃ lambda\ \ text(that satisfies:)`

`1 + lambda`	`= 4\ \ text{… (1)}`
`3-4 lambda`	`= -9\ \ text{… (2)}`
`-2 + 4 lambda`	`= 10\ \ text{… (3)}`

`lambda = 3\ \ text(satisfies all equations)`

`:. (4, –9, 10)\ \ text(lies on the line.)`