Aussie Maths & Science Teachers: Save your time with SmarterEd
Let \(\underset{\sim}{c}=x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}\) be a unit vector that is perpendicular to both \(\underset{\sim}{a}=2 \underset{\sim}{i}+4 \underset{\sim}{j}-3 \underset{\sim}{k}\) and \(\underset{\sim}{b}=-4 \underset{\sim}{i}-5 \underset{\sim}{j}+3 \underset{\sim}{k}\).
Find all possible vectors \(\underset{\sim}{c}\). (4 marks)
--- 10 WORK AREA LINES (style=lined) ---
\(\underset{\sim}{c}= \pm \dfrac{1}{3}(\underset{\sim}{i}-2\underset{\sim}{j}-2 \underset{\sim}{k})\)
\(\underset{\sim}{c} \cdot \underset{\sim}{a}=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)\left(\begin{array}{c}2 \\ 4 \\ -3\end{array}\right)=2 x+4 y-3 z=0\ \ldots\ (1)\)
\(\underset{\sim}{c} \cdot \underset{\sim}{b}=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)\left(\begin{array}{c}-4 \\ -5 \\ 3\end{array}\right)=-4 x-5 y+3 z=0\ \ldots\ (2)\)
\(\text{Given} \ \ \abs{\underset{\sim}{c}}=1:\)
\(x^2+y^2+z^2=1\ \ldots\ (3)\)
\(\text{Add}\ \ (1)+(2):\)
\(-2 x-y=0 \ \ \Rightarrow\ \ y=-2 x\)
\(\text{Substitute}\ \ y=-2 x \ \ \text{into (1):}\)
\(-6 x-3 z=0 \ \ \Rightarrow\ \ z=-2 x\)
\(\text{Substituting into (3):}\)
\(x^2+4 x^2+4 x^2=1\ \ \Rightarrow\ \ x^2=\dfrac{1}{9}\ \ \Rightarrow\ \ x= \pm \dfrac{1}{3}\)
| \(\underset{\sim}{c}\) | \(=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}\) |
| \(= \pm \dfrac{1}{3}(\underset{\sim}{i}-2\underset{\sim}{j}-2 \underset{\sim}{k})\) |
Three unit vectors \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\), in 3 dimensions, are to be chosen so that \(\underset{\sim}{a} \perp \underset{\sim}{b}, \ \underset{\sim}{b} \perp \underset{\sim}{c}\) and the angle \(\theta\) between \(\underset{\sim}{a}\) and \(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\) is as small as possible.
What is the value of \(\cos \theta\) ?
\(D\)
\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}}\)
\(\theta_{\text{min}} \ \Rightarrow \ \cos\,\theta_{\text{max}}:\)
\(\underset{\sim}{a} \cdot\left(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\right)=\abs{\underset{\sim}{a}}^2+\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{a} \cdot \underset{\sim}{c}=1+\underset{\sim}{a} \cdot \underset{\sim}{c}\)
\(\text{Find}\ \ \abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}:\)
| \(\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}^2\) | \(=\abs{\underset{\sim}{a}}^2+\abs{\underset{\sim}{b}}^2+\abs{\underset{\sim}{c}}^2+2\left(\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{c}+\underset{\sim}{a} \cdot \underset{\sim}{c}\right)\) | |
| \(=3+2 \underset{\sim}{a} \cdot \underset{\sim}{c}\) |
\(\cos \theta=\dfrac{1+a \cdot c}{\sqrt{3+2 a \cdot c}}\)
\(\underset{\sim}{a} \cdot\underset{\sim}{c}=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{c}} \cos\, \alpha = \cos\,\alpha \)
\(0 \leqslant \cos\, \alpha \, \leqslant 1\)
\(\therefore \cos \theta_{\text{max}}=\dfrac{2}{\sqrt{5}}\)
\(\Rightarrow D\)
The vector \(\underset{\sim}{a}\) is \(\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) and the vector \(\underset{\sim}{b}\) is \(\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\). --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- 1. \(\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\) ii. \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\) \( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\) \(\therefore\ \text {Vectors are perpendicular.}\) i. \(\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\) \(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\) \( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)
Show Answers Only
Show Worked Solution
ii. \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)
\(\therefore\ \text{Vectors are perpendicular.}\)
Which of the following vectors is perpendicular to \(3 \underset{\sim}{i}+2 \underset{\sim}{j}-5 \underset{\sim}{k}\) ?
\(D\)
\(\text{Consider option}\ D:\)
\[\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c} 3 \\ 2 \\ -5 \end{array}\right) \left(\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right) = 9-4-5=0 \]
\(\Rightarrow D\)
Consider the vectors \(\underset{\sim}{\text{u}}(x)=-\text{cosec}(x) \underset{\sim}{\text{i}}+\sqrt{3} \underset{\sim}{\text{j}}\) and \(\underset{\sim}{\text{v}}(x)=\text{cos}(x) \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\).
If \(\underset{\sim}{\text{u}}(x)\) is perpendicular to \(\underset{\sim}{\text{v}}(x)\), then possible values for \(x\) are
\(A\)
\(\underset{\sim}{u} \perp \underset{\sim}{v} \ \Rightarrow \ \underset{\sim}{u} \cdot \underset{\sim}{v}=0\)
\(-\text{cosec}(x) \cos (x)+\sqrt{3}=0\)
\begin{aligned}
-\dfrac{\cos (x)}{\sin (x)}+\sqrt{3} & =0 \\
\tan (x) & =\dfrac{1}{\sqrt{3}}
\end{aligned}
\(x=\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\)
\(\Rightarrow A\)
Prove that the vectors `4 underset ~i + 5 underset ~j - 2 underset ~k` and ` −5 underset ~i + 6 underset ~j + 5underset ~k`, are perpendicular. (2 marks)
`text{See Worked Solution}`
| `underset ~a ⋅ underset ~b` | `= ((4),(5),(-2))((-5),(6),(5))` |
| `=4 xx (−5) + 5 xx 6 + (−2) xx 5` | |
| `= -20+30+10` | |
| `=0` |
`text(S)text(ince)\ \ underset ~a ⋅ underset ~b =0 \ =>\ \ underset ~a _|_ underset ~b`
Consider the two vectors `underset~u = 2 underset~i - underset~j + 3 underset~k` and `underset~v = p underset~i + underset~j + 2 underset~k`.
For what values of `p` are `underset~u - underset~v` and `underset~u + underset~v` perpendicular? (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`p= ± 3`
`underset~u – underset~v = ((-2),(-1),(3)) – ((p),(1),(2)) = ((-2 – p),(-2),(1))`
`underset~u + underset~v = ((-2),(-1),(3)) + ((p),(1),(2)) = ((p – 2),(0),(5))`
`⊥ \ text{when} \ \ (underset~u – underset~v) · (underset~u + underset~v ) = 0 :`
`((-2 – p),(-2),(1)) · ((p – 2),(0),(5)) = 0`
| `-(p + 2)(p-2) + 5` | `= 0` |
| `-(p^2 – 4) + 5` | `= 0` |
| `-p^2 + 9` | `= 0` |
| `p^2` | `= 9` |
| `p` | `= ± 3` |
Consider the vector `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`. (2 marks)
--- 2 WORK AREA LINES (style=lined) ---
| i. | `|underset ~a|` | `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)` |
| `= sqrt 6` |
| `:. hat underset ~a` | `= underset ~a/|underset ~a|` |
| `= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)` |
Mean mark part (b) 51%.
| ii. | `underset ~a ⋅ underset ~i` | `= sqrt 3 xx 1 = sqrt 3` |
| `underset ~a ⋅ underset ~i` | `= |underset ~a||underset ~i| cos theta` | |
| `= sqrt 6 cos theta` | ||
| `sqrt 3` | `= sqrt 6 cos theta` |
| `cos theta` | `= 1/sqrt 2` |
| `:. theta` | `= pi/4 = 45^@` |
iii. `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`
| `6 – m + 5 sqrt 2` | `=0` | |
| `:. m` | `=6 + 5 sqrt 2` |
The coordinates of three points are `A\ ((– 1), (2), (4)), \ B\ ((1), (0), (5)) and C\ ((3), (5), (2)).`
--- 2 WORK AREA LINES (style=lined) ---
Prove that the triangle has a right angle at `A.` (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
| i. | `vec(AB)` | `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k` |
| `= 2underset~i – 2underset~j + underset~k` |
| ii. | `overset(->)(AC)` | `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k` |
| `= 4underset~i + 3underset~j – 2underset~k` |
| `overset(->)(AB) · overset(->)(AC)` | `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)` |
| `= 8 – 6 – 2` | |
| `= 0` |
`=> overset(->)(AB) ⊥ overset(->)(AC)`
`:. DeltaABC\ text(has a right angle at)\ A.`
| iii. | `overset(->)(BC)` | `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k` |
| `= 2underset~i + 5underset~j – 3underset~k` |
| `|overset(->)(BC)|` | `= sqrt(2^2 + 5^2 + (−3)^2)` |
| `= sqrt(4 + 25 + 9)` | |
| `= sqrt38` |
If `underset ~a = -2 underset ~i - underset ~j + 3 underset ~k` and `underset ~b = -m underset ~i + underset ~j + 2 underset ~k`, where `m` is a real constant, find `m` such that the vector `underset ~a - underset ~b` will be perpendicular to vector `underset ~b`. (2 marks)
`0 or 2`
| `underset ~a – underset ~b` | `= (m – 2) underset ~i – 2 underset ~j + underset ~k ` |
| `(underset ~a – underset ~b) ⋅ underset ~b` | `= -m(m – 2) – 2(1) + 1(2) = 0` |
| `0` | `= -m^2 + 2m – 2 + 2` |
| `0` | `= 2m – m^2` |
| `0` | `= m(2 – m)` |
| `:. m` | `= 0 \ or \ 2` |
Two vectors are given by `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k` and `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where `m`, `n in R^+`.
If `|\ underset ~a\ | = 10` and `underset ~a` is perpendicular to `underset ~b`, then find `m` and `n`. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
`m=5sqrt3, \ n=sqrt3/3`
`text(Using)\ \ |\ underset ~a\ | = 10:`
| `10` | `= sqrt(4^2 + m^2 + (-3)^2)` |
| `100` | `=m^2 +25` |
| `m` | `= 5sqrt3` |
`text(S)text(ince)\ \ underset ~a _|_ underset ~b :`
| `underset ~a ⋅ underset ~b` | `= 0` |
| `0` | `=4 xx (−2) + mn + (−3) xx (−1)` |
| `0` | `= mn-5` |
| `n` | `=5/(5sqrt3)` |
| `=sqrt3/3` |
Find the value(s) of `m` so that the vectors `underset~a = 2underset~i + m underset~j - 3underset~k` and `underset~b = m^2underset~i - underset~j + underset~k` are perpendicular. (2 marks)
`m = 3/2, quad m = -1`
`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`
| `underset ~a ⋅ underset ~b` | `= 2m^2 + m(-1) + (-3)(1)` |
| `0` | `= 2m^2 – m – 3` |
| `0` | `= (2m – 3)(m + 1)` |
`:. m = 3/2, quad m = -1`
Consider the three vectors
`underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k` and `underset ~c = underset ~i + underset ~j - underset ~k`, where `m in R.`
| i. | `|underset~b|` | `= sqrt(1^2 + 2^2 + m^2)` | `= 2sqrt3` |
| `1 + 4 + m^2` | `= 4 xx 3` | ||
| `m^2` | `= 7` | ||
| `m` | `= ±sqrt7` |
| ii. | `underset~a * underset~b` | `= 1 xx 1 + (−1) xx 2 + 2 xx m` | |
| `0` | `= 1 – 2 + 2m\ \ ( underset~a ⊥ underset~b)` | ||
| `2m` | `= 1` | ||
| `m` | `= 1/2` |