If the sum of two unit vectors is a unit vector, then the magnitude of the difference of the two vectors is
- \(0\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(\sqrt{2}\)
- \(\sqrt{3}\)
- \(\sqrt{5}\)
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If the sum of two unit vectors is a unit vector, then the magnitude of the difference of the two vectors is
\(D\)
\(\text{Vectors can be drawn as two sides of an equilateral triangle.}\)
\(\text{Using the cosine rule for the difference between the two vectors:}\)
\(c^2\) | \(=a^2+b^2-2ab\, \cos C\) | |
\(=1+1-2\times -\dfrac{1}{2} \) | ||
\(=3\) | ||
\(c\) | \(=\sqrt{3}\) |
\(\Rightarrow D\)
A plane contains the points \( A(1,3,-2), B(-1,-2,4)\) and \( C(a,-1,5)\), where \(a\) is a real constant. The plane has a \(y\)-axis intercept of 2 at the point \(D\).
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a. \(D(0,2,0)\)
b. \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)
\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)
c. \(4 x+2 y+3 z =4\)
d. \(a=-\dfrac{9}{4}\)
e. \(A=\sqrt{29}\)
a. \(D(0,2,0)\)
b. \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)
\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)
c. \(\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ -2 & -5 & 6 \\ -1 & -1 & 2\end{array}\right|\)
\(\ \quad \quad \quad \quad \ \begin{aligned} & =(-10+6)\underset{\sim}{\text{i}}-(-4+6)\underset{\sim}{\text{j}}+(2-5) \underset{\sim}{\text{k}} \\ & =-4 \underset{\sim}{i}-2 \underset{\sim}{j}-3 \underset{\sim}{k}\end{aligned}\)
\(\text{Plane:}\ \ -4 x-2 y-3 z=k\)
\(\text{Substitute}\ D(0,2,0)\ \text{into equation}\ \ \Rightarrow \ \text{k}=-4\)
\begin{aligned}
\therefore-4 x-2 y-3 z & =-4 \\
4 x+2 y+3 z & =4
\end{aligned}
d. \(\text{Find}\ a\ \Rightarrow \ \text{substitute}\ (a, 1,-5)\ \text{into plane equation:}\)
\begin{aligned}
-4 & =-4a+2-15 \\
4 a & =-9 \\
a & =-\dfrac{9}{4}
\end{aligned}
e. | \(\text{Area}\) | \(=|\overrightarrow{A B} \times \overrightarrow{A D}|\) |
\(=|-4\underset{\sim}{i}-2\underset{\sim}{j}-3\underset{\sim}{k}|\) | ||
\(=\sqrt{16+4+9} \) | ||
\(=\sqrt{29}\) |
A triangle has vertices `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)` and `C(2, –2, sqrt3 + 3)`.
a. `overset(->)(BA) = ((sqrt3 + 1), (-2), (4)) – ((1), (-2), (3)) = ((sqrt3), (0), (1))`
`overset(->)(BC) = ((2), (-2), (sqrt3 + 3)) – ((1), (-2), (3)) = ((1), (0), (sqrt3))`
`cos ∠ABC` | `= (overset(->)(BA) · overset(->)(BC))/ (|overset(->)(BA)| |overset(->)(BC)|` |
`= (2 sqrt3)/(sqrt4 sqrt4)` | |
`= (sqrt3)/(2)` |
`:. \ ∠ABC = (pi)/(6)`
b. | `text(Area)` | `= (1)/(2) · |overset(->)(BA)| |overset(->)(BC)| \ sin ∠ABC` |
`= (1)/(2) xx 2 xx 2 xx sin \ (pi)/(6)` | ||
`= 1 \ text(u²)` |
Vectors `underset ~a, underset ~b` and `underset ~c` are shown below.
From the diagram it follows that
A. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2`
B. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a\ | |\ underset ~b\ |`
C. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a * underset ~b\ |`
D. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a\ | |\ underset ~b\ |`
E. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a * underset ~b\ |`
`D`
`underset~ c + underset~ b = underset~ a\ \ => \ underset~ c = underset~ a – underset~ b`
`:. underset~ c * underset~ c` | `= (underset~ a – underset~ b) * (underset~ a – underset~ b)` |
`= underset~ a * underset~ a – underset~ a * underset~ b -underset~b*underset~a + underset~ b * underset~ b` | |
`= underset~ a * underset~ a + underset~ b * underset~ b-2underset~b*underset~a ` |
`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\ underset~ b\ |^2 – 2 |\ underset~ a\ | |\ underset~ b\ | cos 120^@`
`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\underset~ b\ |^2 + |\ underset~ a\ | |\ underset~ b\ |`
`=> D`
The diagram below shows a rhombus, spanned by the two vectors `underset~a` and `underset~b`.
It follows that
A. `underset~a.underset~b = 0`
B. `underset~a = underset~b`
C. `(underset~a + underset~b).(underset~a - underset~b) = 0`
D. `|\ underset~a + underset~b\ | = |\ underset~a - underset~b|`
E. `2underset~a + 2underset~b = underset~0`
`C`
`text(Consider A:)`
`text(If)\ \ underset~a · underset~b = 0\ \ =>\ \ underset~a ⊥ underset~b\ \ text{(incorrect)}`
`text(Consider B:)`
`underset~a != underset~b\ \ text{(incorrect)}`
`text(Consider C:)`
`underset~a + underset~b` | `= overset(->)(OC)` |
`underset~a – underset~b` | `= overset(->)(BA)` |
`overset(->)(OC) · overset(->)(BA)=0`
`text{The diagonals of a rhombus are perpendicular (correct)}`
`=> C`
The coordinates of three points are `A (– 1, 2, 4), \ B(1, 0, 5) and C(3, 5, 2).`
Prove that the triangle has a right angle at `A.` (2 marks)
a. | `vec(AB)` | `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k` |
`= 2underset~i – 2underset~j + underset~k` |
b. | `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.`
c. | `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` |
Consider the rhombus `OABC` shown below, where `vec (OA) = a underset ~i` and `vec (OC) = underset ~i + underset ~j + underset ~k`, and `a` is a positive real constant.
a. `|\ vec(OA)\ | = |\ vec(OC)\ |,`
`:. a` | `= sqrt (1^2 + 1^2 + 1^2)` |
`= sqrt 3` |
b. `overset(->)(CB) = overset(->)(OA), \ \ overset(->)(AB) = overset(->)(OC)`
`overset(->)(OB) = overset(->)(OC) + overset(->)(CB)`
`= underset~i + underset~j + underset~k + sqrt3 underset~i`
`= (sqrt3 + 1)underset~i + underset~j + underset~k`
`overset(->)(AC) = overset(->)(AO) + overset(->)(OC)`
`= −sqrt3underset~i + underset~i + underset~j + underset~k`
`= (1 – sqrt3)underset~i + underset~j + underset~k`
`overset(->)(AC) · overset(->)(OB)` | `= (1 + sqrt3)(1 – sqrt3) + 1 + 1` |
`= 1 – 3 + 1 + 1` | |
`= 0` |
`:. overset(->)(AC) ⊥ overset(->)(OB)`
In the diagram above, `LOM` is a diameter of the circle with centre `O`.
`N` is a point on the circumference of the circle.
If `underset~r = vec(ON)` and `underset ~s = vec(MN)`, then `vec(LN)` is equal to
`E`
`vec(LN)` | `= vec(LM) + vec(MN)` |
`vec(LN)` | `= vec(LO) + vec(ON)` |
`= 1/2\ vec(LM) + vec(ON)` |
`:. vec(LM) + underset~s` | `= 1/2\ vec(LM) + underset~r` |
`1/2\ vec(LM)` | `= underset ~r – underset~s` |
`vec(LM)` | `= 2 underset~r – 2 underset~s` |
`:. vec(LN)` | `= 2 underset~r – 2 underset~s + underset~s` |
`= 2 underset~r – underset~s` |
`=> E`