Question
Quadrilateral OABC is shown on the following set of axes.
$OABC\text{ is symmetrical about }[OB].$
$A\text{ has coordinates }(6,0)\text{ and }C\text{ has coordinates }(3,3\sqrt{3}).$
$(a)\quad(i)\text{ Write down the coordinates of the midpoint of }[AC].$
$\quad\quad(ii)\text{ Hence or otherwise, find the equation of the line passing through the points }O\text{ and }B.$
$(b)\text{ Given that }[OA]\text{ is perpendicular to }[AB],\text{ find the area of the quadrilateral }OABC.$
▶️Answer/Explanation
Solution:-
(b) substituting $x=6$ into their equation
so at B $y=2\sqrt{3}$
area of triangle $OAB=\frac{1}{2}\times6\times2\sqrt{3}=6\sqrt{3}$
area of quadrilateral $OABC=12\sqrt{3}$
Solution
(a) (i)The mid point of AC given
$A\text{ has coordinates }(6,0)\text{ and }C\text{ has coordinates }(3,3\sqrt{3})$, let the mid point be M
Therefore, the mid point M is $\frac{6+3}{2}, \frac{0+3\sqrt{3}}{2}$
Coordinates of M = $\frac{9}{2}, \frac{3\sqrt{3}}{2}$
(ii) Point M lies on OB, the slope of line OB can be found by Coordinates of point O(0,0) and coordinates of M = $\frac{9}{2}, \frac{3\sqrt{3}}{2}$
Slope $ tan\theta = m = \frac{\frac{3\sqrt{3}}{2}-0}{\frac{9}{2}-0}$ = $\frac{1}{\sqrt{3}}$
Therefore, equation of line OB passing through origin, y=$\frac{1}{\sqrt{3}}$x
(b) Since $OABC\text{ is symmetrical about }[OB]$
Therefore area of the quadrilateral }OABC = $(2\times area of triangle OAB)$
= $(2\times \frac{1}{2} OA. AB)$
$\text{ Given that }[OA]\text{ is perpendicular to }[AB]$, so $ tan\theta$ = $\frac{1}{\sqrt{3}}$ = $\frac{AB}{OA}$ = $\frac{AB}{6}$
Therefore AB =\(\frac{6}{\sqrt{3}}\),Therefore area of the quadrilateral OABC =\( 2 \times\frac{6}{\sqrt{3}}\)
area of quadrilateral OABC=\(12\sqrt{3}\)
Question
Consider the three planes
∏1 : 2x – y + z = 4
∏2 : x – 2y + 3z = 5
∏3 : -9x + 3y – 2z = 32
(a) Show that the three planes do not intersect.
(b) (i) Verify that the point P(1 , -2 , 0) lies on both ∏1 and ∏2 .
(ii) Find a vector equation of L, the line of intersection of ∏1 and ∏2 .
(c) Find the distance between L and ∏3 .
▶️Answer/Explanation
Ans:
(a) METHOD 1
attempt to eliminate a variable
obtain a pair of equations in two variables
EITHER
-3x + z = -3 and
-3x + z = 44
OR
-5x + y = -7 and
-5x + y = 40
OR
3x – z = 3 and
3x – z = \(\frac{79}{5}\)
THEN
the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -\(\frac{79}{5}\)
Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.
hence the three planes do not intersect
METHOD 2
vector product of the two normals = \(\begin{pmatrix}-1\\-5\\-3\end{pmatrix}\) (or equivalent)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) (or equivalent)
Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.
Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏3
-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32
− 15 = 32, a contradiction
hence the three planes do not intersect
(b) (i) ∏1 : 2 + 2 + 0 = 4 and ∏2 : 1 + 4 + 0 = 5
(ii) METHOD 1
attempt to find the vector product of the two normals
\(\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}\)
\(=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}\)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize
lack of “r =” only once.
METHOD 2
attempt to eliminate a variable from ∏1 and ∏2
3x – z = 3 OR 3y – 5z = -6 OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏1 and ∏2 .
(c) METHOD 1
the line connecting L and ∏3 is given by L1
attempt to substitute position and direction vector to form L1
\(s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}\)
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏3
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
\(94t = 47\Rightarrow t=\frac{1}{2}\)
attempt to find distance between (1, -2,0) and their point \(\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}\)
\(=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}\)
\(=\frac{\sqrt{94}}{2}\)
METHOD 2
unit normal vector equation of ∏3 is given by 
\(=\frac{32}{\sqrt{94}}\)
Question
A company is designing a new logo. The logo is created by removing two equal segments from a rectangle, as shown in the following diagram.
The rectangle measures 5cm by 4cm. The points A and B lie on a circle, with centre O and radius 2cm, such that AÔB = θ, where 0 < θ < π. This information is shown in the following diagram.
(a) Find the area of one of the shaded segments in terms of θ.
(b) Given that the area of the logo is 13.4cm2, find the value of θ.
▶️Answer/Explanation
Ans:
(a) valid approach to find area of segment by finding area of sector – area of triangle
(b) EITHER
area of logo = area of rectangle – area of segments
5 × 4 – 2 × (2θ – 2sin θ) = 13.4
OR
area of one segment = \(\frac{20-13.4}{2}(=3.3)\)
2θ – 2sinθ = 3.3
THEN
θ = 2.35672….
θ = 2.36 (do not accept an answer in degrees)
Note: Award (M1)(A1)A0 if there is more than one solution.
Award (M1)(A1FT)A0 if the candidate works in degrees and obtains a final answer of 135.030…
Question
The function f is defined by \(f (y) =\sqrt{r^2-y^2}\) for \( -r\leq y \leq r\). The region enclosed by the graph of x = f ( y) and the y-axis is rotated by 360° about the y-axis to form a solid sphere. The sphere is drilled through along the y-axis, creating a cylindrical hole. The resulting spherical ring has height, h. This information is shown in the following diagrams.
▶️Answer/Explanation
Radius of cylinder , R is \(\sqrt{r^2 – \frac{h^2}{4}}\)
= \(R^2 =r^2 – \frac{h^2}{4}\)
Volume of part of sphere = \(\pi \int_{\frac{h}{2}}^{r} (r^2-y^2)\)dy
On solving and putting the limits , we get,
\(\frac{2r^3}{3} – \frac{r^2h}{2} +\frac{h^3}{24}\)
Volume of the cylindrical hole is \(\pi \int (r^2-y^2) dy +\pi R^2h\) where R\(\neq\) r
\(\pi \int (r^2-y^2) dy +\pi (r^2-\frac{h^2}{4})h= \frac{4}{3}\pi r^3-\pi\)
\(\frac{2r^3}{3} – \frac{r^2h}{2} +\frac{h^3}{24} +\pi (r^2-\frac{h^2}{4})h= \frac{4}{3}\pi r^3-\pi\)
On solving the equation we get ,
\(h=\sqrt[3]{6}\)