Question
The diagram below shows a circle with centre O. The points A, B, C lie on the circumference of the circle and [AC] is a diameter.
Let \(\overrightarrow {{\text{OA}}} = {\boldsymbol{a}}\) and \(\overrightarrow {{\text{OB}}} = {\boldsymbol{b}}\) .
a. Write down expressions for \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{CB}}} \) in terms of the vectors \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) .[2]
b. Hence prove that angle \({\text{A}}\hat {\rm{B}}{\text{C}}\) is a right angle. [3]
▶️Answer/Explanation
Markscheme
a.
\(\overrightarrow {{\text{AB}}} = {\boldsymbol{b}} – {\boldsymbol{a}}\) A1
\(\overrightarrow {{\text{CB}}} = {\boldsymbol{a}} + {\boldsymbol{b}}\) A1 [2 marks]
\(\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{CB}}} = \left( {{\boldsymbol{b}} – {\boldsymbol{a}}} \right) \cdot \left( {{\boldsymbol{b}} + {\boldsymbol{a}}} \right)\) M1
\( = {\left| {\mathbf{b}} \right|^2} – {\left| {\mathbf{a}} \right|^2}\) A1
\( = 0\) since \(\left| {\boldsymbol{b}} \right| = \left| {\boldsymbol{a}} \right|\) R1
Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.
so \(\overrightarrow {{\text{AB}}} \) is perpendicular to \(\overrightarrow {{\text{CB}}} \) i.e. \({\text{A}}\hat {\rm{B}}{\text{C}}\) is a right angle AG
[3 marks]
Question
The points A(1, 2, 1) , B(−3, 1, 4) , C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.
a. Find the vectors \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \).[2]
b. Find the Cartesian equation of the plane \(\prod \) that contains the face ABC.[4]
▶️Answer/Explanation
Markscheme
a.
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{ – 1} \\
3
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 3} \\
1
\end{array}} \right)\) A1A1
Note: Accept row vectors.
[2 marks]
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
{ – 4}&{ – 1}&3 \\
4&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
8 \\
{16} \\
{16}
\end{array}} \right)\) M1A1
normal \({\boldsymbol{n}} = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) so \({\boldsymbol{r}} \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) (M1)
\(x + 2y + 2z = 7\) A1
Note: If attempt to solve by a system of equations:
Award A1 for 3 correct equations, A1 for eliminating a variable and A2 for the correct answer. [4 marks]
Question
In the diagram below, [AB] is a diameter of the circle with centre O. Point C is on the circumference of the circle. Let \(\overrightarrow {{\text{OB}}} = \boldsymbol{b} \) and \(\overrightarrow {{\text{OC}}} = \boldsymbol{c}\) .
a. Find an expression for \(\overrightarrow {{\text{CB}}} \) and for \(\overrightarrow {{\text{AC}}} \) in terms of \(\boldsymbol{b}\) and \(\boldsymbol{c}\) .[2]
b. Hence prove that \({\rm{A\hat CB}}\) is a right angle.[3]
▶️Answer/Explanation
Markscheme
a.
\(\overrightarrow {{\text{CB}}} = \boldsymbol{b} – \boldsymbol{c}\) , \(\overrightarrow {{\text{AC}}} = \boldsymbol{b} + \boldsymbol{c}\) A1A1
Note: Condone absence of vector notation in (a).
[2 marks]
\(\overrightarrow {{\text{AC}}} \cdot \overrightarrow {{\text{CB}}} = \)(b + c)\( \cdot \)(b – c) M1
= \(|\)b\({|^2}\) – \(|\)c\({|^2}\) A1
= 0 since \(|\)b\(|\) = \(|\)c\(|\) R1
Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.
so \(\overrightarrow {{\text{AC}}} \) is perpendicular to \(\overrightarrow {{\text{CB}}} \) i.e. \({\rm{A\hat CB}}\) is a right angle AG
[3 marks]
Question
The vertices of a triangle ABC have coordinates given by A(−1, 2, 3), B(4, 1, 1) and C(3, −2, 2).
a.(i) Find the lengths of the sides of the triangle.
(ii) Find \(\cos {\rm{B\hat AC}}\). [6]
b. (i) Show that \(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \) −7i − 3j − 16k.
(ii) Hence, show that the area of the triangle ABC is \(\frac{1}{2}\sqrt {314} \).[5]
c. Find the Cartesian equation of the plane containing the triangle ABC. [3]
d. Find a vector equation of (AB). [2]
The point D on (AB) is such that \(\overrightarrow {{\text{OD}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) where O is the origin.
e. (i) Find the coordinates of D.
(ii) Show that D does not lie between A and B. [5]
▶️Answer/Explanation
Markscheme
(i) \(\overrightarrow {{\text{AB}}} = \overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OA}}} = \) 5i – j – 2k (or in column vector form) (A1)
Note: Award A1 if any one of the vectors, or its negative, representing the sides of the triangle is seen.
\(\overrightarrow {{\text{AB}}} = \) |5i – j – 2k|= \(\sqrt {30} \)
\(\overrightarrow {{\text{BC}}} = \) |–i – 3j + k|= \(\sqrt {11} \)
\(\overrightarrow {{\text{CA}}} = \) |–4i + 4j + k|= \(\sqrt {33} \) A2
Note: Award A1 for two correct and A0 for one correct.
(ii) METHOD 1
a.
\(\cos {\text{BAC}} = \frac{{20 + 4 + 2}}{{\sqrt {30} \sqrt {33} }}\) M1A1
Note: Award M1 for an attempt at the use of the scalar product for two vectors representing the sides AB and AC, or their negatives, A1 for the correct computation using their vectors.
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
Note: Candidates who use the modulus need to justify it – the angle is not stated in the question to be acute.
METHOD 2
using the cosine rule
\(\cos {\text{BAC}} = \frac{{30 + 33 – 11}}{{2\sqrt {30} \sqrt {33} }}\) M1A1
\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) A1
[6 marks]
\(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \left| {\begin{array}{*{20}{c}}
i&j&k \\
{ – 1}&{ – 3}&1 \\
{ – 4}&4&1
\end{array}} \right|\) A1
\( = \left( {( – 3) \times 1 – 1 \times 4} \right)\)i + \(\left( {1 \times ( – 4) – ( – 1) \times 1} \right)\)j + \(\left( {( – 1) \times 4 – ( – 3) \times ( – 4)} \right)\)k M1A1
= –7i – 3j – 16k AG
(ii) the area of \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} } \right|\) (M1)
\(\frac{1}{2}\sqrt {{{( – 7)}^2} + {{( – 3)}^2} + {{( – 16)}^2}} \) A1
\( = \frac{1}{2}\sqrt {314} \) AG
[5 marks]
attempt at the use of “(r – a)\( \cdot \)n = 0” (M1)
using r = xi + yj + zk, a = \(\overrightarrow {{\text{OA}}} \) and n = –7i – 3j – 16k (A1)
\(7x + 3y + 16z = 47\) A1
Note: Candidates who adopt a 2-parameter approach should be awarded, A1 for correct 2-parameter equations for x, y and z; M1 for a serious attempt at elimination of the parameters; A1 for the final Cartesian equation.
[3 marks]
r =. \(\overrightarrow {{\text{OA}}} + t\overrightarrow {{\text{AB}}} \) (or equivalent) M1
r = (–i + 2j + 3k) + t (5i – j – 2k) A1
Note: Award M1A0 if “r =” is missing.
Note: Accept forms of the equation starting with B or with the direction reversed.
[2 marks]
(i) \(\overrightarrow {{\text{OD}}} = \) (–i + 2j + 3k) + t(5i – j – 2k)
statement that \(\overrightarrow {{\text{OD}}} \cdot \overrightarrow {{\text{BC}}} = 0\) (M1)
\(\left( {\begin{array}{*{20}{c}}
{ – 1 + 5t} \\
{2 – t} \\
{3 – 2t}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
{ – 1} \\
{ – 3} \\
1
\end{array}} \right) = 0\) A1
\( – 2 – 4t = 0{\text{ or }}t = – \frac{1}{2}\) A1
coordinates of D are \(\left( { – \frac{7}{2},\frac{5}{2},4} \right)\) A1
Note: Different forms of \(\overrightarrow {{\text{OD}}} \) give different values of t, but the same final answer.
(ii) \(t < 0 \Rightarrow \) D is not between A and B R1 [5 marks]
Question
In the diagram below, [AB] is a diameter of the circle with centre O. Point C is on the circumference of the circle. Let \(\overrightarrow {{\text{OB}}} = \boldsymbol{b} \) and \(\overrightarrow {{\text{OC}}} = \boldsymbol{c}\) .
a. Find an expression for \(\overrightarrow {{\text{CB}}} \) and for \(\overrightarrow {{\text{AC}}} \) in terms of \(\boldsymbol{b}\) and \(\boldsymbol{c}\) .[2]
b. Hence prove that \({\rm{A\hat CB}}\) is a right angle. [3]
▶️Answer/Explanation
Markscheme
a.
\(\overrightarrow {{\text{CB}}} = \boldsymbol{b} – \boldsymbol{c}\) , \(\overrightarrow {{\text{AC}}} = \boldsymbol{b} + \boldsymbol{c}\) A1A1
Note: Condone absence of vector notation in (a).
[2 marks]
\(\overrightarrow {{\text{AC}}} \cdot \overrightarrow {{\text{CB}}} = \)(b + c)\( \cdot \)(b – c) M1
= \(|\)b\({|^2}\) – \(|\)c\({|^2}\) A1
= 0 since \(|\)b\(|\) = \(|\)c\(|\) R1
Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.
so \(\overrightarrow {{\text{AC}}} \) is perpendicular to \(\overrightarrow {{\text{CB}}} \) i.e. \({\rm{A\hat CB}}\) is a right angle AG [3 marks]