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13 questions
SA-9709-P3-VE4-Q01 1 mark L1 Foundation Calculator Intersections
You solve the x x x y y y λ \lambda λ μ \mu μ z z z
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-Q02 1 mark L1 Foundation Calculator Intersections
Two lines in three dimensions that are not parallel and do not intersect are called
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-002 3 marks L2 Standard Calculator Intersections
l 1 : r = ( 1 1 0 ) + λ ( 1 0 1 ) l_1:\ \mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\1\end{pmatrix} l 1 : r = 1 1 0 + λ 1 0 1 l 2 : r = ( 2 0 1 ) + μ ( 0 1 1 ) l_2:\ \mathbf{r}=\begin{pmatrix}2\\0\\1\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix} l 2 : r = 2 0 1 + μ 0 1 1
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-003 2 marks L2 Standard Calculator Parallel lines
l 1 : r = ( 1 2 3 ) + λ ( 2 − 1 4 ) l_1:\ \mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\4\end{pmatrix} l 1 : r = 1 2 3 + λ 2 − 1 4 l 2 : r = ( 0 1 0 ) + μ ( − 4 2 − 8 ) l_2:\ \mathbf{r}=\begin{pmatrix}0\\1\\0\end{pmatrix}+\mu\begin{pmatrix}-4\\2\\-8\end{pmatrix} l 2 : r = 0 1 0 + μ − 4 2 − 8
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-005 2 marks L2 Standard Calculator Parallel lines
The lines with directions ( 2 a − 3 ) \begin{pmatrix}2\\a\\-3\end{pmatrix} 2 a − 3 ( − 4 6 b ) \begin{pmatrix}-4\\6\\b\end{pmatrix} − 4 6 b a a a b b b
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-008 2 marks L2 Standard Calculator Parallel lines
l 1 : r = ( 1 0 2 ) + λ ( 1 1 − 1 ) l_1:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\-1\end{pmatrix} l 1 : r = 1 0 2 + λ 1 1 − 1 l 2 : r = ( 3 2 0 ) + μ ( − 2 − 2 2 ) l_2:\ \mathbf{r}=\begin{pmatrix}3\\2\\0\end{pmatrix}+\mu\begin{pmatrix}-2\\-2\\2\end{pmatrix} l 2 : r = 3 2 0 + μ − 2 − 2 2
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-G01 3 marks L2 Standard Calculator Intersections
The lines r = ( 1 1 1 ) + λ ( 1 2 0 ) \mathbf{r}=\begin{pmatrix}1\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\0\end{pmatrix} r = 1 1 1 + λ 1 2 0 r = ( 0 5 − 1 ) + μ ( 1 − 1 1 ) \mathbf{r}=\begin{pmatrix}0\\5\\-1\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\1\end{pmatrix} r = 0 5 − 1 + μ 1 − 1 1 λ \lambda λ μ \mu μ
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-001 5 marks L3 Exam Calculator Intersections
Show that the lines r = ( 2 0 1 ) + λ ( 1 − 1 2 ) \mathbf{r}=\begin{pmatrix}2\\0\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\2\end{pmatrix} r = 2 0 1 + λ 1 − 1 2 r = ( 5 0 6 ) + μ ( 1 2 1 ) \mathbf{r}=\begin{pmatrix}5\\0\\6\end{pmatrix}+\mu\begin{pmatrix}1\\2\\1\end{pmatrix} r = 5 0 6 + μ 1 2 1
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-004 4 marks L3 Exam Calculator Unknown constants
The lines r = ( 1 0 2 ) + λ ( 2 1 1 ) \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\1\end{pmatrix} r = 1 0 2 + λ 2 1 1 r = ( 4 k 2 ) + μ ( 1 1 − 1 ) \mathbf{r}=\begin{pmatrix}4\\k\\2\end{pmatrix}+\mu\begin{pmatrix}1\\1\\-1\end{pmatrix} r = 4 k 2 + μ 1 1 − 1 k k k
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-006 3 marks L3 Exam Calculator Intersections
The lines r = ( 1 1 1 ) + λ ( 1 2 0 ) \mathbf{r}=\begin{pmatrix}1\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\0\end{pmatrix} r = 1 1 1 + λ 1 2 0 r = ( 0 5 − 1 ) + μ ( 1 − 1 1 ) \mathbf{r}=\begin{pmatrix}0\\5\\-1\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\1\end{pmatrix} r = 0 5 − 1 + μ 1 − 1 1 ( 2 , 3 , 1 ) (2,3,1) ( 2 , 3 , 1 )
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-008 5 marks L4 Challenging Calculator Intersections
Lines l 1 l_1 l 1 l 2 l_2 l 2 r = ( 2 − 1 0 ) + λ ( 1 2 1 ) \mathbf{r}=\begin{pmatrix}2\\-1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\1\end{pmatrix} r = 2 − 1 0 + λ 1 2 1 r = ( 0 4 4 ) + μ ( 1 − 1 − 1 ) \mathbf{r}=\begin{pmatrix}0\\4\\4\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\-1\end{pmatrix} r = 0 4 4 + μ 1 − 1 − 1
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-009 4 marks L4 Challenging Calculator Intersections
Lines l 1 : r = ( 1 0 2 ) + λ ( 2 1 − 1 ) l_1:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix} l 1 : r = 1 0 2 + λ 2 1 − 1 l 2 : r = ( 3 1 0 ) + μ ( 1 − 1 2 ) l_2:\ \mathbf{r}=\begin{pmatrix}3\\1\\0\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\2\end{pmatrix} l 2 : r = 3 1 0 + μ 1 − 1 2
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-007 5 marks L4 Challenging Calculator Unknown constants
The lines r = ( 1 0 2 ) + λ ( 3 1 p ) \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}3\\1\\p\end{pmatrix} r = 1 0 2 + λ 3 1 p r = ( 3 1 1 ) + μ ( 1 1 1 ) \mathbf{r}=\begin{pmatrix}3\\1\\1\end{pmatrix}+\mu\begin{pmatrix}1\\1\\1\end{pmatrix} r = 3 1 1 + μ 1 1 1 p p p x x x
Salam Academy Original Practice · Proposed mark scheme