3.4 Vectors (HL)

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M17/5/MATHL/HP1/ENG/TZ1/XX/5
Question Text	ABCD is a parallelogram, where $\overrightarrow{A B}=-i+2 j+3 k$ and $\overrightarrow{A D}=4 i-j-2 k$ . (a) The area of the parallelogram ABCD can be written as $a \sqrt{b}$ . Find $a+b$ .
Total Mark	3
Correct Answer	11
Explanation	n/a
Mark Scheme	$\overrightarrow{A B} \times \overrightarrow{A D} =-i+10 j-7 k$ Area = = $\|\overrightarrow{A B} \times \overrightarrow{A D}\|=\sqrt{1^2+10^2+7^2}=5 \sqrt{6}(=\sqrt{150})$ Thus, $a+b$ is 11 .
Question Text	(b) By using a suitable scalar product of two vectors, determine whether $A \hat{B} C$ is acute or obtuse. Type in 'acute' or 'obtuse'.
Total Mark	4
Correct Answer	Acute
Explanation	n/a
Mark Scheme	METHOD 1 $\overrightarrow{A B} \cdot \overrightarrow{A D}=-4-2-6=-12$ considering the sign of the answer $\overrightarrow{A B} \cdot \overrightarrow{A D}<0$ therefore angle $D \hat{A} B$ is obtuse (as it is a parallelogram), $A \hat{B} C$ is acute METHOD 2 $\overrightarrow{B A} \cdot \overrightarrow{B C}=+4+2+6=12$ considering the sign of the answer $\overrightarrow{B A} \cdot \overrightarrow{B C}>0 \Rightarrow A \hat{B} C$ is acute

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N16-TZ0-P1 -4(HL)
Question Text	Consider the vectors $a=i-3 j-2 k, b=-3 j+2 k$ . (a) The vector $a \times b$ can be written as $a i+b j+c k$ . Find $a+b+c$ .
Total Mark	2
Correct Answer	-17
Explanation	n/a
Mark Scheme	$a \times b=-12 i-2 j-3 k$ thus, $a+b+c=-12-2-3=-17$
Question Text	(b) Hence find the Cartesian equation of the plane containing the vectors $a$ and $b$ , and passing through the point $(1,0,-1)$ . (a) $12 x+2 y-3 z=9$ (b) $12 x-2 y-3 z=9$ (c) $-12 x-2 y-3 z=-9$ (d) $-12 x-2 y-3 z=9$
Total Mark	3
Correct Answer	(c)
Explanation	n/a
Mark Scheme	METHOD 1 $-12 x-2 y-3 z=d-12 \times 1-2 \times 0-3(-1)=d \\ \Rightarrow d=-9-12 x-2 y-3 z=-9(\text {or } 12 x+2 y+3 z=9)$ METHOD 2 $(x y z) \cdot(-12-2-3)=(10-1) \cdot(-12-2-3) \\ -12 x-2 y-3 z=-9(\text { or } 12 x+2 y+3 z=9)$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N16-TZ0-P1 -8(HL)
Question Text	Consider the lines $l_1$ and $l_2$ defined by $l_1: r=(-3\quad-2\quad a)+\beta(1\quad4\quad2)$ and $l_2: \frac{6-x}{3}=\frac{y-2}{4}=1-z$ where a is a constant. Given that the lines $l_1$ and $l_2$ intersect at a point $P$ , (a) find the value of $a$ ;
Total Mark	4
Correct Answer	-7
Explanation	n/a
Mark Scheme	METHOD 1 $l_1: r=(-3\quad-2\quad a)+\beta(1\quad4\quad2) \Rightarrow\{x=-3+\beta, y=-2+4 \beta, z=a+2 \beta\} \\ \frac{6-(-3+\beta)}{3}=\frac{(-2+4 \beta)-2}{4} \Rightarrow 4=\frac{4 \beta}{3} \Rightarrow \beta=3, \\ \frac{6-(-3+\beta)}{3}=1-(a+2 \beta) \Rightarrow 2=-5-a \Rightarrow a=-7$ METHOD 2 $-3+\beta=6-3 \lambda,-2+4 \beta=4 \lambda+2, a+2 \beta=1-\lambda$ Attempt to solve $\lambda=2, \beta=3 a=1-\lambda-2 \beta=-7$
Question Text	(b) the coordinates of the point of intersection $P$ is $(p, q, r)$ . Find $p+q+r$ .
Total Mark	2
Correct Answer	9
Explanation	n/a
Mark Scheme	$\overrightarrow{O P}=(-3-2-7)+3 \cdot(1\quad4\quad2) \\ =(0\quad 10\quad-1) \\ \therefore P(0,10,-1)$ Hence, $p+q+r=0+10-1=9$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M16-TZ1-P1-11(HL)
Question Text	Two planes have equations $\Pi_1: 4 x+y+z=8 \text { and } \Pi_2: 4 x+3 y-z=0$ (a) The cosine of the angle between the two planes in the form $\sqrt{\frac{p}{q}}$ (fully simplified) where $p, q \in Z$ . Find $p+q$ .
Total Mark	4
Correct Answer	22
Explanation	n/a
Mark Scheme	angle between planes is equal to the angles between the normal to the planes $(4\quad1\quad1) \cdot(4\quad3\quad-1)=18$ let $\theta$ be the angle between the normal to the planes $\cos \theta=\frac{18}{\sqrt{18} \sqrt{26}}=\sqrt{\frac{18}{26}}=\sqrt{\frac{9}{13}}$ Hence, $a+b=9+13=22$
Question Text	(b) Let $L$ be the line of intersection of the two planes. $B$ is the point on $\Pi_1$ with coordinates $(a, b, 1)$ . Given the vector $\overrightarrow{AB }$ is perpendicular to $L$ find the value of $a+b$ .
Total Mark	5
Correct Answer	4
Explanation	n/a
Mark Scheme	$\overrightarrow{A B}=(a\quad b \quad1)-(1\quad0\quad4)=(a-1\quad b\quad-3) \\ (a-1\quad b\quad-3) \cdot(-1\quad2\quad2)=0 \\ -a+1+2 b-6=0 \Rightarrow a-2 b=-5$ lies on $\Pi_1$ so $4 a+b+1=8 \Rightarrow 4 a+b=7, a=1, b=3$ Hence, $a+b=4$
Question Text	(c) The point $P$ lies on $L$ and $A B P=45^{\circ}$ . Find the coordinates of the two possible positions of $P$ . (a) ( $1-\sqrt{2}, 2 \sqrt{2}, 4+2 \sqrt{2})$ (b) $(1+\sqrt{2}, 2 \sqrt{2}, 4+2 \sqrt{2})$ (c) $(1-\sqrt{2},-2 \sqrt{2}, 4-2 \sqrt{2})$ (d) $(1+\sqrt{2},-2 \sqrt{2}, 4-2 \sqrt{2})$ (e) $(-\sqrt{2},-2 \sqrt{2}, 2 \sqrt{2})$
Total Mark	5
Correct Answer	(a), (d)
Explanation	n/a
Mark Scheme	METHOD 1 $\|\overrightarrow{A B}\|=\|\overrightarrow{A P}\|=3 \sqrt{2} \\ \overrightarrow{A P}=t(-1\quad2\quad2) \\ \|3 t\|=3 \sqrt{2} \Rightarrow t= \pm \sqrt{2} P(1-\sqrt{2}, 2 \sqrt{2}, 4+2 \sqrt{2}) \text { and }(1+\sqrt{2},-2 \sqrt{2}, 4-2\sqrt{2})$ METHOD 2 let $P$ have coordinates $(1-\lambda, 2 \lambda, 4+2 \lambda)$ $\overrightarrow{B A}=(0\quad-3\quad3), \overrightarrow{B P}=(-\lambda \quad2 \lambda-3\quad3+2 \lambda) \\ \cos 45^{\circ}=\frac{\overrightarrow{B A} \cdot \overrightarrow{B P}}{\|B A \\| B P\|} \\ \overrightarrow{B A} \cdot \overrightarrow{B P}=18 \\ \frac{1}{\sqrt{2}}=\frac{18}{\sqrt{18} \sqrt{9 \lambda^2+18}}$ $\lambda= \pm \sqrt{2} P(1-\sqrt{2}, 2 \sqrt{2}, 4+2 \sqrt{2}) \text { and }(1+\sqrt{2},-2 \sqrt{2}, 4-2 \sqrt{2})$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M16-TZ2-P1-10(HL)
Question Text	A line $L$ has equation $\frac{x-2}{p}=\frac{y-q}{2}=z-1$ where $p, q \in R$ . A plane $\Pi$ has equation $x+y+3 z=9$ . (a) Given that $L$ lies in the plane $\Pi$ , find the value of $p$ and the value of $q$ . Hence find $p+q$
Total Mark	4
Correct Answer	-1
Explanation	n/a
Mark Scheme	METHOD 1 $(2+p \lambda)+(q+2 \lambda)+3(1+\lambda)=9(q+5)+(p+5) \lambda=9 p=-5 \text { and } q=4$ METHOD 2 direction vector of line is perpendicular to plane, so $(p\quad2\quad1) \cdot(1\quad1\quad3)=0 \quad$ & $\hspace{0.2cm} p=-5$ $(2, q, 1)$ is common to both $L$ and $\Pi$ $q=4$ Hence, $p+q=-5+4=-1$
Question Text	(b) Consider the different case where the acute angle between $L$ and $\Pi$ is $\theta$ . where $\theta=\arcsin \left(\frac{1}{\sqrt{11}}\right)$ (i) Find $p$ Total Mark: 4 Correct Answer: -2 Explanation: n/a Mark Scheme: METHOD 1 $\alpha$ is the acute angle between $n$ and $L$ if $\sin \theta=\frac{1}{\sqrt{11}}$ then $\cos \alpha=\frac{1}{\sqrt{11}}$ attempting to use $\cos \alpha=\frac{n \cdot d}{\|n\|\|d\|}$ or $\sin \theta=\frac{n \cdot d}{\|n\|\|d\|}$ $\frac{p+5}{\sqrt{11} \times \sqrt{p^2+5}}=\frac{1}{\sqrt{11}}$ $(p+5)^2=p^2+5$ $10 p=-20$ (or equivalent) $p=-2$ METHOD 2 $\alpha$ is the angle between $n$ and $L$ if $\sin \theta=\frac{1}{\sqrt{11}}$ then $\sin \alpha=\frac{\sqrt{10}}{\sqrt{11}}$ attempting to use $\sin \alpha=\frac{\|n \times d\|}{\|n\|\|d\|}$ $\frac{\sqrt{(-5)^2+(3 p-1)^2+(2-p)^2}}{\sqrt{11} \times \sqrt{p^2+5}}=\frac{\sqrt{10}}{\sqrt{11}}$ $p^2-p+3=p^2+5$ $-p+3=5$ (or equivalent) $p=-2$ (ii) If $L$ intersects $\Pi$ at $z=-1$ , find the value of $q$ . Total Mark: 7 Correct Answer: 10 Explanation: n/a Mark Scheme: $p=-2$ and z=-1 $\Rightarrow \frac{x-2}{-2}=\frac{y-q}{2}=-2$ $x=6$ and $y=q-4$ this satisfies $\Pi$ so $6+q-4-3=9$ thus, $q=10$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N14/5/MATHL/HP1/ENG/TZ0/XX/3
Question Text	A point P , relative to an origin O , has position vector $\overrightarrow{O P}=(1+s\hspace{0.2cm}3+2s\hspace{0.2cm}1-s), s \in R$ . Find $s$ which gives the minimum length of $\overrightarrow{O P}$ .
Total Mark	7
Correct Answer	-1
Explanation	n/a
Mark Scheme	$\|\overrightarrow{O P}\|=\sqrt{(1+s)^2+(3+2 s)^2+(1-s)^2} \quad\left(=\sqrt{6 s^2+12 s+11}\right)$ EITHER attempt to differentiate: $\frac{d}{d s}\|\overrightarrow{O P}\|^2(=12 s+12)$ attempt to solve $\frac{d}{d s}\|\overrightarrow{O P}\|^2=0$ for $s$ $s=-1$ OR attempt to differentiate: $\frac{d}{d s}\|\overrightarrow{O P}\|\left(=\frac{6 s+6}{\sqrt{6 s^2+12 s+11}}\right)$ attempt to solve $\frac{d}{d s}\|\overrightarrow{O P}\|=0$ for $s$ $s=-1$ OR attempt at completing the square: $\left(\|\overrightarrow{O P}\|^2=6(s+1)^2+5\right)$ minimum value occurs at $s=-1$ THEN the minimum length of $\overrightarrow{O P}$ is a minimum when $\overrightarrow{O P}$ is perpendicular to $(1\hspace{0.2cm}2\hspace{0.2cm}-1)$ $(1+s\quad3+2s\quad1-s) \cdot(1\quad2\quad-1)=0$ attempt to solve $1+s+6+4 s-1+s=0(6 s+6=0)$ for $s$ $s=-1$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M14/5/MATHL/HP1/ENG/TZ1/X/12
Question Text	(a) Find the coordinates of M , the mid-point of [OB]. (a) $(3, \sqrt{6}, \sqrt{3})$ (b) $(3,-\sqrt{6}, \sqrt{3}) \quad$ (c) $(3,-\sqrt{6},-\sqrt{3})$ (d) $(-3,-\sqrt{6},-\sqrt{3})$
Total Mark	2
Correct Answer	(b)
Explanation	n/a
Mark Scheme	$M\left(3,-\frac{\sqrt{24}}{2}, \frac{\sqrt{12}}{2}\right)=(3,-\sqrt{6}, \sqrt{3}).$
Question Text	(b) Find the equation of the plane $\Pi$ , containing the square OABC. (a) $y+2 z=0$ (b) $y+\sqrt{3} z=0$ (c) $y+4 z=0$ (d) $y+\sqrt{2} z=0$
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	METHOD 1 $\overrightarrow{O A} \times \overrightarrow{O C}=(6\quad0\quad0) \times(0\quad-\sqrt{24}\quad\sqrt{12})=(0\quad-6\quad\sqrt{12}-6 \sqrt{24})(=0\quad{0.2cm}-12 \sqrt{3}\quad-12 \sqrt{6})$ equation of plane is $-6 \sqrt{12} y-6 \sqrt{24} z=d$ any valid method showing that $d=0$ $\Pi: y+\sqrt{2} z=0$ METHOD 2 Equation of plane is $a x+b y+c z=d$ SubstitutingO to find $d=0$ Substituting two points(A, B, C or M) eg $6 a=0,-\sqrt{24} b+\sqrt{12} c=0$ $\Pi: y+\sqrt{2} z=0$
Question Text	(c) Find a vector equation of the line $L$ , through $M$ , perpendicular to the plane $\Pi$ . (a) $(3\hspace{0.2cm}-\sqrt{6}\hspace{0.2cm}\sqrt{3})+\lambda\left(0\hspace{0.2cm}1 \hspace{0.2cm}\sqrt{3}\right)$ (b) $(3\hspace{0.2cm}-\sqrt{6}\hspace{0.2cm}\sqrt{3})+\lambda\left(0\hspace{0.2cm}1 \hspace{0.2cm}-\sqrt{2}\right)$ (c) $(3\quad-\sqrt{6}\quad\sqrt{3})+\lambda\left(0\quad1 \quad2\right)$ (d) $(3\quad-\sqrt{6}\quad\sqrt{3})+\lambda\left(0\quad1 \quad\sqrt{2}\right)$
Total Mark	4
Correct Answer	(d)
Explanation	n/a
Mark Scheme	$r=$ $(3\quad-\sqrt{6}\quad\sqrt{3})+\lambda\left(0\quad1 \quad\sqrt{2}\right)$
Question Text	(d) Find the coordinates of $D$ , the point of intersection of the $L$ with the plane whose equation is $y=0$ . (a) $(3,0, \sqrt{3})$ (b) $(-3,0,3 \sqrt{3})$ (c) $(3,0,3 \sqrt{3})$ (d) $(3,0,3 \sqrt{2})$
Total Mark	3
Correct Answer	(c)
Explanation	n/a
Mark Scheme	Using $y=0$ to find $\lambda$ Substitute their $\lambda$ into their equation from part(c) $D$ has coordinates $(3,0,3 \sqrt{3})$
Question Text	(e) Find the coordinates of $E$ , the reflection of the point $D$ in the plane $\Pi$ . (a) $(3,2 \sqrt{6}, \sqrt{3})$ (b) $(3,-2 \sqrt{6},-\sqrt{3})$ (c) $(-3,2 \sqrt{6},-\sqrt{3})$ (d) $(3,-2 \sqrt{6},-\sqrt{2})$
Total Mark	5
Correct Answer	(b)
Explanation	n/a
Mark Scheme	$\lambda$ for point $E$ is the negative of the $\lambda$ for point $D$ $E$ has coordinates $(3, -2\sqrt6,-\sqrt3)$
Question Text	(f) Find the angle $O \hat{D} A$ (in degrees).
Total Mark	4
Correct Answer	60
Explanation	n/a
Mark Scheme	$\overrightarrow{D A} \cdot \overrightarrow{D O}=(3\quad0\quad-3 \sqrt{3}) \cdot(-3\quad0\quad-3 \sqrt{3})=18$ $\cos O D A=\frac{18}{\sqrt{36} \sqrt{36}}=\frac{1}{2}$ Hence $O \hat{D} A=60^{\circ}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N13-TZ0-P1-11(HL)
Question Text	Consider the points $\mathrm{A}(1,0,0), \mathrm{B}(2,2,2)$ , and $\mathrm{C}(0,2,1)$ . (a) The vector $\overrightarrow{C A} \times \overrightarrow{C B}$ can be expressed as $(a, b, c)$ . Find $a+b+c$ .
Total Mark	4
Correct Answer	-1
Explanation
Mark Scheme	$\overrightarrow{C A}=(1\quad-2\quad-1) \\ \overrightarrow{C B}=(2\quad0\quad1) \\ \overrightarrow{C A} \times \overrightarrow{C B}=\left(\begin{matrix} i & j & k\\ 1 & -2 & c\\ 2 & 0 & 1 \end{matrix}\right)$ $=(-2\quad-3\quad4)$ Thus, $a+b+c=-1$
Question Text	(b) Find an exact value for the area of the triangle $A B C$ . (a) $\frac{1}{\sqrt{30}}$ (b) $\frac{1}{2}$ (c) $\frac{\sqrt{29}}{\sqrt{30}}$ (d) $\frac{\sqrt{29}}{2}$
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	METHOD 1 $\frac{1}{2}\|\overrightarrow{C A} \times \overrightarrow{C B}\|=\frac{1}{2} \sqrt{(-2)^2+(-3)^2+4^2}=\frac{\sqrt{29}}{2}$ METHOD 2 attempt to apply $\frac{1}{2}\|C A\|\|C B\| \sin C$ $C A \cdot C B=\sqrt{5} \cdot \sqrt{6} \cos C \Rightarrow \cos C=\frac{1}{\sqrt{30}} \Rightarrow \sin C=\frac{\sqrt{29}}{\sqrt{30}}$ area $=\frac{\sqrt{29}}{2}$
Question Text	(c) What is the Cartesian equation of the plane $\Pi_1$ , containing the triangle $A B C$ . (a) $2 x+3 y-4 z=3$ (b) $2 x-3 y-4 z=2$ (c) $2 x+3 y-4 z=0$ (d) $2 x+3 y-4 z=2$
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	METHOD 1 $\begin{aligned} & r \cdot(-2\quad-3\quad4)=(1\quad0\quad0) \cdot(-2\quad-3\quad4) \& \Rightarrow-2 x-3 y+4 z=-2 \\ & \& \Rightarrow 2 x+3 y-4 z=2\end{aligned}$ METHOD 2 $-2 x-3 y+4 z=d$ substituting a point in the plane $d=-2 \Rightarrow-2 x-3 y+4 z=-2 \Rightarrow 2 x+3 y-4 z=2$
Question Text	(d) A second plane $\Pi_2$ is defined by the Cartesian equation $\Pi_2: 4 x-y-z=4$ . $L_1$ is the line of intersection of the planes $\Pi_1$ and $\Pi_2$ . Find a vector equation for $L_1$ . (a) $L_1: r=(-1\quad0\quad0)+\lambda(1\quad2\quad2)$ (b) $L_1: r=(-1\quad0\quad0)+\lambda(-1\quad2\quad2)$ (c) $L_1: r=(1\quad0\quad0)+\lambda(1\quad-2\quad-2)$ (d) $L_1: r=(1\quad0\quad0)+\lambda(1\quad2\quad2)$
Total Mark	5
Correct Answer	(d)
Explanation	n/a
Mark Scheme	METHOD 1 $\left\|\begin{matrix} i & j & k\\ 1 & -2 & c\\ 2 & 0 & 1 \end{matrix}\right\| = (-7\quad-14\ quad-14)$ $\begin{aligned} & n=\left(1\quad 2\quad 2\right) \\ & z=0 \Rightarrow y=0, x=1\end{aligned}$ $L_1: r=(1\quad0\quad0)+\lambda(1\quad2\quad2)$ METHOD 2 eliminate 1 of the variables, e.g. $x$ $-7 y+7 z=0$ introduce a parameter $\Rightarrow z=\lambda$ $y=\lambda, x=1+\frac{\lambda}{2}$ $L_1: r=(1\quad0\quad0)+\lambda(1\quad2\quad2)$ METHOD 3 $z=t$ write $x$ and $y$ in terms of $t \Rightarrow 4 x-y=4+t, 2 x+3 y=2+4 t$ or equivalent attempt to eliminate $x$ or $y$ $x, y, z$ expressed in parameters $\Rightarrow z=t$ $y=t, x=1+\frac{t}{2}$ $L_1: r=(1\quad0\quad0)+\lambda(1\quad2\quad2)$
Question Text	(e) A third plane $\Pi_3$ is defined by the Cartesian equation $16 x+\alpha y-3 z=\beta$ . Find the value of $\alpha$ if all three planes contain $L_1$ .
Total Mark	3
Correct Answer	-5
Explanation	n/a
Mark Scheme	METHOD 1 direction of the line is perpendicular to the normal of the plane $(16 \alpha-3) \cdot(1\quad2\quad2)=0 \quad\&\quad 16+2 \alpha-6=0 \Rightarrow \alpha=-5$ METHOD 2 solving line/plane simultaneously $16(1+\lambda)+2 \alpha \lambda-6 \lambda=\beta 16+(10+2 \alpha) \lambda=\beta \Rightarrow \alpha=-5$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M13-TZ1-P1-2(HL)
Question Text	Consider the points $A(1,2,3), B(1,0,5)$ and $C(2,-1,4)$ . (a) $\overrightarrow{A B} \times \overrightarrow{A C}$ can be expressed as $(a \quad b \quad c)$ . Find $a+b+c$
Total Mark	4
Correct Answer	8
Explanation	n/a
Mark Scheme	$\overrightarrow{A B}=(1\quad0\quad5)-(1\quad2\quad3)=(0\quad-2\quad2), \overrightarrow{A C}=(2\quad-1\quad4)-(1\quad2\quad3)=(1\quad-3\quad1)$ $\overrightarrow{A B} \times \overrightarrow{A C}$ $= \left\|\begin{matrix} i & j & k\\0&-2&2\\1&-3&1\end{matrix}\right\|=(4\quad2\quad2)$ thus, $a+b+c=8$
Question Text	(b) The area of the triangle $ABC$ can be expressed fully into $\sqrt{x}$ . Find $x$
Total Mark	2
Correct Answer	6
Explanation	n/a
Mark Scheme	area = $\frac{1}{2}\|(\overrightarrow{A B} \times \overrightarrow{A C})\| \\ =\frac{1}{2} \sqrt{4^2+2^2+2^2}=\frac{1}{2} \sqrt{24}(=\sqrt{6})$ Thus, $x=6$ .

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M13-TZ2-P1-11(HL)
Question Text	The vertices of a triangle $ABC$ has coordinates given by $A(-1,2,3), B(4,1,1)$ and $C(3,-2,2)$ . (a) Find the lengths of the sides of the triangle.
Total Mark	4
Correct Answer	$\sqrt{33}, \sqrt{30}, \sqrt{11}$
Explanation	n/a
Mark Scheme	$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=5 i-j-2 k \\ \|\overrightarrow{A B}\|=\|5 i-j-2 k\|=\sqrt{30}\\\|\overrightarrow{B C}\|=\|-i-3 j+k\|=\sqrt{11}\\\|\overrightarrow{C A}\|=\|-4 i+4 j+k\|=\sqrt{33}$
Question Text	(b) What is $\overrightarrow{B C} \times \overrightarrow{C A}$ . (a) $-7 i-3 j+16 k$ (b) $-7 i-3 j-16 k$ (c) $7 i+3 j-16 k$ (d) $7 i-3 j+16 k$
Total Mark	3
Correct Answer	(b)
Explanation	n/a
Mark Scheme	$\overrightarrow{B C} \times \overrightarrow{C A}= \left\|\begin{matrix}i&j&k\\-1&-3&1\\-4&4&1\end{matrix}\right\| \\ =((-3) \times 1-1 \times 4) i+(1 \times(-4)-(-1) \times 1) j+((-1) \times 4-(-3) \times(-4)) k=-7 i-3 j-16 k$
Question Text	(c) Find the Cartesian equation of the plane containing the triangle $A B C$ . (a) $7 x-3 y-16 z=47$ (b) $-7 x-3 y+16 z=47$ (c) $7 x+3 y-16 z=-47$ (d) $7 x+3 y+16 z=47$
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	attempt at the use of $(r-a) \cdot n=0$ using $r=x i+y i+z k, a=\overrightarrow{O A}$ and $n=-7 i-3 i-16 k$ $7 x+3 y+16 z=47$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N19/5/MATHL/HP1/ENG/TZ1/XX/4
Question Text	$A$ and $B$ are acute angles such that $\cos A=\frac{2}{3}$ and $\sin B=\frac{1}{3}$ . What is $\cos (2 A+B)$ . (a) $\frac{2 \sqrt{2}}{27}+\frac{4 \sqrt{5}}{27}$ (b) $-\frac{2 \sqrt{2}}{27}+\frac{4 \sqrt{5}}{27}$ (c) $\frac{2 \sqrt{2}}{27}-\frac{4 \sqrt{5}}{27}$ (d) $-\frac{2 \sqrt{2}}{27}-\frac{4 \sqrt{5}}{27}$
Total Mark	7
Correct Answer	(d)
Explanation	n/a
Mark Scheme	attempt to use $\cos (2 A+B)=\cos 2 A \cos B-\sin 2 A \sin B$ (may be seen later) attempt to use any double angle formulae (seen anywhere) attempt to find either $\sin A$ or $\cos B$ (seen anywhere) $\cos A=\frac{2}{3} \Rightarrow \sin A\left(=\sqrt{1-\frac{4}{9}}\right)=\frac{\sqrt{5}}{3} \\ \sin B=\frac{1}{3} \Rightarrow \cos B\left(=\sqrt{1-\frac{1}{9}}=\frac{\sqrt{8}}{3}\right)=\frac{2 \sqrt{2}}{3} \\ \cos 2 A\left(=2 \cos ^2 A-1\right)=-\frac{1}{9} \\ \sin 2 A(=2 \sin A \cos A)=\frac{4 \sqrt{5}}{9} \\ \operatorname{So} \cos (2 A+B)=\left(-\frac{1}{9}\right)\left(\frac{2 \sqrt{2}}{3}\right)-\left(\frac{4 \sqrt{5}}{9}\right)\left(\frac{1}{3}\right) \\ =-\frac{2 \sqrt{2}}{27}-\frac{4 \sqrt{5}}{27}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N19/5/MATHL/HP1/ENG/TZ1/XX/11
Question Text	Points $A(0,0,5), B(0,5,0), C(5,0,0), V(p, p, p)$ form the vertices of a tetrahedron. Consider the case where the faces $A B V$ and $A C V$ are perpendicular. Find the two possible values of $p$ . (a) $-\frac{1}{3}$ (b) 0 (c) $\frac{1}{3}$ (d) $2$ (e) $\frac{10}{3}$
Total Mark	8
Correct Answer	(b), (e)
Explanation	n/a
Mark Scheme	Take the normal of the two planes Normal of $ABV$ $\left(N_{A B V}\right)$ : $\overrightarrow{A B} \times \overrightarrow{B V}=\left(\begin{array}{c} 0 \\ 5 \\ -5 \end{array}\right) \times\left(\begin{array}{c} p \\ p-5 \\ p \end{array}\right)=\left(\begin{array}{c} 10 p-25 \\ -5 p \\ -5 p \end{array}\right)$ Normal $\left(N_{A C V}\right)$ $\overrightarrow{A C} \times \overrightarrow{C V}=\left(\begin{array}{c} 5 \\ 0 \\ -5 \end{array}\right) \times\left(\begin{array}{c} p-5 \\ p \\ p \end{array}\right)=\left(\begin{array}{c} 5 p \\ -10 p+25 \\ 5 p \end{array}\right)$ As the planes are perpendicular $N_{A B V} \cdot N_{A C V}=0 \\ (10 p-25) 5 p-5 p(-10 p+25)-5 p(5 p)=0 \\ 2 p^2-5 p+2 p^2-5 p-p^2=3 p^2-10 p=p(3 p-10)=0 \\ p=0, \frac{10}{3}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M19/5/MATHL/HP1/ENG/TZ1/XX/1
Question Text	Let $a=\left(\begin{array}{lll}k & -k & -1\end{array}\right)$ and $b=\left(\begin{array}{lll}2 k & 5 & -3\end{array}\right), k \in R$ . Given that $a$ and $b$ are perpendicular, find the possible values of $k$ . (select all that apply) (a) $\frac{1}{2}$ (b) $1$ (c) $\frac{3}{2}$ (d) $2$ (e) $\frac{5}{2}$
Total Mark	4
Correct Answer	(b), (c)
Explanation	n/a
Mark Scheme	$a \cdot b=0 \\ 2 k^2-5 k+3=(2 k-3)(k-1)=0 \\ k=1, \frac{3}{2}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M19/5/MATHL/HP1/ENG/TZ1/XX/11
Question Text	Two distinct lines, $l_1$ and $l_2$ , intersect at a point $P$ . The line $l_1$ has vector equation $r_1=\left(1\quad3\quad1\right)+\lambda(1\quad-1\quad1)$ , $\lambda \in R$ and the line $l_2$ has vector equation $r_2 = \left(-1\quad1\quad1\right) + \mu\left(4\quad3\quad2\right), \mu\in R$ . (a) The point A has coordinates $(3,1,3)$ and lies on $l_1$ . Write down the value of $\lambda$ corresponding to the point $A$ .
Total Mark	1
Correct Answer	2
Explanation	n/a
Mark Scheme	$\lambda = 2$
Question Text	(b) The point $B$ has coordinates $(3,4,3)$ and lies on $l_2$ . Let $C$ be the point on $l_1$ with coordinates $(0,4,0)$ and $D$ be the point on $l_2$ with parameter $\mu=-1$ . Find the area of the quadrilateral $CDBA$ . (a) $\frac{1}{2}(\sqrt{193}+3 \sqrt{2})$ (b) $\frac{1}{2}(\sqrt{383}+3 \sqrt{2})$ (c) $\frac{1}{2}(\sqrt{193}+9 \sqrt{2})$ (d) $\frac{1}{2}(\sqrt{517}+9 \sqrt{2})$
Total Mark	8
Correct Answer	(d)
Explanation	n/a
Mark Scheme	Area of $CDBA$ $= \frac{1}{2}\|C D \times D B\|+\frac{1}{2}\|B A \times A C\|$ $\frac12\times\left(\left\|\left(\begin{matrix}-5\\-6\\-1\end{matrix}\right)\times\left(\begin{matrix}8\\6\\3\end{matrix}\right) \right\| +\left\|\left(\begin{matrix}0\\-3\\0\end{matrix}\right)\times\left(\begin{matrix}-3\\3\\-3\end{matrix}\right) \right\|\right) = \frac12\times\left(\left\|\left(\begin{matrix}-12\\7\\18\end{matrix}\right)\times\left(\begin{matrix}9\\0\\9\end{matrix}\right) \right\|\right)$ $\frac{1}{2}(\sqrt{144+49+324}+\sqrt{81+81})=\frac{1}{2}(\sqrt{11 \times 47}+9 \sqrt{2}) \\ \frac{1}{2}(\sqrt{517}+9 \sqrt{2})$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M19/5/MATHL/HP1/ENG/TZ2/XX/2 a
Question Text	Three points in three-dimensional space have coordinates $A(0,0,1), B(0,3,0)$ and $C(3,2,0)$ . If the area of the triangle $ABC$ can be written as $\frac{\sqrt{a}}{2}$ where $a \in Z^{+}$ , find the value of $a$
Total Mark	5
Correct Answer	91
Explanation	n/a
Mark Scheme	$area=\frac12\left\|\overrightarrow{AB}\times\overrightarrow{AC}\right\|$ $\left\|\left(\begin{matrix}0\\-3\\-1\end{matrix}\right)\times\left(\begin{matrix}3\\2\\-1\end{matrix}\right) \right\|=\left\|\left(\begin{matrix}-1\\-3\\-9\end{matrix}\right)\right\|$ $\frac12\left\|\overrightarrow{AB}\times\overrightarrow{AC}\right\| = \frac12\times\sqrt{91}$ $\frac{\sqrt{91}}{2}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N18/5/MATHL/HP1/ENG/TZ0/XX/5
Question Text	The vectors $a$ and $b$ are defined by $a=i-t k, b=i-t j+4 t k$ where $t \in R$ . Find the values of $t$ for which the angle between $a$ and $b$ is obtuse. (a) $t<-\frac{1}{2}, t>\frac{1}{2}$ (b) $t<-\frac{1}{4}, t>\frac{1}{4}$ (c) $-\frac{1}{2}<t<\frac{1}{2}$ (d) $-\frac{1}{4}<t<\frac{1}{4}$
Total Mark	6
Correct Answer	(a)
Explanation	n/a
Mark Scheme	Use of $a \cdot b=\|a \\| b\| \cos \theta$ $a \cdot b=(1 \times 1)+(0 \times-t)+(-t \times 4 t)=1-4 t^2$ $\theta$ is obtuse when $\cos \theta<0$ $1-4 t^2<0 \\ t<-\frac{1}{2}, t>\frac{1}{2}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	N18/5/MATHL/HP1/ENG/TZ0/XX/9
Question Text	Consider a triangle $OAB$ such that $O$ has coordinates $(0,0,0)$ , $A$ has coordinates $(0,2,1)$ and $B$ has coordinates $(a, 0,2 a-1)$ where $b<0$ . Find, in terms of $b$ , a Cartesian equation of the plane $\Pi$ containing this triangle. (a) $(2 a-1) x+a y-2 a z=0$ (b) $(4 a-2) x+a y-2 a z=0$ (c) $(2 a-1) x+a y+2 a z=0$ (d) $(4 a-2) x+a y+2 a z=0$
Total Mark	5
Correct Answer	(b)
Explanation	n/a
Mark Scheme	Normal vector is obtained by $\overrightarrow{O B} \times \overrightarrow{O A}$ $\overrightarrow{O B} \times \overrightarrow{O A} = \left(\begin{matrix}0\\2\\1\end{matrix}\right)\times\left(\begin{matrix}a\\0\\2a-1\end{matrix}\right)=\left(\begin{matrix}4a-2\\a\\-2a\end{matrix}\right)$ $П: (4 a-2) x+a y-2 a z=0$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M18/5/MATHL/HP1/ENG/TZ1/XX/10
Question Text	A square based pyramid has vertices at $O (0,0,0), A(2,0,0), B(2,2,0), C(0,2,0)$ and $D(0,0,2)$ . (a) The Cartesian equation of the plane $\Pi_1$ , passing through the points $A, B$ and $D$ can be written in the form $a x+b y+z y=2$ where $a, b, c \in Z$ . Find the value of $a+b+c$ .
Total Mark	3
Correct Answer	2
Explanation	n/a
Mark Scheme	Take the cross product of $\overrightarrow{A B}$ and $\overrightarrow{A D}$ $\overrightarrow{A B} \times \overrightarrow{A D}=\left(\begin{matrix}0\\2\\0\end{matrix}\right)\times\left(\begin{matrix}-2\\0\\2\end{matrix}\right)=\left(\begin{matrix}4\\0\\4\end{matrix}\right)$ $\Pi_1: 4 x+4 z=8 \\$ $\Pi_1: x+z=2$
Question Text	(b) The Cartesian equation of the plane $\Pi_2$ , passing through the points $B, C$ and $D$ , is $y+z=2$ Find the angle in degrees between the faces $ABD$ and $BCD$ .
Total Mark	4
Correct Answer	60
Explanation	n/a
Mark Scheme	Take the normal to the two planes, and find the angle between them $\left(\begin{matrix}1\\0\\1\end{matrix}\right)\cdot\left(\begin{matrix}0\\1\\1\end{matrix}\right)=\left\|\left(\begin{matrix}1\\0\\1\end{matrix}\right)\right\|\left\|\cdot\left(\begin{matrix}0\\1\\1\end{matrix}\right)\right\|\cos\theta$ $1=\sqrt{2} \times \sqrt{2} \times \cos \theta$ $\cos \theta=\frac{1}{2}\\$ $\theta=60^{\circ}$
Question Text	(c) The plane $\Pi_3$ passes through $O$ and is normal to the line $BD$ . Find the Cartesian equation of $\Pi_3$ . (a) $x+y+z=0$ (b) $x+y-z=0$ (c) $x-y+z=0$ (d) $x-y-z=0$
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	$\overrightarrow{B D}$ has a vector slope of $i-j-k$ Thus the equation of the plane of $\Pi_3$ can be written as $x-y-z=n$ . As it passes through $O$ , $x-y-z=0$
Question Text	(d) $\Pi_3$ cuts $AD$ and $BD$ at the points $P$ and $Q$ respectively. Given that $P$ is the midpoint of $AD$ .Find the area of the triangle $OPQ$ . (a) $\frac{\sqrt{3}}{3}$ (b) $\frac{5 \sqrt{3}}{6}$ (c) $\frac{\sqrt{2}}{2}$ (d) $\frac{3 \sqrt{2}}{4}$
Total Mark	7
Correct Answer	(a)
Explanation	n/a
Mark Scheme	The angle $\mathrm{OPQ}=90$ Area of $O P Q=\frac{1}{2} \times O P \times P Q$ Noticing that triangle $O Q D$ is similar to triangle $ODB$ . $\frac{O Q}{O D}=\frac{O B}{D B} \\ O Q=\frac{2 \sqrt{2}}{2 \sqrt{3}} \times 2=\frac{2 \sqrt{6}}{3}$ Since $P$ is the midpoint of $A D$ , $P=(1,0,1) \\ O P=\sqrt{1+0+1}=\sqrt{2} \\ P Q^2=O Q^2-O P^2=\frac{8}{3}-2=\frac{2}{3}$ So, $\frac{1}{2} \times P Q \times O Q=\frac{1}{2} \times \frac{\sqrt{2}}{\sqrt{3}} \times \sqrt{2}=\frac{\sqrt{3}}{3}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M18/5/MATHL/HP1/ENG/TZ2/XX/1
Question Text	The acute angle between the vectors $2i-3 j-6k$ and $4i-3 j+k$ is denoted by $\theta$ . Find $\cos \theta$ . (a) $\frac{7}{13}$ (b) $\frac{26}{91}$ (c) $\frac{\sqrt{26}}{13}$ (d) $\frac{9 \sqrt{26}}{91}$
Total Mark	4
Correct Answer	(d)
Explanation	n/a
Mark Scheme	Using $\cos \theta=\frac{v \cdot w}{\|v\|\|w\|}$ $\cos \theta=\frac{(2 i-3 j-6 k) \cdot(4 i-3 j+k)}{\|2 i-3 j-6 k \\| 4 i-3 j+k\|} \\ =\frac{18}{\sqrt{49} \sqrt{26}}=\frac{18}{7 \sqrt{26}} \\ =\frac{9 \sqrt{26}}{91}$

Topic	3.4 Vectors (HL)
Tag	Vectors; Line; Plane; Intersection; Distance; Angle; Dot product; Cross product; Cartesian; Parametric
Source	M18/5/MATHL/HP1/ENG/TZ2/XX/9
Question Text	The points $A, B, C$ and $D$ have position vectors $a, b, c$ and $d$ , relative to the origin $O$ . It is given that $\overrightarrow{A B}=\overrightarrow{D C}$ . The position vectors $\overrightarrow{O A}, \overrightarrow{O B}, \overrightarrow{O C}$ and $\overrightarrow{O D}$ are given by $a=i+2 j-3 k \quad b=3 i-j+k \quad c=q i+j+2 k \quad d=-i+r j-2 k$ Where $p, q$ , and $r$ are constants. (a) Find the value of (i) $q$ Total Mark: 3 Correct Answer: 1 Explanation: n/a Mark Scheme: Use of $\overrightarrow{A B}=\overrightarrow{D C}$ $(2 i-3 j+4 k)=((q+1) i+(1-r) j+4 k)$ so $q=1, r=4$ (ii) r Total Mark: 2 Correct Answer: 4 Explanation: n/a Mark Scheme: n/a
Question Text	(b) Find the area of $ABCD$ .
Total Mark	5
Correct Answer	15
Explanation	n/a
Mark Scheme	Use of $\overrightarrow{A D}=\overrightarrow{B C}$ (this can be deduced from the fact that $\overrightarrow{A B}=\overrightarrow{D C}$ ) It can be noticed that $A B C D$ is a parallelogram so the area is equivalent to $\|\overrightarrow{A B} \times \overrightarrow{A D}\|$ $\overrightarrow{A B}=2 i-3 j+4 k$ $\overrightarrow{A D}=-2 i+2 j+k$ $Area = \|\overrightarrow{A B} \times \overrightarrow{A D}\|=\|-11 i-10 j-2 k\|=\sqrt{225}=15$

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