Spheres, lines and planes

The plane containing the $x$-axis and the $y$-axis is the $xy$-plane, where $z=0$. The centre is 6 units above it. The sphere just touches the plane, so the radius is the distance from the centre straight down to the plane.

(C) is the distance from the centre to the origin in the $xy$-plane, $\sqrt{3^2+4^2}=5$, which is a different distance. Try the 2025 preset in the interactive above.

The radius is the distance from $A$ to $B$.

$$\begin{aligned}r^2&=(1-1)^2+(-3-0)^2\\&\quad+(-5+1)^2\\&=0+9+16=25\end{aligned}$$

Write the sphere with centre $(1,0,-1)$, then expand to match the options.

$$\begin{aligned}(x-1)^2+y^2+(z+1)^2&=25\\x^2-2x+y^2+z^2+2z+2&=25\\x^2-2x+y^2+z^2+2z&=23\end{aligned}$$

(C) forgets to move the two 1s across. (B) and (D) have the signs of the centre the wrong way round.

Substitute $h=-2$, $k=3$, $l=-4$ and $r=9$. Subtracting a negative gives a plus, and the right-hand side is $9^2=81$.

$$\begin{aligned}&(x-(-2))^2+(y-3)^2\\&\quad+(z-(-4))^2=9^2\end{aligned}$$

Complete the square in $x$ and in $y$. There is no $z$ term, so $z^2$ stays as it is.

$$\begin{aligned}(x+1)^2+(y-1)^2+z^2-2&=7\\(x+1)^2+(y-1)^2+z^2&=9\end{aligned}$$

The centre is $(-1,1,0)$ and $r^2=9$, so $r=3$. This is the first preset in the interactive above.

$$\overrightarrow{AB}=\begin{pmatrix}3\\-2\\1\end{pmatrix}-\begin{pmatrix}2\\4\\5\end{pmatrix}=\begin{pmatrix}1\\-6\\-4\end{pmatrix}$$

Check each option for a direction that is a multiple of $(1,-6,-4)$ and a point on the line. (A) uses $B$ as the direction. (B) has the right direction, but $(-3,2,-1)$ is $-B$, which is not on the line. (C) uses the differences as the point and $A$ as the direction.

(D) has numerators that give the point $(3,-2,1)$, which is $B$, and denominators $(-1,6,4)=-\overrightarrow{AB}$, which is parallel to the line.

Part a)

$$x=2-k,\quad y=2k$$

Part b)

Make $k$ the subject of the first equation and substitute into the second.

$$\begin{aligned}k&=2-x\\y&=2(2-x)\\&=4-2x\end{aligned}$$

Part c)

$$x=2-5=-3,\quad y=2(5)=10$$

Part d)

On the $y$-axis, $x=0$.

$$0=2-k\ \Rightarrow\ k=2$$

There is no published marking guide for the sample paper, so this is my own solution.

The points on the line are $(4k,\ 7+16k,\ 13-2k)$. Use the $x$-coordinate to find $k$, then check the other two.

$$\begin{aligned}\text{(B)}\quad 4k=2&\Rightarrow k=\tfrac{1}{2}\\y&=7+16\left(\tfrac{1}{2}\right)=15\\z&=13-2\left(\tfrac{1}{2}\right)=12\end{aligned}$$

All three coordinates agree on $k=\frac{1}{2}$. (A) needs $k=0$ but then $z=13$. (C) needs $k=1$ but then $y=23$. (D) needs $k=1$ but then $z=11$.

Take $O$ as the origin, so $\overrightarrow{OP}=\boldsymbol{a}$, $\overrightarrow{OQ}=\boldsymbol{a}+\boldsymbol{b}$, $\overrightarrow{OR}=\boldsymbol{b}+\boldsymbol{c}$ and $\overrightarrow{OS}=\boldsymbol{c}$.

$$\overrightarrow{PR}=\boldsymbol{b}+\boldsymbol{c}-\boldsymbol{a}$$

$$\overrightarrow{QS}=\boldsymbol{c}-\boldsymbol{a}-\boldsymbol{b}$$

Each diagonal is a line through its starting point, in the direction of the diagonal.

$$\begin{aligned}PR:\ \boldsymbol{r}&=\boldsymbol{a}+t_1(\boldsymbol{b}+\boldsymbol{c}-\boldsymbol{a})\\QS:\ \boldsymbol{r}&=\boldsymbol{a}+\boldsymbol{b}+t_2(\boldsymbol{c}-\boldsymbol{a}-\boldsymbol{b})\end{aligned}$$

Where the diagonals meet, the two position vectors are equal. Take $\boldsymbol{a}$ from both sides.

$$t_1(\boldsymbol{b}+\boldsymbol{c}-\boldsymbol{a})=\boldsymbol{b}+t_2(\boldsymbol{c}-\boldsymbol{a}-\boldsymbol{b})$$

$\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{c}$ point in different directions, so the coefficients of each one must match.

$$\begin{aligned}\boldsymbol{a}:&\quad -t_1=-t_2\\\boldsymbol{b}:&\quad t_1=1-t_2\end{aligned}$$

So $t_1=t_2=\frac{1}{2}$. The lines meet half way along each diagonal, so the diagonal from $P$ to $R$ and the diagonal from $Q$ to $S$ bisect each other.

Going round the cycle $\hat{i}\to\hat{j}\to\hat{k}\to\hat{i}$ gives a positive answer, so $\hat{j}\times\hat{k}=\hat{i}$.

(A) is wrong twice: a cross product is a vector, and $\hat{i}\times\hat{i}$ is the zero vector because a vector is parallel to itself. (B) should be $-\hat{j}$. (C) goes against the cycle, so it should be $-\hat{k}$.

Write the missing components as zeros first: $\boldsymbol{a}=(0,1,1)$ and $\boldsymbol{b}=(2,0,1)$.

$$\boldsymbol{a}\times\boldsymbol{b}=\begin{pmatrix}1(1)-1(0)\\1(2)-0(1)\\0(0)-1(2)\end{pmatrix}=\begin{pmatrix}1\\2\\-2\end{pmatrix}$$

This is the first preset in the interactive above.

The plane contains the origin, so the position vectors of the two points lie in the plane. Their cross product is a normal.

$$\begin{aligned}&\begin{pmatrix}1\\2\\3\end{pmatrix}\times\begin{pmatrix}3\\2\\1\end{pmatrix}\\&\quad=\begin{pmatrix}2(1)-3(2)\\3(3)-1(1)\\1(2)-2(3)\end{pmatrix}=\begin{pmatrix}-4\\8\\-4\end{pmatrix}\end{aligned}$$

That is not one of the options, but any multiple of a normal is also a normal. Multiplying by $-1$ gives $(4,-8,4)$.

Only the $\hat{k}$ component survives, because both vectors have a zero $\hat{k}$ component and $\boldsymbol{b}$ has a zero $\hat{i}$ component. That component is $a_1b_2-a_2b_1$.

$$\begin{aligned}(3n+2)(n-2)-2(0)&=1-2n\\3n^2-4n-4&=1-2n\\3n^2-2n-5&=0\\(3n-5)(n+1)&=0\end{aligned}$$

$$\begin{aligned}&\begin{pmatrix}1\\3\\0\end{pmatrix}\times\begin{pmatrix}1\\0\\2\end{pmatrix}\\&\quad=\begin{pmatrix}3(2)-0(0)\\0(1)-1(2)\\1(0)-3(1)\end{pmatrix}=\begin{pmatrix}6\\-2\\-3\end{pmatrix}\end{aligned}$$

Check: $(6,-2,-3)\cdot(1,3,0)=6-6+0=0$ and $(6,-2,-3)\cdot(1,0,2)=6+0-6=0$. This is the second preset in the interactive above.

Part a)

The cube has sides of length 2, because $|\overrightarrow{OA}|=2$. $B$ is 2 along the $y$-axis and 2 up.

$$\overrightarrow{OB}=2\hat{j}+2\hat{k}$$

Part b)

$$\begin{aligned}\overrightarrow{OA}\times\overrightarrow{OB}&=\begin{pmatrix}2\\0\\0\end{pmatrix}\times\begin{pmatrix}0\\2\\2\end{pmatrix}\\&=\begin{pmatrix}0(2)-0(2)\\0(0)-2(2)\\2(2)-0(0)\end{pmatrix}\\&=\begin{pmatrix}0\\-4\\4\end{pmatrix}\end{aligned}$$

Part c)

$$\text{Area}=\tfrac{1}{2}\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|$$

$$\begin{aligned}&=\tfrac{1}{2}\sqrt{0^2+(-4)^2+4^2}\\&=\tfrac{1}{2}\sqrt{32}\\&=2\sqrt{2}\text{ units}^2\end{aligned}$$

Part d)

$$\begin{aligned}\overrightarrow{OM}&=\hat{i}\\\overrightarrow{ON}&=\hat{j}+\hat{k}\\\overrightarrow{MN}&=\overrightarrow{ON}-\overrightarrow{OM}\\&=-\hat{i}+\hat{j}+\hat{k}\end{aligned}$$

Part e)

$$\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\&=-2\hat{i}+2\hat{j}+2\hat{k}\\&=2\overrightarrow{MN}\end{aligned}$$

So the length of $AB$ is twice the length of $MN$.

The shaded face is triangle $ABC$. Take $A$ as the corner.

$$\begin{aligned}\overrightarrow{AB}&=\begin{pmatrix}-1\\2\\-2\end{pmatrix}-\begin{pmatrix}1\\2\\5\end{pmatrix}=\begin{pmatrix}-2\\0\\-7\end{pmatrix}\\[6pt]\overrightarrow{AC}&=\begin{pmatrix}0\\5\\2\end{pmatrix}-\begin{pmatrix}1\\2\\5\end{pmatrix}=\begin{pmatrix}-1\\3\\-3\end{pmatrix}\end{aligned}$$

$$\begin{aligned}&\overrightarrow{AB}\times\overrightarrow{AC}\\&=\begin{pmatrix}0(-3)-(-7)(3)\\(-7)(-1)-(-2)(-3)\\(-2)(3)-0(-1)\end{pmatrix}\\&=\begin{pmatrix}21\\1\\-6\end{pmatrix}\end{aligned}$$

$$\begin{aligned}\text{Area}&=\tfrac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|\\&=\tfrac{1}{2}\sqrt{21^2+1^2+(-6)^2}\end{aligned}$$

$$=\tfrac{1}{2}\sqrt{478}\approx10.93\text{ units}^2$$

$$\begin{aligned}\overrightarrow{OA}\times\overrightarrow{OB}&=\begin{pmatrix}1\\2\\5\end{pmatrix}\times\begin{pmatrix}-3\\4\\0\end{pmatrix}\\&=\begin{pmatrix}2(0)-5(4)\\5(-3)-1(0)\\1(4)-2(-3)\end{pmatrix}\\&=\begin{pmatrix}-20\\-15\\10\end{pmatrix}\end{aligned}$$

$$\begin{aligned}\text{Area}&=\tfrac{1}{2}\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|\\&=\tfrac{1}{2}\sqrt{(-20)^2+(-15)^2+10^2}\\&=\tfrac{1}{2}\sqrt{725}\approx13.46\text{ units}^2\end{aligned}$$

This matches part e), so the result is verified. In exact form both are $\frac{5\sqrt{29}}{2}$, because $\sqrt{725}=5\sqrt{29}$.

$$\begin{aligned}\overrightarrow{OP}\times\overrightarrow{OQ}&=\begin{pmatrix}2\\-3\\1\end{pmatrix}\times\begin{pmatrix}2\\2\\-4\end{pmatrix}\\&=\begin{pmatrix}(-3)(-4)-1(2)\\1(2)-2(-4)\\2(2)-(-3)(2)\end{pmatrix}\\&=\begin{pmatrix}10\\10\\10\end{pmatrix}\end{aligned}$$

$$\begin{aligned}A&=\tfrac{1}{2}\left|10\hat{i}+10\hat{j}+10\hat{k}\right|\\&=\tfrac{1}{2}\sqrt{300}\\&=5\sqrt{3}\approx8.66\text{ units}^2\end{aligned}$$

There is no published marking guide for the sample paper, so this is my own solution and my own split of the marks.

Part a)

Parallelogram ABGH has two sides from A: $\overrightarrow{AB}=\boldsymbol{r}$ and $\overrightarrow{AH}=\boldsymbol{q}$.

$$\text{Area}=|\boldsymbol{q}\times\boldsymbol{r}|$$

Part b)

With A as the origin, B has position vector $\boldsymbol{r}$ and D has position vector $\boldsymbol{p}$.

$$\overrightarrow{BD}=\boldsymbol{p}-\boldsymbol{r}$$

$$\boldsymbol{x}=\boldsymbol{r}+t(\boldsymbol{p}-\boldsymbol{r}),\quad t\in\mathbb{R}$$

I have used $\boldsymbol{x}$ for a point on the line, because the question already uses $\boldsymbol{r}$ for $\overrightarrow{AB}$. $\boldsymbol{p}+t(\boldsymbol{r}-\boldsymbol{p})$ is equally correct.

The vector equation of a plane uses dot products, which rules out (C) and (D). The normal $\boldsymbol{n}=(1,1,2)$ appears on both sides of $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{a}\cdot\boldsymbol{n}$, and the point $\boldsymbol{a}=(1,3,1)$ appears only on the right.

(A) swaps the roles of the point and the normal.

Part a)

$$\overrightarrow{AB}=\begin{pmatrix}1\\1\\6\end{pmatrix}-\begin{pmatrix}3\\-1\\3\end{pmatrix}=\begin{pmatrix}-2\\2\\3\end{pmatrix}$$

Part b)

Use $A$ for the point and $\overrightarrow{AB}$ for the direction. This is the line in the animation in Part 2.

$$\frac{x-3}{-2}=\frac{y+1}{2}=\frac{z-3}{3}$$

Part c)

$\overrightarrow{AB}$ is perpendicular to the plane, so it is the normal. The point is $A$.

$$\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}-2\\2\\3\end{pmatrix}=\begin{pmatrix}3\\-1\\3\end{pmatrix}\cdot\begin{pmatrix}-2\\2\\3\end{pmatrix}$$

$$-2x+2y+3z=-6-2+9=1$$

Part a)

The prism is 1 along $x$, 3 along $y$ and 2 along $z$.

$$A(1,3,0),\quad B(0,3,2),\quad C(1,0,2)$$

Part b)

$$\overrightarrow{AB}=\begin{pmatrix}-1\\0\\2\end{pmatrix},\quad \overrightarrow{AC}=\begin{pmatrix}0\\-3\\2\end{pmatrix}$$

$$\begin{aligned}\boldsymbol{n}&=\overrightarrow{AB}\times\overrightarrow{AC}\\&=\begin{pmatrix}0(2)-2(-3)\\2(0)-(-1)(2)\\(-1)(-3)-0(0)\end{pmatrix}\\&=\begin{pmatrix}6\\2\\3\end{pmatrix}\end{aligned}$$

$$\begin{aligned}\hat{\boldsymbol{n}}&=\frac{1}{\sqrt{6^2+2^2+3^2}}\begin{pmatrix}6\\2\\3\end{pmatrix}\\&=\frac{1}{7}\begin{pmatrix}6\\2\\3\end{pmatrix}\end{aligned}$$

Part c)

Perpendicular vectors have a dot product of 0.

$$\hat{\boldsymbol{n}}\cdot\overrightarrow{AB}=\frac{1}{7}\begin{pmatrix}6\\2\\3\end{pmatrix}\cdot\begin{pmatrix}-1\\0\\2\end{pmatrix}$$

$$=\frac{1}{7}(-6+0+6)=0$$

Part d)

$$\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}6\\2\\3\end{pmatrix}=\begin{pmatrix}1\\3\\0\end{pmatrix}\cdot\begin{pmatrix}6\\2\\3\end{pmatrix}$$

$$6x+2y+3z=12$$

Check with B and C: $0+6+6=12$ and $6+0+6=12$. Load the 2020 preset in the interactive above to see this plane.

On the $z$-axis, $x=0$ and $y=0$.

$$2(0)-4z-8=0\ \Rightarrow\ z=-2$$

Write the missing $y$ term as $0y$: $1x+0y-2z=5$. The normal is the coefficients of $x$, $y$ and $z$. The 5 tells you where the plane is, and plays no part in its normal.

Part a)

The $x$-coordinate gives $\frac{-1+1}{-2}=0$, so every fraction equals 0, including $z-2$.

$$a-2=0\ \Rightarrow\ a=2$$

Part b)

The direction is in the denominators. $z-2$ has a hidden denominator of 1.

$$\boldsymbol{d}=\begin{pmatrix}-2\\2\\1\end{pmatrix}$$

Part c)

$$\begin{aligned}b-b+4(-2b)&=8\\-8b&=8\\b&=-1\end{aligned}$$

Part d)

$$\boldsymbol{n}=\begin{pmatrix}1\\-1\\4\end{pmatrix}$$

Part e)

$$\begin{aligned}\boldsymbol{d}\cdot\boldsymbol{n}&=-2(1)+2(-1)+1(4)\\&=-2-2+4\\&=0\end{aligned}$$

Part f)

The statement is not reasonable. $\boldsymbol{d}\cdot\boldsymbol{n}=0$ shows the line is perpendicular to the normal of the plane. A line at right angles to the normal runs parallel to the plane, so it cannot be perpendicular to the plane.

For the line to be perpendicular to the plane, $\boldsymbol{d}$ would need to be parallel to $\boldsymbol{n}$, a scalar multiple of it, as in 2021 Paper 1, Question 12 below.

Part a)

$$\boldsymbol{n}=\begin{pmatrix}1\\-1\\-2\end{pmatrix}$$

Part b)

The line is perpendicular to the plane, so its direction is the normal.

$$\boldsymbol{r}=\begin{pmatrix}-2\\1\\3\end{pmatrix}+k\begin{pmatrix}1\\-1\\-2\end{pmatrix},\quad k\in\mathbb{R}$$

Part c)

$$\begin{aligned}x&=-2+k\\y&=1-k\\z&=3-2k\end{aligned}$$

Part d)

S is on the plane, so substitute the parametric equations into the plane equation. This is the animation above.

$$\begin{aligned}&(-2+k)-(1-k)\\&\quad-2(3-2k)=15\end{aligned}$$

$$6k-9=15\ \Rightarrow\ k=4$$

$$\begin{aligned}S&=(-2+4,\ 1-4,\ 3-8)\\&=(2,-3,-5)\end{aligned}$$

The marking guide also accepts working backwards: show $(2,-3,-5)$ satisfies the plane equation, then show the same parameter value comes from all three parametric equations.

Part e)

$$\overrightarrow{AS}=\begin{pmatrix}2\\-3\\-5\end{pmatrix}-\begin{pmatrix}-2\\1\\3\end{pmatrix}=\begin{pmatrix}4\\-4\\-8\end{pmatrix}$$

Part f)

$$\overrightarrow{AS}=\begin{pmatrix}4\\-4\\-8\end{pmatrix}=4\begin{pmatrix}1\\-1\\-2\end{pmatrix}=4\boldsymbol{n}$$

$\overrightarrow{AS}$ is a scalar multiple of $\boldsymbol{n}$, so they are parallel.