CP-6 Further vectors — printable coverage pack

CP-6.1 · Understand and use the vector and Cartesian forms of an equation of a straight line in 3-D.

A line through the point with position vector $\mathbf a$ and parallel to the non-zero direction vector $\mathbf d$ has vector equation $\mathbf r=\mathbf a+\lambda\mathbf d$ .
If $\mathbf a=(a,b,c)$ and $\mathbf d=(l,m,n)$ with non-zero components, the Cartesian form is $(x-a)/l=(y-b)/m=(z-c)/n$ .
To form a line through points $A$ and $B$ , use $\overrightarrow{AB}=\mathbf b-\mathbf a$ as its direction; any non-zero scalar multiple gives the same line.
A zero component of the direction vector becomes a fixed coordinate, not a zero denominator. A common error is to use $\mathbf a-\mathbf b$ inconsistently with the stated parameterisation.

Tier 1 · Easy

1. Write a vector equation of the line through $(2,-1,4)$ with direction vector $3\mathbf i+2\mathbf j-\mathbf k$ .[2 marks]
Answer
- $\mathbf r=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\-1\end{pmatrix}$
Method: Use $\mathbf r=\mathbf a+\lambda\mathbf d$ with $\mathbf a=(2,-1,4)$ and $\mathbf d=(3,2,-1)$ .

Tier 2 · Standard

1. A line has Cartesian equation $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z-4}{3}$ . Write it in vector form and find the point for which the common parameter is $2$ .[4 marks]
Answer
- $\mathbf r=\begin{pmatrix}1\\-3\\4\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}$
- Point: $(5,-5,10)$
Method: Set each Cartesian ratio equal to $\lambda$ . Then $x=1+2\lambda$ , $y=-3-\lambda$ and $z=4+3\lambda$ , giving the vector equation. At $\lambda=2$ the point is $(1+4,-3-2,4+6)=(5,-5,10)$ .

Tier 3 · Hard

1. The line $l$ passes through $A(1,2,-1)$ and $B(4,-2,5)$ . Find vector and Cartesian equations of $l$ , and determine the point on $l$ whose $x$ -coordinate is $7$ .[6 marks]
Answer
- $\mathbf r=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}3\\-4\\6\end{pmatrix}$
- $\dfrac{x-1}{3}=\dfrac{y-2}{-4}=\dfrac{z+1}{6}$
- Point: $(7,-6,11)$
Method: $\overrightarrow{AB}=(4-1,-2-2,5-(-1))=(3,-4,6)$ , which gives both forms. On the line, $x=1+3\lambda$ . Setting this equal to $7$ gives $\lambda=2$ , so $y=2-8=-6$ and $z=-1+12=11$ .

CP-6.2 · Understand and use the vector and Cartesian forms of the equation of a plane.

A plane through position vector $\mathbf a$ and parallel to independent vectors $\mathbf b$ and $\mathbf c$ has vector equation $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c$ .
A plane with normal vector $\mathbf n=(p,q,r)$ has Cartesian equation $px+qy+rz=d$ , where $d=\mathbf n\cdot\mathbf a$ for any point $\mathbf a$ on the plane.
To convert a vector plane to Cartesian form, let $\mathbf n=(p,q,r)$ and solve $\mathbf n\cdot\mathbf b=0$ and $\mathbf n\cdot\mathbf c=0$ for a non-zero normal vector.
The two spanning vectors must not be parallel. A common error is to use a direction lying in the plane as though it were the plane's normal.

Tier 1 · Easy

1. Find a Cartesian equation of the plane through $(1,2,0)$ with normal vector $2\mathbf i-\mathbf j+3\mathbf k$ .[3 marks]
Answer
- $2x-y+3z=0$
Method: Use $2(x-1)-(y-2)+3(z-0)=0$ . Expanding gives $2x-y+3z=0$ .

Tier 2 · Standard

1. The plane $\Pi$ has vector equation $\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\0\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix}$ . Find a Cartesian equation of $\Pi$ .[5 marks]
Answer
- $2x-y+z=4$
Method: Let a normal be $(p,q,r)$ . Perpendicularity to $(1,2,0)$ and $(0,1,1)$ gives $p+2q=0$ and $q+r=0$ . Taking $q=-1$ gives the normal $(2,-1,1)$ . Using the point $(1,0,2)$ gives $2(x-1)-y+(z-2)=0$ , hence $2x-y+z=4$ .

Tier 3 · Hard

1. A plane passes through $A(1,1,0)$ , $B(3,0,1)$ and $C(0,2,2)$ . Find both a vector equation and a Cartesian equation of the plane.[6 marks]
Answer
- $\mathbf r=\begin{pmatrix}1\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\1\end{pmatrix}+\mu\begin{pmatrix}-1\\1\\2\end{pmatrix}$
- $3x+5y-z=8$
Method: $\overrightarrow{AB}=(2,-1,1)$ and $\overrightarrow{AC}=(-1,1,2)$ give the vector form. For a normal $(p,q,r)$ , the equations $2p-q+r=0$ and $-p+q+2r=0$ have the solution $(p,q,r)=(3,5,-1)$ . Through $A$ , the Cartesian equation is $3(x-1)+5(y-1)-z=0$ , which simplifies to $3x+5y-z=8$ .

CP-6.3 · Calculate the scalar product and use it to express the equation of a plane, and to calculate the angle between two lines, two planes and between a line and a plane.

For vectors $\mathbf a$ and $\mathbf b$ , $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$ , so the acute angle uses the absolute value of the scalar product when orientation is irrelevant.
The angle between two lines is the angle between their direction vectors, while the angle between two planes is the angle between their normal vectors.
If $\alpha$ is the acute angle between a line of direction $\mathbf d$ and a plane of normal $\mathbf n$ , then $\sin\alpha=|\mathbf d\cdot\mathbf n|/(|\mathbf d||\mathbf n|)$ .
A plane through $\mathbf a$ with normal $\mathbf n$ is $\mathbf n\cdot(\mathbf r-\mathbf a)=0$ . A common error is to report the complementary angle for a line and a plane.

Tier 1 · Easy

1. The plane $\Pi$ passes through $(1,-1,2)$ and has normal vector $(2,-1,2)$ . Use a scalar-product equation to find a Cartesian equation of $\Pi$ .[3 marks]
Answer
- $2x-y+2z=7$
Method: The scalar-product form is $(2,-1,2)\cdot((x,y,z)-(1,-1,2))=0$ . Thus $2(x-1)-(y+1)+2(z-2)=0$ , so $2x-y+2z=7$ .

Tier 2 · Standard

1. Planes $\Pi_1$ and $\Pi_2$ have normals $(1,2,2)$ and $(2,-1,2)$ respectively. Find the exact cosine of the acute angle between the planes.[4 marks]
Answer
- $\cos\theta=\dfrac49$
- $\theta=\arccos\left(\dfrac49\right)$
Method: The scalar product of the normals is $1(2)+2(-1)+2(2)=4$ . Each normal has magnitude $3$ , so $\cos\theta=|4|/(3\times3)=4/9$ .

Tier 3 · Hard

1. A line has direction vector $(2,-1,2)$ and the plane $3x+5y-z=8$ has normal $(3,5,-1)$ . Determine the exact sine of the acute angle $\alpha$ between the line and the plane.[5 marks]
Answer
- $\sin\alpha=\dfrac{1}{3\sqrt{35}}$
- $\alpha=\arcsin\left(\dfrac{1}{3\sqrt{35}}\right)$
Method: The scalar product is $(2,-1,2)\cdot(3,5,-1)=6-5-2=-1$ . The magnitudes are $3$ and $\sqrt{35}$ . Therefore $\sin\alpha=|-1|/(3\sqrt{35})=1/(3\sqrt{35})$ .

CP-6.4 · Check whether vectors are perpendicular by using the scalar product.

Two non-zero vectors are perpendicular exactly when their scalar product is zero.
To test whether two direction vectors are perpendicular, calculate their scalar product; the initial points from which the vectors are drawn do not affect the test.
An unknown component can be found by setting the scalar product equal to zero and solving the resulting linear equation.
Do not infer perpendicularity from a diagram or from one pair of components. Every component product must be included in the scalar-product sum.

Tier 1 · Easy

1. Show that the vectors $(1,2,-1)$ and $(3,-1,1)$ are perpendicular.[2 marks]
Answer
- $(1,2,-1)\cdot(3,-1,1)=0$ , so the vectors are perpendicular.
Method: Their scalar product is $1(3)+2(-1)+(-1)(1)=3-2-1=0$ . Since both vectors are non-zero, they are perpendicular.

Tier 2 · Standard

1. Find the value of $k$ for which $(k,2,-1)$ is perpendicular to $(3,1,5)$ .[3 marks]
Answer
- $k=1$
Method: Perpendicularity requires $3k+2(1)+(-1)(5)=0$ . Thus $3k-3=0$ , giving $k=1$ .

Tier 3 · Hard

1. Points $A(1,0,2)$ , $B(3,-1,5)$ , $C(-2,4,1)$ and $D(-1,9,2)$ define the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ . Prove that these vectors are perpendicular.[4 marks]
Answer
- $\overrightarrow{AB}=(2,-1,3)$ and $\overrightarrow{CD}=(1,5,1)$
- $\overrightarrow{AB}\perp\overrightarrow{CD}$
Method: $\overrightarrow{AB}=B-A=(2,-1,3)$ and $\overrightarrow{CD}=D-C=(1,5,1)$ . Their scalar product is $2(1)+(-1)(5)+3(1)=2-5+3=0$ , so the two vectors are perpendicular.

CP-6.5 · Find the intersection of a line and a plane. Calculate the perpendicular distance between two lines, from a point to a line and from a point to a plane.

To intersect a line with a plane, substitute the line's parametric coordinates into the plane equation, solve for the parameter, and then recover all three coordinates.
The distance from $P(x_0,y_0,z_0)$ to $ax+by+cz+d=0$ is $|ax_0+by_0+cz_0+d|/\sqrt{a^2+b^2+c^2}$ .
For a point-to-line distance, find the point $Q$ on the line for which $\overrightarrow{PQ}$ is perpendicular to the line direction. For two skew lines, make the connector between general points perpendicular to both directions.
Distances are non-negative and require a perpendicular separation. A common error is to use the distance between arbitrary given points on the two objects.

Tier 1 · Easy

1. The line $\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}$ meets the plane $x+y+z=6$ . Find the point of intersection.[4 marks]
Answer
- Point of intersection: $\left(4,\dfrac32,\dfrac12\right)$
Method: On the line, $x=1+2t$ , $y=t$ and $z=2-t$ . Substitution into the plane gives $1+2t+t+2-t=6$ , so $2t=3$ and $t=3/2$ . The coordinates are $(4,3/2,1/2)$ .

Tier 2 · Standard

1. Find the perpendicular distance from $P(2,-1,4)$ to the plane $2x-y+2z-9=0$ .[3 marks]
Answer
- Distance $=\dfrac43$
Method: Substitute $P$ into the left side and take the absolute value: $|2(2)-(-1)+2(4)-9|=|4|=4$ . The normal has magnitude $\sqrt{2^2+(-1)^2+2^2}=3$ , so the distance is $4/3$ .

Tier 3 · Hard

1. The skew lines $l_1$ and $l_2$ have equations $\mathbf r=t(1,0,1)$ and $\mathbf r=(1,2,0)+s(0,1,1)$ . Calculate the perpendicular distance between them.[6 marks]
Answer
- Distance $=\sqrt3$
Method: Take $P=(t,0,t)$ on $l_1$ and $Q=(1,2+s,s)$ on $l_2$ . Then $\overrightarrow{PQ}=(1-t,2+s,s-t)$ . Perpendicularity to $(1,0,1)$ and $(0,1,1)$ gives $1+s-2t=0$ and $2+2s-t=0$ . Solving gives $t=0$ and $s=-1$ , so the shortest connector is $(1,1,-1)$ and its length is $\sqrt{1^2+1^2+(-1)^2}=\sqrt3$ .