True or False? Two vector equations of a plane that share no direction vectors at all can still describe the same plane.

True. Any point of the plane may be used as the starting point, and any two non-parallel directions lying in it may be used as the direction vectors. That is the same freedom a line has, with a second direction vector to choose as well as the first.

How do you test whether a point lies on a plane given in vector form?

Set the point's position vector equal to the plane's equation and solve two of the three component equations for the two parameters. The point lies on the plane only if that same pair of values also satisfies the third equation.

Define a normal vector to a plane.

A normal vector to a plane is a vector perpendicular to that plane, and so perpendicular to every direction vector lying in it. Its scalar product with any direction vector in the plane is therefore zero.

How do you test whether a line is parallel to a plane?

Check whether the line's direction vector is perpendicular to the plane's normal vector , which means testing whether their scalar product is zero. A direction lying flat in the plane must be at right angles to the normal, so a zero scalar product is exactly the condition.

A line is known to be parallel to a plane. What further check tells you whether it lies in the plane?

Test whether any single point of the line satisfies the plane's equation. A line parallel to a plane either misses it entirely or lies wholly within it, so one shared point settles which of the two it is.

True or False? A line that is not parallel to a plane meets it at exactly one point.

True. The substitution leaves a single linear equation in the parameter, and such an equation has exactly one solution. If the line were parallel the parameter would cancel out instead, leaving either no solution at all or infinitely many.

Why is the angle between a line and a plane found using the plane's normal?

The direction lying in the plane that makes the smallest angle with the line is unknown, whereas the normal is known straight away. The line, the normal and the plane form a right-angled triangle, and that is what connects the angle you can find to the angle you want.

What are the only two possibilities when two distinct planes are considered together?

They are either parallel , and never meet at all, or they intersect along a whole line . Two planes can never meet at just a single point, because each of them is infinite in two directions.

How do you find the line of intersection of two planes given in cartesian form?

Set one of the variables equal to a parameter, then solve the two equations simultaneously for the other two in terms of it. The three resulting expressions are the parametric equations of the line, which can then be written in vector form.

What can you do if a line of intersection comes out with fractions in its direction vector?

Multiply every component by the common denominator, which clears the fractions without changing the direction. Scaling a direction vector leaves the line completely unchanged, so the tidier vector describes exactly the same line.

How do you find the angle between two planes?

Find the angle between their normal vectors , using the scalar product. The two planes and their two normals form a quadrilateral containing two right angles, which is what ties the angle between the normals to the angle between the planes.

Why can the angle between two planes come out as either of two values?

The angle between the normals and the angle between the planes are supplementary in one arrangement and equal in the other, so an acute and an obtuse value both describe the same pair of planes. Taking the absolute value of the scalar product is what picks out the acute one.

What does a singular coefficient matrix tell you about three planes?

It rules out the single-point case, so the planes must instead be parallel or coincident, meet along a line, or form a triangular prism. Which of those it is has to be settled by looking at the equations themselves.

Define a sheaf of planes.

A sheaf is three planes that all intersect along one common line. Their three equations are consistent but have infinitely many solutions, one for each point of that line.

In a triangular prism arrangement, how do the three planes actually meet?

Each pair of planes meets in a line, but those three lines are parallel to one another and never come together. So no point lies on all three planes at once, even though every pair of them does intersect.

Once that perpendicular line from a point to a plane is written down, what are the remaining two steps?

Find where the line meets the plane, by substituting it into the plane's equation and solving for the parameter. Then work out the distance between the original point and that point of intersection.

Why is the normal the right direction to travel in from a point to a plane?

Any other direction reaches the plane further away, because it forms the hypotenuse of a right-angled triangle whose shorter side runs along the normal. Travelling along the normal is the only route with no sideways movement wasted.

If a question gives the acute angle between a line and a plane, how does that shorten finding the distance from a point on the line to the plane?

Right-angled trigonometry replaces the whole construction: the distance from the plane is the distance measured along the line, multiplied by the sine of that angle. The line is the hypotenuse of the triangle and the perpendicular to the plane is the side opposite the angle.

How do you find the shortest distance between a plane and a line parallel to it?

Take any point on the line and find its distance to the plane, exactly as for a single point. Every point of the line gives the same answer, because a parallel line neither approaches nor recedes from the plane.

How do you find the shortest distance between two parallel planes?

Take any point on one plane and find its distance to the other along the normal direction. Where one plane is given in vector form, its own position vector is a ready-made point to start from.

True or False? Two planes that are not parallel have a shortest distance of zero between them.

True. Two planes that are not parallel intersect along a whole line, and every point of that line lies on both of them. The same holds for a line and a plane that are not parallel, which is why these questions only have any content when the two never meet.

A LevelFurther MathsEdexcelCore PureFlashcardsFurther VectorsVector Planes

Vector Planes (Edexcel A Level Further Maths: Core Pure): Flashcards

Exam code: 9FM0

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FrontEquations of planes
In the plane equation $r = a + s b + t c$ , what must be true of $b$ and $c$ ?

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Cards in this collection (41)

In the plane equation $r = a + s b + t c$ , what must be true of $b$ and $c$ ?
They must both be parallel to the plane and not parallel to each other.
Two non-parallel directions are what let $s$ and $t$ between them reach every point of a two-dimensional surface, whereas two parallel ones would only ever generate a line.
How do you write a vector equation of the plane through three non-collinear points?
Take one point's position vector as $a$ and use the displacements to the other two as the direction vectors, giving $r = a + λ (b - a) + μ (c - a)$ .
The points must not be collinear, or those two displacements would be parallel and would fail to span a plane at all.
True or False?
Two vector equations of a plane that share no direction vectors at all can still describe the same plane.
True.
Any point of the plane may be used as the starting point, and any two non-parallel directions lying in it may be used as the direction vectors.
That is the same freedom a line has, with a second direction vector to choose as well as the first.
How do you test whether a point lies on a plane given in vector form?
Set the point's position vector equal to the plane's equation and solve two of the three component equations for the two parameters.
The point lies on the plane only if that same pair of values also satisfies the third equation.
Points $A$ , $B$ and $C$ have position vectors $3 i + 2 j - k$ , $i - 2 j + 4 k$ and $4 i - j + 3 k$ . Find a vector equation of the plane through them.
The displacements from $A$ are $[\begin{matrix} - 2 \\ - 4 \\ 5 \end{matrix}]$ and $[\begin{matrix} 1 \\ - 3 \\ 4 \end{matrix}]$ , which are not parallel.
One valid answer is $r = [\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}] + λ [\begin{matrix} - 2 \\ - 4 \\ 5 \end{matrix}] + μ [\begin{matrix} 1 \\ - 3 \\ 4 \end{matrix}]$ .
Define a normal vector to a plane.
A normal vector to a plane is a vector perpendicular to that plane, and so perpendicular to every direction vector lying in it.
Its scalar product with any direction vector in the plane is therefore zero.
How do you read a normal vector straight off the cartesian equation of a plane?
The coefficients are its components, so $a x + b y + c z = d$ has normal vector $[\begin{matrix} a \\ b \\ c \end{matrix}]$ .
So $2 x - y + 3 z = 7$ has normal vector $[\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}]$ , read straight from the left-hand side.
A plane has normal vector $n$ and passes through the point with position vector $a$ . Fill in the perpendicularity statement that gives its equation:
$n \cdot (r -______) = 0$
The completed statement is:
$n \cdot (r - a) = 0$
The displacement $r - a$ runs from the known point to any other point of the plane, so it lies in the plane and is perpendicular to $n$ .
What does $n \cdot (r - a) = 0$ rearrange into?
Expanding the bracket gives $n \cdot r - n \cdot a = 0$ , so $n \cdot r = a \cdot n$ .
Writing $r$ as $[\begin{matrix} x \\ y \\ z \end{matrix}]$ turns the left-hand side into $n_{1} x + n_{2} y + n_{3} z$ , which is the cartesian form.
True or False?
$2 x - y + 3 z = 7$ and $4 x - 2 y + 6 z = 14$ are different planes.
False.
The second equation is exactly twice the first, so every point satisfying one satisfies the other.
Multiplying a plane's equation through by a non-zero constant changes nothing, which is why a normal vector is only ever fixed up to a scalar multiple.
How do you convert a plane from vector form to cartesian form without finding a normal vector?
Write the three component equations, then eliminate the two parameters by combining the equations in pairs.
For $r = [\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}] + λ [\begin{matrix} - 2 \\ - 4 \\ 5 \end{matrix}] + μ [\begin{matrix} 1 \\ - 3 \\ 4 \end{matrix}]$ that gives $x - 13 y - 10 z + 13 = 0$ .
How do you test whether a line is parallel to a plane?
Check whether the line's direction vector is perpendicular to the plane's normal vector, which means testing whether their scalar product is zero.
A direction lying flat in the plane must be at right angles to the normal, so a zero scalar product is exactly the condition.
A line is known to be parallel to a plane. What further check tells you whether it lies in the plane?
Test whether any single point of the line satisfies the plane's equation.
A line parallel to a plane either misses it entirely or lies wholly within it, so one shared point settles which of the two it is.
How do you find where a line meets a plane given in cartesian form?
Write the line in parametric form and substitute the three expressions for $x$ , $y$ and $z$ into the plane's equation.
That leaves one equation in the parameter alone, and putting its value back into the line gives the point.
Find where $r = [\begin{matrix} 1 \\ - 3 \\ 2 \end{matrix}] + λ [\begin{matrix} 2 \\ - 1 \\ - 1 \end{matrix}]$ meets $3 x - 4 y + z = 8$ .
Substituting $x = 1 + 2 λ$ , $y = - 3 - λ$ and $z = 2 - λ$ gives $17 + 9 λ = 8$ , so $λ = - 1$ .
The point of intersection is therefore $(- 1, - 2, 3)$ .
True or False?
A line that is not parallel to a plane meets it at exactly one point.
True.
The substitution leaves a single linear equation in the parameter, and such an equation has exactly one solution.
If the line were parallel the parameter would cancel out instead, leaving either no solution at all or infinitely many.
Why is the angle between a line and a plane found using the plane's normal?
The direction lying in the plane that makes the smallest angle with the line is unknown, whereas the normal is known straight away.
The line, the normal and the plane form a right-angled triangle, and that is what connects the angle you can find to the angle you want.
A line makes an angle $α$ with the normal to a plane. Fill in the angle $θ$ between the line and the plane itself:
$θ =______- α$
The completed relationship is:
$θ = 90 ° - α$
In radians it reads $θ = \frac{π}{2} - α$ , and the subtraction is needed because $α$ is measured to the normal rather than to the plane.
Why is the absolute value of the scalar product used when finding the angle between a line and a plane?
A negative scalar product would give an obtuse angle to the normal, and subtracting that from $90 °$ would produce a negative answer.
Taking the absolute value gives the acute angle to the normal instead, which is the one the geometry requires.
Find, in radians, the angle between $r = [\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}] + λ [\begin{matrix} - 1 \\ 1 \\ - 2 \end{matrix}]$ and the plane $x - 3 y + 2 z = 5$ .
The scalar product of the direction with the normal $[\begin{matrix} 1 \\ - 3 \\ 2 \end{matrix}]$ is $- 8$ , and the magnitudes are $\sqrt{6}$ and $\sqrt{14}$ .
The angle to the normal is $\arccos (\frac{8}{\sqrt{6} \sqrt{14}})$ , so the angle to the plane is $\frac{π}{2}$ minus that, giving $1.06$ radians.
What are the only two possibilities when two distinct planes are considered together?
They are either parallel, and never meet at all, or they intersect along a whole line.
Two planes can never meet at just a single point, because each of them is infinite in two directions.
True or False?
$3 x + y - 2 z = 5$ and $3 x + y - 2 z = 11$ are parallel planes.
True.
Their normal vectors are identical, so the two planes face the same way and can never meet.
Only the constant on the right differs, and changing it slides the plane to a new position without turning it.
How do you find the line of intersection of two planes given in cartesian form?
Set one of the variables equal to a parameter, then solve the two equations simultaneously for the other two in terms of it.
The three resulting expressions are the parametric equations of the line, which can then be written in vector form.
What can you do if a line of intersection comes out with fractions in its direction vector?
Multiply every component by the common denominator, which clears the fractions without changing the direction.
Scaling a direction vector leaves the line completely unchanged, so the tidier vector describes exactly the same line.
How do you find the angle between two planes?
Find the angle between their normal vectors, using the scalar product.
The two planes and their two normals form a quadrilateral containing two right angles, which is what ties the angle between the normals to the angle between the planes.
Why can the angle between two planes come out as either of two values?
The angle between the normals and the angle between the planes are supplementary in one arrangement and equal in the other, so an acute and an obtuse value both describe the same pair of planes.
Taking the absolute value of the scalar product is what picks out the acute one.
Find the acute angle between $2 x - y + 3 z = 7$ and $x + 2 y - z = 20$ .
The normals are $[\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}]$ and $[\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}]$ , with scalar product $- 3$ and magnitudes $\sqrt{14}$ and $\sqrt{6}$ .
So $\cos θ = \frac{3}{2 \sqrt{21}}$ , giving $θ = 1.24$ radians.
Three planes have coefficient matrix $M$ and constants $d_{1}$ , $d_{2}$ and $d_{3}$ . When $\det M \neq 0$ , fill in their single point of intersection:
$[\begin{matrix} x \\ y \\ z \end{matrix}] =______[\begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$
The completed result is:
$[\begin{matrix} x \\ y \\ z \end{matrix}] = M^{- 1} [\begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$
Three planes meeting at one point is the same thing as three simultaneous equations having a unique solution.
What does a singular coefficient matrix tell you about three planes?
It rules out the single-point case, so the planes must instead be parallel or coincident, meet along a line, or form a triangular prism.
Which of those it is has to be settled by looking at the equations themselves.
Define a sheaf of planes.
A sheaf is three planes that all intersect along one common line.
Their three equations are consistent but have infinitely many solutions, one for each point of that line.
How do you tell a sheaf from a triangular prism when the coefficient matrix is singular?
Eliminate variables between the three equations and look at what is left.
A statement that is always true, such as $0 = 0$ , means a sheaf, while an impossible one such as $0 = 1$ means a triangular prism.
In a triangular prism arrangement, how do the three planes actually meet?
Each pair of planes meets in a line, but those three lines are parallel to one another and never come together.
So no point lies on all three planes at once, even though every pair of them does intersect.
To find the shortest distance from the point with position vector $p$ to a plane with normal vector $n$ , a new line is used. Fill it in:
$r = p + λ______$
The completed line is:
$r = p + λ n$
Setting off from $p$ in the direction of the normal is the only route that reaches the plane at right angles.
Once that perpendicular line from a point to a plane is written down, what are the remaining two steps?
Find where the line meets the plane, by substituting it into the plane's equation and solving for the parameter.
Then work out the distance between the original point and that point of intersection.
Why is the normal the right direction to travel in from a point to a plane?
Any other direction reaches the plane further away, because it forms the hypotenuse of a right-angled triangle whose shorter side runs along the normal.
Travelling along the normal is the only route with no sideways movement wasted.
Find the shortest distance from $P (- 2, 11, - 15)$ to the plane $r \cdot [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}] = 6$ .
The line $r = [\begin{matrix} - 2 \\ 11 \\ - 15 \end{matrix}] + λ [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$ meets the plane where $6 λ - 30 = 6$ , so $λ = 6$ and the foot is $(10, 5, - 9)$ .
The distance from $P$ to that point is $\sqrt{144 + 36 + 36} = 6 \sqrt{6}$ .
If a question gives the acute angle between a line and a plane, how does that shorten finding the distance from a point on the line to the plane?
Right-angled trigonometry replaces the whole construction: the distance from the plane is the distance measured along the line, multiplied by the sine of that angle.
The line is the hypotenuse of the triangle and the perpendicular to the plane is the side opposite the angle.
How do you find the shortest distance between a plane and a line parallel to it?
Take any point on the line and find its distance to the plane, exactly as for a single point.
Every point of the line gives the same answer, because a parallel line neither approaches nor recedes from the plane.
How do you show that a plane given in vector form is parallel to a plane given by $r \cdot n = d$ ?
Show that $n$ is perpendicular to both direction vectors of the plane in vector form, by checking that two scalar products are zero.
That makes $n$ a normal to that plane as well, and two planes sharing a normal direction are parallel.
How do you find the shortest distance between two parallel planes?
Take any point on one plane and find its distance to the other along the normal direction.
Where one plane is given in vector form, its own position vector is a ready-made point to start from.
True or False?
Two planes that are not parallel have a shortest distance of zero between them.
True.
Two planes that are not parallel intersect along a whole line, and every point of that line lies on both of them.
The same holds for a line and a plane that are not parallel, which is why these questions only have any content when the two never meet.

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