True or False? A scalar triple product is a vector, because a vector product appears inside it.

False. The vector product does produce a vector, but that vector is then dotted with a third one, and a scalar product always gives an ordinary number. The operation is named for what comes out of it, not for what happens in the middle.

What does a scalar triple product of zero tell you about the three vectors?

They are coplanar , meaning all three lie in one plane. It happens automatically whenever two of the three are parallel, since the vector product of those two is then the zero vector and everything collapses.

In the formula for the distance from a point to a plane, what do the numerator and the denominator each do?

The numerator is the plane's equation evaluated at the point's coordinates, taken in modulus so the result cannot come out negative. The denominator is the magnitude of the plane's normal vector, which scales that number down to an actual distance rather than a multiple of one.

Three of the distance cases have no formula of their own. Which formula does each of them reduce to?

Two parallel planes, and a plane with a line parallel to it, both reduce to the point-to-plane formula. Two parallel lines reduce to the point-to-line formula. In every case a point is taken on one of the two objects and the formula is applied to the other.

You are given the three vertices of a triangle as coordinates. What is the first thing to do before any area formula can be used?

Form two displacement vectors that both start from the same vertex , by subtracting that vertex's position vector from each of the other two. Taking them from different vertices would give sides that do not meet, and the formula needs two sides meeting at a corner.

What is the direction cosine identity actually used for?

It finds a missing direction cosine when a question supplies the angles to only two of the three axes. Square-rooting gives two values differing in sign, so such a question usually has two lines as its answer rather than one.

A LevelFurther MathsEdexcelFurther Pure 1FlashcardsFurther VectorsFurther Vectors

Further Vectors (Edexcel A Level Further Maths: Further Pure 1): Flashcards

Exam code: 9FM0

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FrontVector (Cross) Product
Define the vector product of two vectors.

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Cards in this collection (41)

Define the vector product of two vectors.
The vector product $a \times b$ is an operation on two vectors whose result is another vector, one that is perpendicular to both of them.
It is also called the cross product, after the symbol used to write it.
Which of the two directions perpendicular to $a$ and $b$ does $a \times b$ point in?
The one given by the right-hand rule: point the index finger of your right hand along $a$ and your middle finger along $b$ , and your thumb then points along $a \times b$ .
There are always two perpendicular directions to choose between, and the rule is what settles which of them the vector product takes.
What is the relationship between $a \times b$ and $b \times a$ ?
They are the negatives of one another, $a \times b = - b \times a$ , so the vector product is not commutative.
Swapping the order leaves the magnitude alone and reverses the direction, which is exactly what the right-hand rule predicts.
True or False?
If $a \times b = 0$ for two non-zero vectors, then $a$ and $b$ are parallel.
True.
Since $| a \times b | = | a | | b | \sin θ$ and neither magnitude is zero, the only way to reach zero is $\sin θ = 0$ .
That makes $θ$ either $0 °$ or $180 °$ , and in both cases the two vectors lie along the same line.
What is $| a \times b |$ when $a$ and $b$ are perpendicular?
It is simply the product of their magnitudes, $| a | | b |$ .
With $θ = 90 °$ the factor $\sin θ$ is $1$ , and the unit normal has magnitude $1$ as well, so there is nothing left to reduce it.
Which ordinary algebraic rules does the vector product still obey?
Brackets expand as usual, so $a \times (b + c) = a \times b + a \times c$ .
A scalar factor may also be moved anywhere in the product, so $(k a) \times b = k (a \times b)$ .
Given $a = 2 i - 5 k$ and $b = 3 i - 2 j - k$ , find $a \times b$ .
The vector product is $- 10 i - 13 j - 4 k$ .
With $a$ having components $2$ , $0$ and $- 5$ and $b$ having $3$ , $- 2$ and $- 1$ , the three entries work out as $0 \times (- 1) - (- 5) \times (- 2)$ , then $(- 5) \times 3 - 2 \times (- 1)$ , then $2 \times (- 2) - 0 \times 3$ .
Define the scalar triple product.
The scalar triple product of $a$ , $b$ and $c$ is $a \cdot (b \times c)$ : the vector product of two of them, followed by the scalar product of the third with that result.
The brackets are essential, since the vector product has to be worked out before anything can be dotted with it.
What are the two ways of calculating a scalar triple product?
Either work out the vector $b \times c$ first and then take its scalar product with $a$ .
Or evaluate a single $3 \times 3$ determinant whose three rows are the components of $a$ , $b$ and $c$ in that order.
True or False?
A scalar triple product is a vector, because a vector product appears inside it.
False.
The vector product does produce a vector, but that vector is then dotted with a third one, and a scalar product always gives an ordinary number.
The operation is named for what comes out of it, not for what happens in the middle.
The dot and the cross can be swapped without changing the value. Complete the right-hand side:
$a \cdot (b \times c) = (______\times______) \cdot______$
The completed identity is:
$a \cdot (b \times c) = (a \times b) \cdot c$
The three vectors stay in the same order and only the two operation symbols trade places, which is why this is described as swapping the dot and the cross.
What does a scalar triple product of zero tell you about the three vectors?
They are coplanar, meaning all three lie in one plane.
It happens automatically whenever two of the three are parallel, since the vector product of those two is then the zero vector and everything collapses.
Cycling $a$ , $b$ and $c$ round in order leaves the scalar triple product unchanged. Why does swapping just two of them not?
A cycle moves every vector along one place and so preserves the order they run in, whereas a single swap reverses it.
Reversing the vector product changes its sign, since $b \times c = - c \times b$ , and that sign carries straight through the scalar product.
Find $a \cdot (b \times c)$ for $a = 2 i + 3 j - k$ , $b = i - 2 j$ and $c = - 5 i + j + 2 k$ .
The value is $- 5$ .
First $b \times c$ comes out as $- 4 i - 2 j - 9 k$ , remembering that the missing $k$ term of $b$ means its third component is $0$ .
Dotting that with $a$ gives $2 \times (- 4) + 3 \times (- 2) + (- 1) \times (- 9)$ .
A line passes through the point with position vector $a$ in the direction $b$ . Complete its vector-product form:
$(r -______) \times______= 0$
The completed form is:
$(r - a) \times b = 0$
Unlike $r = a + λ b$ , this version contains no parameter at all, so it describes the line as a single condition on $r$ .
How is the vector-product form of a line obtained from $r = a + λ b$ ?
Rearrange it to $r - a = λ b$ , then take the vector product of both sides with $b$ .
The right-hand side becomes $λ (b \times b)$ , which is the zero vector, so the parameter disappears entirely.
True or False?
In the vector-product form of a line, the right-hand side is the number zero.
False.
It is the zero vector, since a vector product always produces a vector and both sides of the equation must therefore be vectors.
Writing it as an ordinary $0$ is a common slip, and the distinction matters because it says the vector product vanishes in all three components.
A plane is given as $r = a + s b + t c$ . How do you find a normal vector to it?
Take the vector product of the two direction vectors, $n = b \times c$ .
It is perpendicular to both of them by construction, and anything perpendicular to two non-parallel directions lying in a plane is perpendicular to the plane itself.
Two non-parallel planes have normals $n_{1}$ and $n_{2}$ . What direction does their line of intersection run in?
The direction $n_{1} \times n_{2}$ .
The line lies inside both planes, so it is perpendicular to both normals at once, and the vector product is exactly the vector with that property.
The line of intersection of two planes is $r = a + λ (n_{1} \times n_{2})$ . How do you find $a$ if the question does not give it?
Set one of $x$ , $y$ and $z$ to zero, substitute into both plane equations, and solve the resulting pair of simultaneous equations for the other two.
Only a variable whose matching component of $n_{1} \times n_{2}$ is non-zero may be set to zero, because otherwise that coordinate is constant along the line and may never take the value $0$ .
In what direction does the line of intersection of $2 x - y + 3 z = 7$ and $x - 3 y + 4 z = 11$ run?
The normals are $2 i - j + 3 k$ and $i - 3 j + 4 k$ , read straight off the coefficients.
Their vector product is $5 i - 5 j - 5 k$ , and that is the direction the line of intersection runs in.
A point has position vector $p$ and a line has equation $r = a + λ b$ . Complete the formula for the shortest distance between them:
$distance = \frac{|______\times b |}{______}$
The completed formula is:
$distance = \frac{| (p - a) \times b |}{| b |}$
The numerator uses the displacement from the point $a$ on the line to the point $p$ , and dividing by $| b |$ removes the effect of however long the direction vector happens to be.
Two skew lines have equations $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ . Complete the formula for the shortest distance between them:
$distance = \frac{| (a_{1} - a_{2}) \cdot (______) |}{|______|}$
The completed formula is:
$distance = \frac{| (a_{1} - a_{2}) \cdot (b_{1} \times b_{2}) |}{| b_{1} \times b_{2} |}$
The same vector $b_{1} \times b_{2}$ appears above and below, so the whole expression is a scalar triple product divided by a magnitude.
In the formula for the distance from a point to a plane, what do the numerator and the denominator each do?
The numerator is the plane's equation evaluated at the point's coordinates, taken in modulus so the result cannot come out negative.
The denominator is the magnitude of the plane's normal vector, which scales that number down to an actual distance rather than a multiple of one.
True or False?
The point-to-line formula gives the same distance whichever point of the line is used as $a$ .
True.
Moving $a$ along the line changes $p - a$ only by some multiple of $b$ , and a vector parallel to $b$ contributes nothing to a vector product with $b$ .
So the numerator is unaffected, and you may take whichever point on the line is most convenient.
Why does $b_{1} \times b_{2}$ appear in the skew-lines formula?
It is perpendicular to both direction vectors at once, which is the direction the shortest segment between two skew lines has to run in.
The numerator measures how far the displacement between the two lines reaches in that direction, and dividing by its magnitude turns it into a length.
Three of the distance cases have no formula of their own. Which formula does each of them reduce to?
Two parallel planes, and a plane with a line parallel to it, both reduce to the point-to-plane formula.
Two parallel lines reduce to the point-to-line formula.
In every case a point is taken on one of the two objects and the formula is applied to the other.
A triangle and a parallelogram each have two sides from the same vertex given by the vectors $a$ and $b$ . Complete both areas:
$triangle: A =______, parallelogram: A =______$
The completed areas are:
$triangle: A = \frac{1}{2} | a \times b |, parallelogram: A = | a \times b |$
The parallelogram is made of two copies of the triangle, so its area is simply the triangle's doubled.
You are given the three vertices of a triangle as coordinates. What is the first thing to do before any area formula can be used?
Form two displacement vectors that both start from the same vertex, by subtracting that vertex's position vector from each of the other two.
Taking them from different vertices would give sides that do not meet, and the formula needs two sides meeting at a corner.
How does $A = \frac{1}{2} | a \times b |$ connect to the trigonometric area rule $A = \frac{1}{2} a b \sin C$ ?
Taking the magnitude of $| a | | b | \sin θ \hat{n}$ leaves $| a | | b | | \sin θ |$ , because the unit normal has magnitude $1$ .
In a triangle $θ$ lies strictly between $0 °$ and $180 °$ , so $\sin θ$ is positive and the modulus can be dropped, leaving the trigonometric rule with $| a |$ and $| b |$ as the two side lengths.
True or False?
The vector-product method for the area of a triangle only works if the vertices are given in three dimensions.
False.
A triangle given by two-dimensional coordinates is handled by writing each vertex with a third coordinate of $0$ , so $P (1, 2)$ becomes $P (1, 2, 0)$ .
The vector product then works as usual, and the answer comes out with only a $k$ component.
A tetrahedron and a parallelepiped each have three edges from one vertex given by $a$ , $b$ and $c$ . Complete both volumes:
$tetrahedron: V =______, parallelepiped: V =______$
The completed volumes are:
$tetrahedron: V = \frac{1}{6} | c \cdot (a \times b) |, parallelepiped: V = | c \cdot (a \times b) |$
Both are built on the same scalar triple product, and the tetrahedron takes a sixth of it.
Where does the $\frac{1}{6}$ in the tetrahedron's volume come from?
A pyramid's volume is $\frac{1}{3}$ of its base area times its vertical height, and the triangular base contributes a further factor of $\frac{1}{2}$ .
Those two fractions multiply together to give $\frac{1}{6}$ , which is why the parallelepiped, whose base is a whole parallelogram, has no fraction at all.
Does it matter which of the three edges you call $a$ , $b$ and $c$ ?
No, provided all three run from the same vertex.
Reordering them can change the sign of $c \cdot (a \times b)$ , but the modulus signs remove that, so every ordering returns the same positive volume.
Define the direction ratio of a line.
The direction ratio of a line with direction vector $b = b_{1} i + b_{2} j + b_{3} k$ is $b_{1} : b_{2} : b_{3}$ .
It is written with colons rather than as a vector because it records only the proportions between the three components.
A line has direction vector $b$ , and $α$ , $β$ and $γ$ are the angles it makes with the positive $x$ -, $y$ - and $z$ -axes. Complete the three direction cosines:
$l = \frac{b_{1}}{______} = \cos α, m = \frac{______}{| b |} = \cos β, n = \frac{b_{3}}{| b |} =______$
The completed direction cosines are:
$l = \frac{b_{1}}{| b |} = \cos α, m = \frac{b_{2}}{| b |} = \cos β, n = \frac{b_{3}}{| b |} = \cos γ$
Each one takes its own component of $b$ over the same magnitude, and each is paired with the angle to its own axis.
For the line $r = i + 2 j + 3 k + λ (- i + 5 j - 3 k)$ , find the direction cosine for the angle made with the positive $y$ -axis.
The direction cosine is $m = \frac{\sqrt{35}}{7}$ .
The direction vector has magnitude $\sqrt{{(- 1)}^{2} + 5^{2} + {(- 3)}^{2}} = \sqrt{35}$ , so $m = \frac{5}{\sqrt{35}}$ , which rationalises to $\frac{5 \sqrt{35}}{35}$ .
The three direction cosines are not independent of one another. Complete the identity they always satisfy:
$l^{2} +______+______\equiv______$
The completed identity is:
$l^{2} + m^{2} + n^{2} \equiv 1$
Squaring and adding the three cosines puts $b_{1}^{2} + b_{2}^{2} + b_{3}^{2}$ over $| b |^{2}$ , and that numerator is exactly $| b |^{2}$ .
True or False?
A line can be drawn making an angle of $45 °$ with all three coordinate axes.
False.
All three direction cosines would be $\cos 45 ° = \frac{\sqrt{2}}{2}$ , so their squares would add to $\frac{3}{2}$ rather than to $1$ .
Equal angles to all three axes are possible, but each must be about $54.7 °$ instead.
What is the direction cosine identity actually used for?
It finds a missing direction cosine when a question supplies the angles to only two of the three axes.
Square-rooting gives two values differing in sign, so such a question usually has two lines as its answer rather than one.
How is the cartesian equation of a line written in terms of its direction cosines?
It becomes $\frac{x - a_{1}}{l} = \frac{y - a_{2}}{m} = \frac{z - a_{3}}{n}$ , with the direction cosines standing in the denominators in place of the components.
It comes from the ordinary cartesian form by multiplying every part by $| b |$ , which turns each denominator $b_{i}$ into $\frac{b_{i}}{| b |}$ .

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