4.04g - Finding a common perpendicular | OCR A-Level Further Mathematics A

Find a vector perpendicular to two given vectors using the vector product, then verify it with scalar products. You will use the supplied component formula carefully, recognise the parallel-vector case and solve problems where another perpendicularity condition determines a parameter.

A direction perpendicular to both

Can one non-zero vector be perpendicular to both $(2,1,0)$ and $(-1,2,1)$ ? Try $(1,-2,5)$ :

(2,1,0)\cdot(1,-2,5)=2-2=0 \qquad (-1,2,1)\cdot(1,-2,5)=-1-4+5=0

It works. How can we find such a vector systematically, without guessing? The vector product will build it directly from the two given vectors.

You already know

For non-zero vectors, a scalar product of zero tests perpendicularity. The scalar product is a number: multiply matching components and add. Multiplying a non-zero vector by a non-zero scalar preserves its line of direction.

Two non-parallel vectors placed tail to tail determine a flat surface of directions. A vector perpendicular to both points out of that surface. There are two opposite perpendicular directions, and any non-zero length works. A useful surprise: the two vectors can have a very small angle between them, yet the required perpendicular direction still exists. They do not have to form a right angle with each other.

[DIAGRAM: asset_name: One vector perpendicular to two given vectors; asset_slug: h245_4_04g_common_perpendicular; caption: Both scalar products are zero, so n is perpendicular to a and b; description: Three-dimensional coordinate diagram with arrows a=(2,1,0), b=(-1,2,1) and n=(1,-2,5) from the origin, accompanied by the two scalar-product calculations 2-2+0=0 and -1-4+5=0.]

Vector product

The vector product $\mathbf{a}\times\mathbf{b}$ is a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ . For non-parallel non-zero vectors it is non-zero. For parallel vectors it is the zero vector, which has no direction and cannot itself serve as a perpendicular direction vector.

Do not confuse the symbols: $\mathbf{a}\cdot\mathbf{b}$ gives a scalar; $\mathbf{a}\times\mathbf{b}$ gives a vector. For a request to find a common perpendicular, any non-zero scalar multiple of a valid answer is valid. For a request to calculate the vector product, give that product itself.

Key idea

Use the vector product to construct a common perpendicular; use two scalar products to check it.

The distinction between a product and an arbitrary perpendicular matters when an answer is simplified.

Using the component formula

Calculate the three components separately; the order of the subtractions matters, especially in the middle component.

Vector product in components

For $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$ ,

\mathbf{a}\times\mathbf{b} =\begin{pmatrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1 \end{pmatrix}

In the formula booklet.

OCR allows this supplied formula or a calculator when a vector product is required. Reading the component expression carefully is safer than trying to remember a half-familiar sign pattern. The geometric check is available even if you use a calculator.

Why does the formula give a perpendicular? Taking its scalar product with $\mathbf{a}$ gives

a_1a_2b_3-a_1a_3b_2+a_2a_3b_1-a_1a_2b_3+a_1a_3b_2-a_2a_3b_1=0

Every term cancels with its opposite. The same cancellation occurs with $\mathbf{b}$ . Also, reversing the order reverses every subtraction, so $\mathbf{b}\times\mathbf{a}=-(\mathbf{a}\times\mathbf{b})$ . This reversal identity is not in the formula booklet: learn it.

Worked example 1: calculating and simplifying a direction

Question. Calculate $\mathbf{a}\times\mathbf{b}$ for $\mathbf{a}=(1,2,3)$ , $\mathbf{b}=(2,-1,1)$ . Hence give a common perpendicular with the smallest possible integer components.

Plan. Substitute into all three component differences, then divide by the common factor only when giving the direction.

\begin{aligned} \mathbf{a}\times\mathbf{b} &=\begin{pmatrix}2(1)-3(-1)\\3(2)-1(1)\\1(-1)-2(2)\end{pmatrix} && \text{component formula}\\ &=\begin{pmatrix}5\\5\\-5\end{pmatrix} && \text{evaluate}\\ \mathbf{n}&=\begin{pmatrix}1\\1\\-1\end{pmatrix} && \text{divide by }5 \end{aligned}

Why this step? The question asks for two things. The product is fixed, but the common perpendicular may be shortened without changing its direction.

Answer. $\mathbf{a}\times\mathbf{b}=(5,5,-5)$ ; a common perpendicular is $(1,1,-1)$ .

Check. $\mathbf{a}\cdot\mathbf{n}=1+2-3=0$ and $\mathbf{b}\cdot\mathbf{n}=2-1-1=0$ .

Worked example 2 changes one feature: the last component of $\mathbf{b}$ changes sign. Recalculate each component; the new product need not share the old direction.

Worked example 2: a negative component

Question. Calculate $\mathbf{a}\times\mathbf{b}$ , where $\mathbf{a}=(1,2,3)$ , $\mathbf{b}=(2,-1,-1)$ , and verify that it is perpendicular to both vectors.

Plan. Keep negative components in brackets in the supplied formula, then test the answer twice.

\begin{aligned} \mathbf{a}\times\mathbf{b} &=\begin{pmatrix}2(-1)-3(-1)\\3(2)-1(-1)\\1(-1)-2(2)\end{pmatrix} && \text{substitute signs}\\ &=\begin{pmatrix}1\\7\\-5\end{pmatrix} && \text{evaluate}\\ \mathbf{a}\cdot(\mathbf{a}\times\mathbf{b})&=1+14-15=0 && \text{first test}\\ \mathbf{b}\cdot(\mathbf{a}\times\mathbf{b})&=2-7+5=0 && \text{second test} \end{aligned}

Why this step? The middle component is $a_3b_1-a_1b_3$ ; the second product here is negative, so subtracting it adds 1.

Answer. $\mathbf{a}\times\mathbf{b}=(1,7,-5)$ , perpendicular to both given vectors.

Check. Reversing the order gives $(-1,-7,5)$ , the negative of the displayed product.

Checking the two scalar products catches an isolated sign error that a second quick glance at the formula may miss.

Watch out

Wrong: using $a_1b_3-a_3b_1$ for the middle component while keeping the first and third components unchanged. That reverses only one sign. Right: the middle component is $a_3b_1-a_1b_3$ ; copy the supplied formula before substituting.

If the product is zero for two non-zero vectors, their directions are parallel. There are still common perpendiculars, but the zero vector does not select one. For example, $(1,2,0)\times(2,4,0)=\mathbf{0}$ , while $(0,0,1)$ is a valid common perpendicular. Use the scalar-product condition to choose such a direction in this special case.

A further perpendicularity condition

When a question says that $\mathbf{a}\times\mathbf{b}$ is perpendicular to a third vector, calculate the vector product first, then set its scalar product with that third vector equal to zero.

This is a chain of two familiar operations, with different outputs: the first makes a vector; the second makes an equation. Check that the vector product is non-zero for any parameter values you find.

Worked example 3: determining one parameter

Question. Let $\mathbf{a}=(1,t,2)$ , $\mathbf{b}=(2,1,1)$ , $\mathbf{c}=(1,2,1)$ . Determine $t$ , given that $\mathbf{a}\times\mathbf{b}$ is perpendicular to $\mathbf{c}$ .

Plan. Calculate the vector product in terms of $t$ , then apply the scalar-product test.

\begin{aligned} \mathbf{a}\times\mathbf{b} &=\begin{pmatrix}t-2\\3\\1-2t\end{pmatrix} && \text{component formula}\\ (\mathbf{a}\times\mathbf{b})\cdot\mathbf{c} &=(t-2)+6+(1-2t) && \text{matching components}\\ 5-t&=0 && \text{perpendicularity}\\ t&=5 && \text{solve} \end{aligned}

Why this step? Taking $\mathbf{a}\cdot\mathbf{b}$ would test a different pair of vectors. The statement names the whole vector $\mathbf{a}\times\mathbf{b}$ .

Answer. $t=5$ .

Check. The product is $(3,3,-9)$ , which is non-zero, and $3+6-9=0$ .

Worked example 4 changes one feature: the last component of $\mathbf{c}$ also becomes $t$ . The final scalar product now contains a quadratic term.

Worked example 4: a quadratic parameter condition

Question. Let $\mathbf{a}=(1,t,2)$ , $\mathbf{b}=(2,1,1)$ , $\mathbf{c}=(1,2,t)$ . Determine the possible values of $t$ , given that $\mathbf{a}\times\mathbf{b}$ is perpendicular to $\mathbf{c}$ .

Plan. Reuse the product expression and multiply its last component by $t$ in the scalar product.

\begin{aligned} \mathbf{a}\times\mathbf{b} &=(t-2,3,1-2t) && \text{same first two vectors}\\ (t-2)+6+t(1-2t)&=0 && \text{perpendicularity}\\ -2t^2+2t+4&=0 && \text{expand}\\ (t-2)(t+1)&=0 && \text{divide by }-2\\ t&=2\text{ or }t=-1 && \text{solve} \end{aligned}

Why this step? The term $t(1-2t)$ comes from the third pair of components. It must be expanded before the equation can be factorised.

Answer. $t=2$ or $t=-1$ .

Check. For $t=2$ , the product is $(0,3,-3)$ and the scalar product with $(1,2,2)$ is $6-6=0$ . For $t=-1$ , the product is $(-3,3,3)$ and its scalar product with $(1,2,-1)$ is $-3+6-3=0$ . Both products are non-zero.

This is also a useful way to organise longer OCR questions: keep the product as a single named vector until the second operation is required.

Watch out

Wrong: setting each component of $\mathbf{a}\times\mathbf{b}$ equal to zero because it is perpendicular to $\mathbf{c}$ . That would make the vector product zero. Right: set the sum of matching products with $\mathbf{c}$ equal to zero.

The opening puzzle now has a systematic answer. The component formula gives

(2,1,0)\times(-1,2,1)=(1,-2,5)

and the two zero scalar products prove it is perpendicular to both. No guessing, and no angle calculation.

Quick Recap

The vector product gives a vector perpendicular to both inputs; the scalar product gives a number.
Use the supplied component formula and preserve the middle component's subtraction order.
Any non-zero multiple of a common perpendicular is valid, but the vector product itself has a fixed value.
For an additional perpendicularity condition, take the scalar product of the calculated vector with the third vector and set it equal to zero.

The opening common perpendicular is $(1,-2,5)$ , found directly as the vector product and checked by two scalar products.

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4.04g - Finding a common perpendicular

A direction perpendicular to both

Using the component formula

A further perpendicularity condition

Quick Recap

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