I think my coaching taught us some out of syllabus content for vectors- they said that for completeness of this topic, learning the cross product is necessary. I'd still love to know how it would be done though, but I'm not too fussed about it since it won't be tested

Also, could someone explain this proofs question: **Q10d** from the cambridge textbook? I've forgotten how roots work- am I meant to use the sum and product of roots?

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thanks so much

Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:

1. Take one point on each line and find a direction vector through these points. So for example, (1,3,1) to (0,2,1). This vector is then a vector going from one of the lines, to the other. (Doesn't matter which line it goes from and to, since ultimately we'll be considering a distance, and hence ignore the direction.) Call this vector \( \vec{a}\).

2. Find the direction vectors of each of the lines.

3. Take the cross product of the vectors in step 2 to find a vector \( \vec{n}\) perpendicular to

*both* of the lines.

4. The projection of the vector joining the two lines \(\vec{a}\), onto the normal vector \(\vec{n}\), that is the vector

\[ \operatorname{proj}_{\vec{n}}\vec{a}, \]

will have the shortest distance between the two lines.

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Within the scope of the new syllabus, sums and products of roots would've been the first thing I thought of. You should find that the equivalence is indeed true, but I only prove the reverse implication. The forward implication is left as your exercise.

\[ \text{Let }\frac1\alpha\text{ and }\frac1\beta\text{ be the roots of}\\ cx^2 + bx + a. \]

\[ \text{Then it follows from the sums and products of roots that}\\ \begin{align*} \frac{\alpha+\beta}{\alpha\beta} = \frac1\alpha+\frac1\beta &= -\frac{b}{c},\\ \frac{1}{\alpha\beta} &= \frac{a}{c} \end{align*} \]

\[ \text{The second equation immediately rearranges to }\alpha\beta = \frac{c}{a}.\\ \text{Subbing into the first equation,}\\ \frac{\alpha+\beta}{\frac{c}{a}} = -\frac{b}{c} \implies \alpha+\beta = -\frac{b}{a}. \]

\[ \text{Thus a quadratic equation with roots }\alpha\text{ and }\beta\text{ is}\\ \begin{align*}(x-\alpha)(x-\beta) &= 0\\

\implies x^2-(\alpha+\beta)x + \alpha\beta &= 0\\

\implies x^2 + \frac{b}{a}x + \frac{c}{a} &= 0\\

\implies ax^2+bx+c &= 0 \end{align*}\\ \text{as required.} \]