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I am reading Understanding Pure Mathematics by A.J. Sadler & D.W.S. Thorning and in Chapter 2 (Vectors) one of the solved examples has got me stuck. Either I need to sleep or there is a typo somewhere in the text.

Please help me understand. Here is the text on page 66 of the textbook:

EXAMPLE 21

Find the vector equation of the straight line that passes through the point with position vector 2i + 3j and which is perpendicular to the line r = 3i + 2j + $\lambda$(i - 2j).

The given line is parallel to the vector i - 2j. If the required line is parallel to a vector ai + bj, it follows that

(ai + bj) $\cdot$ (i - 2j) = 0 [for the two lines to be perpendicular]

so, a - 2b = 0

hence, a = 2b

The required line will therefore be parallel to any vector of the form b(2i + j).

Taking 2i + j as one such vector, the required vector equation will be:

r = 2i + 3j + $\lambda$(2i + j)

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    Could you please take the time to explain? – 2017-01-29
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    I thought It's your solution, sorry. OK. of course. – 2017-01-29
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    What part you are stuck – 2017-01-29
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    Which line is perpendicular to which and parallel to which. And how is "The required line will therefore be parallel to any vector of the form b(2i + j)"? The author hasn't given a diagram for this solved example, so I don't know which line(s) or point(s) he is taking about. – 2017-01-29
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    you know the in line $r=a+\lambda\vec{v}$, the $\vec{v}$ is line vector. the line span with this vector. – 2017-01-29
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    In this line $\vec{v}=i-2j$, this not it.? – 2017-01-29
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    Every vector which is parallel with $\vec{v}$ has the form $k\vec{v}$ here $k(i-2j)$. right – 2017-01-29
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    Yes, every vector parallel to $\vec{v}$ has the form $k\vec{v}$. And the vector perpendicular (say $\vec{p}$) to $\vec{v}$ will have a dot product with it as 0. That is, $\vec{p} \cdot \vec{v}$ = 0 – 2017-01-29
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    Well, let $ai+bj$ is the required line vector, which is **perpendicular** to the given line so $(ai + bj).(i-2j)=0$, with **dot** product. – 2017-01-29
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    And how do we get the following statements?: The required line will therefore be parallel to any vector of the form b(2i + j). Taking 2i + j as one such vector, the required vector equation will be: r = 2i + 3j + λλ(2i + j) – 2017-01-29
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    We said every line has the form $r=a+\lambda\vec{v}$, the $\vec{v}$ is the line vetor and $a$ is a point on line (not different which that), the $\lambda$ is a parameter for spanning whole line. – 2017-01-29
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    Now, The required line vector is $\vec{v}=b(2i + j)$, it's no metter we take $2i + j$ or $-5(2i+j)$ or $1000(2i+j)$. The point $a=2i+3j$ has given in example statement. – 2017-01-29
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    AHHHHHHHH! We have to substitute a = 2b, and b becomes the scalar quantity that multiplied with (2i +j) to show that it is parallel. – 2017-01-29
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    Thank you so much, @MyGlasses. You are quite the professor. – 2017-01-29
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    You're welcome. – 2017-01-29

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