Hi!

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Is my working out right for this? it says answers may vary, but how did they get the answer below?:

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Also, I'm really struggling with these questions:

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Thanks!

Your working makes perfect sense to me. But just take note that \( \frac{\lambda}{\frac12 \lambda - \frac32} \) reads like a fraction: "lambda over (half lambda - 3 halves)". Try to keep to vector notation: \( \begin{pmatrix} \lambda\\ \frac12 \lambda - \frac32\end{pmatrix} \).

\[ \text{Their working out looks as though they parametrised }y\text{ instead.}\\ \text{That is, they set }y=\lambda.\quad (\lambda \in \mathbb{R}) \]

\[ \text{Then }x=3+2y\text{, so }x=3+2\lambda.\\ \text{Continue with the same approach to get to their answer.} \]

In general, it doesn't matter which variable you choose to parametrise.

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You're expected to know and understand that in the vector equation \( \vec{u} = \begin{pmatrix}a\\b\end{pmatrix} + \lambda \begin{pmatrix}x\\y\end{pmatrix} \) that \( \begin{pmatrix}a\\b\end{pmatrix} \)

**is** the position vector of a point on the line. With that in mind, for both questions 3 and 4 you should be able to figure out the \( \begin{pmatrix}a\\b\end{pmatrix} \) vector.

As for the direction vector \( \begin{pmatrix}x\\y\end{pmatrix} \):

- For Q3, you've already begun to use \(m=\tan\theta\). In general, if \(m\) is the gradient of the line, then a direction vector for the line is \( \begin{pmatrix}1\\m\end{pmatrix}\). Think about why! (Or if you can't think about why intuitively, prove it by finding the vector equation of the line itself.)

- For Q4, note that the angle is not made with respect to the positive \(x\)-axis this time. Rather, it's perpendicular to some random line, in particular the line \(x+3y+1=0\). Using only vector techniques, you would probably want to convert this line into its vector equation form.

Once you've done this, apply the scalar product formula \( \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \), where \(\theta = 45^\circ\), to find the possible direction vectors for the lines you seek. Set \(\vec{a}\) to be the direction vector of \(x+3y+1=0\), and in the same spirit as Q3, set \(\vec{b} = \begin{pmatrix}1\\m\end{pmatrix} \).

(The syllabus is still new, so I'm not aware of any concrete rules of thumb in doing these problems. I'm relying on what makes sense based off what I know, and what the syllabus allows for.)

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The last problem is the shortest distance between a point in a line problem, but necessitating vector methods. When vectors are required for shortest distances, always think

**projections**.

Suppose that you want the shortest distance from a point \( P \) to a line. Let \( \vec{p}\) be the position vector of the point. Furthermore, suppose that the equation of the line is \( \vec{u}=\vec{a}+\lambda\vec{b}\).

1. Then the vector \( \vec{a} - \vec{p}\) is a vector that joins the point \(P\), to some random point on the line.

2. Now, \( \operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) \) represents the projection of the vector you found above, onto this

**line**.

3. Therefore, \( \operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) - \vec{p}\) is another vector joining \(P\), to a point on the line. However, this vector is special, in that it is

**perpendicular** to the line.

4. The shortest distance between a point and a line is always the perpendicular distance between them. Hence, the answer you seek can be found by taking the magnitude of this vector, i.e. \(|\operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) - \vec{p}|\).

Note: All of the above is hard to visualis in your head. You should use a diagram when going through all of that.

In your case, \( \vec{p}= \begin{pmatrix}2\\3\end{pmatrix}\). You have to convert your line into its vector equation to write down your \(\vec{a}\) and \(\vec{b}\).