Answer: \(t=8\), and the meeting point has position vector \(195\mathbf{i}+78\mathbf{j}\).

Answer: \(t=8\), and the meeting point has position vector \(195\mathbf{i}+78\mathbf{j}\).

The direction vector for particle \(A\) is \(12\mathbf{i}+5\mathbf{j}\), whose magnitude is

\(\sqrt{12^2+5^2}=13.\)

So the unit direction vector is

\(\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}.\)

Since the speed of \(A\) is \(26\), its velocity is

\(26\left(\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}\right)=24\mathbf{i}+10\mathbf{j}.\)

Thus the position of \(A\) after \(t\) seconds is

\((3+24t)\mathbf{i}+(-2+10t)\mathbf{j}.\)

For particle \(B\), \(\tan\alpha=\frac34\), so the direction ratios are \(4:3\), giving unit direction

\(\frac45\mathbf{i}+\frac35\mathbf{j}.\)

Since the speed of \(B\) is \(20\), its velocity is

\(20\left(\frac45\mathbf{i}+\frac35\mathbf{j}\right)=16\mathbf{i}+12\mathbf{j}.\)

Thus the position of \(B\) after \(t\) seconds is

\((67+16t)\mathbf{i}+(-18+12t)\mathbf{j}.\)

At the meeting point, the \(\mathbf{i}\)-components are equal:

\(3+24t=67+16t.\)

So

\(8t=64,\qquad t=8.\)

Using particle \(A\), the position vector at this time is

\((3+24(8))\mathbf{i}+(-2+10(8))\mathbf{j}=195\mathbf{i}+78\mathbf{j}.\)