Question

1. write the vector v in the form ai + bj, given its magnitude ||v|| and the angle α it makes with the positive x - axis. ||v|| = 15, α = 315°

Explanation:

Step1: Calculate the \(x\) - component (\(a\))

The formula for the \(x\) - component of a vector \(\vec{v}\) is \(a = \|\vec{v}\|\cos\alpha\).
Given \(\|\vec{v}\| = 15\) and \(\alpha=315^{\circ}\), and \(\cos(315^{\circ})=\cos(360^{\circ} - 45^{\circ})=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\) (since \(\cos(A - B)=\cos A\cos B+\sin A\sin B\) and \(\cos(360^{\circ}-\theta)=\cos\theta\)).
So, \(a = 15\times\cos(315^{\circ})=15\times\frac{\sqrt{2}}{2}=\frac{15\sqrt{2}}{2}\)

Step2: Calculate the \(y\) - component (\(b\))

The formula for the \(y\) - component of a vector \(\vec{v}\) is \(b=\|\vec{v}\|\sin\alpha\).
Given \(\|\vec{v}\| = 15\) and \(\alpha = 315^{\circ}\), and \(\sin(315^{\circ})=\sin(360^{\circ}-45^{\circ})=-\sin(45^{\circ})=-\frac{\sqrt{2}}{2}\) (since \(\sin(A - B)=\sin A\cos B-\cos A\sin B\) and \(\sin(360^{\circ}-\theta)=-\sin\theta\)).
So, \(b = 15\times\sin(315^{\circ})=15\times(-\frac{\sqrt{2}}{2})=-\frac{15\sqrt{2}}{2}\)

Answer:

The vector \(\vec{v}\) in the form \(a\vec{i}+b\vec{j}\) is \(\frac{15\sqrt{2}}{2}\vec{i}-\frac{15\sqrt{2}}{2}\vec{j}\)