Question

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two displacement vectors, \\(\vec{a}\\) and \\(\vec{b}\\), are shown in the figure. the magnitudes of the displacements are a = 10.0 m and b = 5.00 m.
what is the x-component of vector \\(\vec{a}\\)?
hint: review section 1.8 in textbook (component method of vector addition), example 9 page 16 in textbook, and example 1-2 from our lecture video.
figure: coordinate system with +x right, +y up, -x left, -y down. vector \\(\vec{a}\\) is at 42.0° from +y towards -x. vector \\(\vec{b}\\) is at 31.0° from -x towards -y.
options: -7.40 m, -8.24 m, -6.69 m, -5.25 m

Explanation:

Step1: Determine the angle for x - component

Vector \(\vec{A}\) makes an angle of \(42.0^{\circ}\) with the \(+y\) axis towards the \(-x\) direction. So the angle with the \(-x\) axis (or the angle we use to find the x - component) is \(90^{\circ}- 42.0^{\circ}=48.0^{\circ}\)? Wait, no. Wait, the x - component of a vector \(\vec{A}\) with magnitude \(A\) and angle \(\theta\) from the \(y\) - axis towards the negative \(x\) direction: the x - component \(A_x=-A\sin(42.0^{\circ})\)? Wait, let's think again. The standard way: if we have a vector, the x - component is \(A\cos(\theta)\) where \(\theta\) is the angle with the \(x\) - axis. But here, the vector \(\vec{A}\) is at an angle of \(42.0^{\circ}\) above the \(-x\) axis? Wait, no, looking at the diagram: the angle between \(\vec{A}\) and \(+y\) is \(42.0^{\circ}\), so the angle between \(\vec{A}\) and \(-x\) is \(90 - 42=48^{\circ}\)? No, wait, let's draw mentally: \(+y\) to \(\vec{A}\) is \(42^{\circ}\), so \(\vec{A}\) is in the second quadrant (since it's between \(+y\) and \(-x\)). So the angle with the \(-x\) axis (the angle between \(\vec{A}\) and the negative \(x\) - axis) is \(90 - 42 = 48^{\circ}\)? No, actually, the x - component of a vector in the second quadrant (with angle \(\alpha\) from the \(y\) - axis towards the negative \(x\)): the x - component is \(-A\sin(\alpha)\), because the adjacent side to the angle with the \(y\) - axis is the y - component, and the opposite side is the x - component (but in the negative x direction). So \(\alpha = 42.0^{\circ}\), so \(A_x=-A\sin(42.0^{\circ})\).

Step2: Calculate the x - component

Given \(A = 10.0\space m\) and \(\alpha=42.0^{\circ}\). So \(A_x=-10.0\times\sin(42.0^{\circ})\). We know that \(\sin(42.0^{\circ})\approx0.6691\). So \(A_x=-10.0\times0.6691=- 6.691\space m\approx - 6.69\space m\). Wait, let's check the formula again. Alternatively, if we consider the angle with the \(x\) - axis: the vector \(\vec{A}\) is at an angle of \(90 + 42=132^{\circ}\) from the positive \(x\) - axis (since from positive \(x\), go 90 degrees to positive \(y\), then 42 degrees towards negative \(x\), so total \(90 + 42 = 132^{\circ}\) from positive \(x\)). Then the x - component is \(A\cos(132^{\circ})\). \(\cos(132^{\circ})=\cos(180 - 48^{\circ})=-\cos(48^{\circ})\approx - 0.6691\). And \(A\cos(132^{\circ})=10.0\times(- 0.6691)=-6.69\space m\). Yes, that's correct. So the x - component of \(\vec{A}\) is \(A_x = A\cos(\theta)\), where \(\theta = 132^{\circ}\) (angle from positive x - axis), or equivalently, since it's \(42^{\circ}\) from positive y - axis towards negative x - axis, \(A_x=-A\sin(42^{\circ})\). Calculating \(10.0\times\sin(42^{\circ})\): \(\sin(42^{\circ})\approx0.6691\), so \(10\times0.6691 = 6.691\), and with the negative sign (because it's in the negative x direction), we get \(- 6.69\space m\).

Answer:

\(-6.69\space m\)