Question

the magnitude and direction of two vectors are shown in the diagram. what is the magnitude of their sum? 6, 2√5, 8, 20

Explanation:

Step1: Resolve vectors into components

For the vector with magnitude 2 and angle \(135^\circ\) from the positive x - axis:
The x - component \(v_{1x}=2\cos135^\circ\), and \(\cos135^\circ =-\frac{\sqrt{2}}{2}\), so \(v_{1x}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}\)
The y - component \(v_{1y}=2\sin135^\circ\), and \(\sin135^\circ=\frac{\sqrt{2}}{2}\), so \(v_{1y}=2\times\frac{\sqrt{2}}{2}=\sqrt{2}\)

For the vector with magnitude 4 and angle \(45^\circ\) from the positive x - axis:
The x - component \(v_{2x}=4\cos45^\circ\), and \(\cos45^\circ=\frac{\sqrt{2}}{2}\), so \(v_{2x}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\)
The y - component \(v_{2y}=4\sin45^\circ\), and \(\sin45^\circ=\frac{\sqrt{2}}{2}\), so \(v_{2y}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\)

Step2: Find the resultant components

The resultant x - component \(V_x = v_{1x}+v_{2x}=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\)
The resultant y - component \(V_y = v_{1y}+v_{2y}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)

Wait, there is a mistake above. Let's re - calculate the angle for the first vector. The angle of the first vector with respect to the negative x - axis is \(135^\circ - 90^\circ = 45^\circ\), so the angle with the positive x - axis is \(180^\circ - 45^\circ=135^\circ\), but maybe a better way is to use the angle between the two vectors. The angle between the two vectors: the first vector makes \(135^\circ\) with the positive x - axis, the second makes \(45^\circ\) with the positive x - axis, so the angle between them is \(135^\circ - 45^\circ = 90^\circ\)? Wait, no. Wait, the first vector: if we consider the angle between the two vectors, let's see the direction. The first vector (magnitude 2) is in the second quadrant, making \(135^\circ\) with the positive x - axis, the second (magnitude 4) is in the first quadrant, making \(45^\circ\) with the positive x - axis. The angle between the two vectors is \(135^\circ - 45^\circ=90^\circ\)? Wait, no, \(180 - 135 + 45=90^\circ\). So the two vectors are perpendicular? Wait, no, let's recalculate the components correctly.

Wait, the first vector: angle with positive x - axis is \(135^\circ\), so \(v_{1x}=2\cos135^\circ = 2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}\), \(v_{1y}=2\sin135^\circ=2\times\frac{\sqrt{2}}{2}=\sqrt{2}\)

Second vector: angle with positive x - axis is \(45^\circ\), so \(v_{2x}=4\cos45^\circ = 4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\), \(v_{2y}=4\sin45^\circ=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}\)

Now, resultant x - component: \(V_x=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\)

Resultant y - component: \(V_y=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)

Wait, that can't be right. Wait, maybe the angle of the first vector is measured from the negative y - axis? No, the diagram shows the first vector (magnitude 2) has an angle of \(135^\circ\) with respect to the second vector? Wait, no, the diagram has a \(135^\circ\) angle between the two vectors? Wait, the angle between the two vectors: the first vector (magnitude 2) and the second (magnitude 4). Let's look at the angles with the x - axis. The first vector: the angle between it and the negative x - axis is \(45^\circ\) (since \(135^\circ\) from the positive x - axis, so \(180 - 135 = 45^\circ\) from negative x - axis), and the second vector is \(45^\circ\) from positive x - axis. So the angle between the two vectors is \(45^\circ+45^\circ = 90^\circ\)? Wait, no, if one is \(135^\circ\) from positive x - axis and the other is \(45^\circ\) from positive x - axis, the angle between them is \(135 - 45=90^\circ\). So the two vectors are perpendicular? Wait, no, \(135 - 45 = 90\), so the angle between them is \(90^\circ\). Wait, but if that's the case, we can use the Pythagorean theorem for the resultant? Wait, no, because we need to check the components again.

Wait, maybe I made a mistake in the angle of the first vector. Let's re - express the first vector: the vector with magnitude 2, the angle between it and the positive y - axis is \(45^\circ\) (since \(135^\circ\) from positive x - axis, so \(135 - 90 = 45^\circ\) from positive y - axis towards negative x - axis). So its x - component is \(-2\sin45^\circ=-\sqrt{2}\), y - component is \(2\cos45^\circ=\sqrt{2}\), which is the same as before.

The second vector: magnitude 4, angle \(45^\circ\) from positive x - axis, so x - component \(4\cos45^\circ = 2\sqrt{2}\), y - component \(4\sin45^\circ = 2\sqrt{2}\)

Now, resultant x: \(-\sqrt{2}+2\sqrt{2}=\sqrt{2}\)

Resultant y: \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)

Then the magnitude of the resultant is \(\sqrt{V_x^2 + V_y^2}=\sqrt{(\sqrt{2})^2+(3\sqrt{2})^2}=\sqrt{2 + 18}=\sqrt{20}=2\sqrt{5}\). Wait, that's one of the options.

Wait, let's do it again. Let's find the angle between the two vectors. The first vector has an angle of \(135^\circ\) with the positive x - axis, the second has \(45^\circ\) with the positive x - axis. So the angle between them is \(135 - 45 = 90^\circ\)? No, \(180 - 135 + 45=90^\circ\), so the angle between the two vectors is \(90^\circ\)? Wait, no, if vector A is at \(135^\circ\) and vector B is at \(45^\circ\), the angle between them is \(135 - 45 = 90^\circ\). So the two vectors are perpendicular? Wait, no, \(135 - 45 = 90\), so the angle between them is \(90^\circ\). Then the magnitude of the resultant should be \(\sqrt{2^2 + 4^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\). Ah! That's a simpler way. Because if the angle between two vectors \(\vec{A}\) and \(\vec{B}\) is \(\theta\), the magnitude of the resultant \(\vec{R}=\vec{A}+\vec{B}\) is given by \(R=\sqrt{A^2 + B^2+2AB\cos\theta}\). Here, \(A = 2\), \(B = 4\), and \(\theta = 90^\circ\) (since the angle between them is \(90^\circ\), because \(135^\circ - 45^\circ=90^\circ\)). So \(\cos\theta=\cos90^\circ = 0\). Then \(R=\sqrt{2^2+4^2 + 2\times2\times4\times0}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\).

Step1: Determine the angle between vectors

The first vector makes \(135^\circ\) with the positive x - axis, the second makes \(45^\circ\) with the positive x - axis. The angle between them \(\theta=135^\circ - 45^\circ = 90^\circ\).

Step2: Use the law of cosines for vector addition

The formula for the magnitude of the sum of two vectors \(\vec{A}\) and \(\vec{B}\) is \(|\vec{A}+\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}\), where \(|\vec{A}| = 2\), \(|\vec{B}| = 4\), and \(\theta = 90^\circ\) (so \(\cos\theta = 0\)).

Substitute the values:
\[ \begin{align*} |\vec{A}+\vec{B}|&=\sqrt{2^2 + 4^2+2\times2\times4\times\cos90^\circ}\\ &=\sqrt{4 + 16+0}\\ &=\sqrt{20}\\ &=2\sqrt{5} \end{align*} \]

Answer:

\(2\sqrt{5}\)