Question

the magnitude and direction of two vectors are shown in the diagram. what is the magnitude of their sum? 8 6 $2sqrt{5}$ 20

Explanation:

Step1: Find x and y components of first vector (magnitude 2, angle 135° from x-axis)

The x - component \( V_{1x}=2\cos(135^{\circ}) \), and \( \cos(135^{\circ}) = -\frac{\sqrt{2}}{2} \), so \( V_{1x}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2} \).
The y - component \( V_{1y}=2\sin(135^{\circ}) \), and \( \sin(135^{\circ})=\frac{\sqrt{2}}{2} \), so \( V_{1y}=2\times\frac{\sqrt{2}}{2}=\sqrt{2} \).

Step2: Find x and y components of second vector (magnitude 4, angle 45° from x - axis)

The x - component \( V_{2x}=4\cos(45^{\circ}) \), and \( \cos(45^{\circ})=\frac{\sqrt{2}}{2} \), so \( V_{2x}=4\times\frac{\sqrt{2}}{2} = 2\sqrt{2} \).
The y - component \( V_{2y}=4\sin(45^{\circ}) \), and \( \sin(45^{\circ})=\frac{\sqrt{2}}{2} \), so \( V_{2y}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2} \).

Step3: Find the sum of x - components (\( V_x \)) and y - components (\( V_y \))

\( V_x=V_{1x}+V_{2x}=-\sqrt{2}+2\sqrt{2}=\sqrt{2} \)
\( V_y=V_{1y}+V_{2y}=\sqrt{2}+2\sqrt{2}=3\sqrt{2} \)? Wait, no, wait. Wait, maybe a better way: Wait, actually, the angle between the two vectors: the first vector is at 135° from positive x - axis, the second at 45°, so the angle between them is \( 135 - 45=90^{\circ} \)? Wait, no, 135° from x - axis (second quadrant) and 45° from x - axis (first quadrant), so the angle between the two vectors is \( 135 - 45 = 90^{\circ} \)? Wait, no, 180 - 135=45° above negative x - axis, and 45° above positive x - axis, so the angle between the two vectors is \( 45 + 45=90^{\circ} \). Wait, so the two vectors are perpendicular? Wait, magnitude 2 and 4, perpendicular. Then the magnitude of the sum is \( \sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5} \)? Wait, no, wait, if they are perpendicular, then the magnitude of the resultant is \( \sqrt{A^{2}+B^{2}} \). Wait, but maybe my component calculation was wrong. Let's recalculate components:

First vector (magnitude 2, angle 135°):
\( \cos(135^{\circ})=\cos(180 - 45)=-\cos(45)=-\frac{\sqrt{2}}{2} \), so \( V_{1x}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2} \)
\( \sin(135^{\circ})=\sin(180 - 45)=\sin(45)=\frac{\sqrt{2}}{2} \), so \( V_{1y}=2\times\frac{\sqrt{2}}{2}=\sqrt{2} \)

Second vector (magnitude 4, angle 45°):
\( \cos(45^{\circ})=\frac{\sqrt{2}}{2} \), so \( V_{2x}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2} \)
\( \sin(45^{\circ})=\frac{\sqrt{2}}{2} \), so \( V_{2y}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2} \)

Now, sum of x - components: \( V_x=-\sqrt{2}+2\sqrt{2}=\sqrt{2} \)
Sum of y - components: \( V_y=\sqrt{2}+2\sqrt{2}=3\sqrt{2} \)? Wait, that can't be, because if the angle between them is 90°, the y - components should add? Wait, no, the first vector is in the second quadrant (x negative, y positive), the second in the first (x positive, y positive). So the angle between the two vectors: the angle of the first vector with positive x - axis is 135°, the second is 45°, so the angle between them is \( 135 - 45 = 90^{\circ} \). So they are perpendicular. Then why is the y - component sum \( 3\sqrt{2} \) and x - component sum \( \sqrt{2} \)? Wait, no, I think I made a mistake in the angle. Wait, 135° from positive x - axis: so the angle with negative x - axis is 45°, and the second vector is 45° from positive x - axis. So the angle between the two vectors is \( 45 + 45=90^{\circ} \), so they are perpendicular. Then the magnitude of the resultant should be \( \sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5} \). Wait, but when we calculate components:

Wait, first vector: magnitude 2, angle 135°:
x - component: \( 2\cos(135^{\circ})=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2} \)
y - component: \( 2\sin(135^{\circ})=2\times\frac{\sqrt{2}}{2}=\sqrt{2} \)

Second vector: magnitude 4, angle 45°:
x - component: \( 4\cos(45^{\circ})=4\times\frac{\sqrt{2}}{2}=2\sqrt{2} \)
y - component: \( 4\sin(45^{\circ})=4\times\frac{\sqrt{2}}{2}=2\sqrt{2} \)

Now, sum of x - components: \( -\sqrt{2}+2\sqrt{2}=\sqrt{2} \)
Sum of y - components: \( \sqrt{2}+2\sqrt{2}=3\sqrt{2} \)

Now, magnitude of resultant: \( \sqrt{(\sqrt{2})^{2}+(3\sqrt{2})^{2}}=\sqrt{2 + 18}=\sqrt{20}=2\sqrt{5} \). Ah, there we go. So that's correct.

Answer:

\( 2\sqrt{5} \)