Question

Vector A+vector B+vector C=0 and the magnitude if A is equal to magnitude of B and magnitude of c is equal to √2 times of magnitude A then the angle between vectors?

Answer 1

90°, 135°, 135°

Answer 2

Let the magnitudes of the vectors be $A = |\vec{A}|$, $B = |\vec{B}|$, and $C = |\vec{C}|$.

Given:

1. $\vec{A} + \vec{B} + \vec{C} = 0$
2. $|\vec{A}| = |\vec{B}|$
3. $|\vec{C}| = \sqrt{2} |\vec{A}|$

Let $|\vec{A}| = k$. Then $|\vec{B}| = k$ and $|\vec{C}| = \sqrt{2} k$.

We need to find the angles between pairs of vectors. We can use the dot product relation for each pair.

1. Angle between $\vec{A}$ and $\vec{B}$ ($\theta_{AB}$):

From the given condition, $\vec{A} + \vec{B} = -\vec{C}$.

Squaring both sides (taking the dot product of each side with itself):

$(\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = (-\vec{C}) \cdot (-\vec{C})$

$|\vec{A}|^2 + |\vec{B}|^2 + 2 \vec{A} \cdot \vec{B} = |\vec{C}|^2$

Using the definition of the dot product, $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta_{AB}$:

$|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta_{AB} = |\vec{C}|^2$

Substitute the magnitudes:

$k^2 + k^2 + 2 (k)(k) \cos \theta_{AB} = (\sqrt{2} k)^2$

$2k^2 + 2k^2 \cos \theta_{AB} = 2k^2$

$2k^2 \cos \theta_{AB} = 0$

Since $k \neq 0$, we have $\cos \theta_{AB} = 0$.

Therefore, $\theta_{AB} = 90^\circ$ or $\frac{\pi}{2}$ radians.

2. Angle between $\vec{B}$ and $\vec{C}$ ($\theta_{BC}$):

From the given condition, $\vec{B} + \vec{C} = -\vec{A}$.

Squaring both sides:

$(\vec{B} + \vec{C}) \cdot (\vec{B} + \vec{C}) = (-\vec{A}) \cdot (-\vec{A})$

$|\vec{B}|^2 + |\vec{C}|^2 + 2 \vec{B} \cdot \vec{C} = |\vec{A}|^2$

Using $\vec{B} \cdot \vec{C} = |\vec{B}| |\vec{C}| \cos \theta_{BC}$:

$|\vec{B}|^2 + |\vec{C}|^2 + 2 |\vec{B}| |\vec{C}| \cos \theta_{BC} = |\vec{A}|^2$

Substitute the magnitudes:

$k^2 + (\sqrt{2} k)^2 + 2 (k)(\sqrt{2} k) \cos \theta_{BC} = k^2$

$k^2 + 2k^2 + 2\sqrt{2} k^2 \cos \theta_{BC} = k^2$

$3k^2 + 2\sqrt{2} k^2 \cos \theta_{BC} = k^2$

$2\sqrt{2} k^2 \cos \theta_{BC} = k^2 - 3k^2$

$2\sqrt{2} k^2 \cos \theta_{BC} = -2k^2$

$\sqrt{2} \cos \theta_{BC} = -1$

$\cos \theta_{BC} = -\frac{1}{\sqrt{2}}$

Therefore, $\theta_{BC} = 135^\circ$ or $\frac{3\pi}{4}$ radians.

3. Angle between $\vec{A}$ and $\vec{C}$ ($\theta_{AC}$):

From the given condition, $\vec{A} + \vec{C} = -\vec{B}$.

Squaring both sides:

$(\vec{A} + \vec{C}) \cdot (\vec{A} + \vec{C}) = (-\vec{B}) \cdot (-\vec{B})$

$|\vec{A}|^2 + |\vec{C}|^2 + 2 \vec{A} \cdot \vec{C} = |\vec{B}|^2$

Using $\vec{A} \cdot \vec{C} = |\vec{A}| |\vec{C}| \cos \theta_{AC}$:

$|\vec{A}|^2 + |\vec{C}|^2 + 2 |\vec{A}| |\vec{C}| \cos \theta_{AC} = |\vec{B}|^2$

Substitute the magnitudes:

$k^2 + (\sqrt{2} k)^2 + 2 (k)(\sqrt{2} k) \cos \theta_{AC} = k^2$

$k^2 + 2k^2 + 2\sqrt{2} k^2 \cos \theta_{AC} = k^2$

$3k^2 + 2\sqrt{2} k^2 \cos \theta_{AC} = k^2$

$2\sqrt{2} k^2 \cos \theta_{AC} = k^2 - 3k^2$

$2\sqrt{2} k^2 \cos \theta_{AC} = -2k^2$

$\sqrt{2} \cos \theta_{AC} = -1$

$\cos \theta_{AC} = -\frac{1}{\sqrt{2}}$

Therefore, $\theta_{AC} = 135^\circ$ or $\frac{3\pi}{4}$ radians.

The angles between the vectors are $90^\circ$, $135^\circ$, and $135^\circ$.