If a, b, c are vectors show that a+b+c=0 and |a|=7, |b|=5, |c|=3 then angle between vector b and c i

Q: Wait, why is the angle between $\vec{b}$ and $\vec{c}$ not the interior angle of the triangle?

This is a classic trap! The angle between two vectors is strictly defined when they are placed tail-to-tail. In a closed triangle, vectors are arranged head-to-tail. If you look at the vertex where $\vec{b}$ and $\vec{c}$ meet, the interior angle is $180^\circ - \theta$. By extending the line of $\vec{b}$, you reveal the true angle $\theta$ between the vectors, which is the exterior angle.

Q: Why do we square the vector equation? Can't we just use the Law of Cosines?

You absolutely can use the Law of Cosines! In fact, the Law of Cosines is derived from this exact vector operation. Squaring the vector equation $(\vec{b} + \vec{c}) = -\vec{a}$ is the fundamental, 'first-principles' way to prove the Law of Cosines. It is a robust method that works even when you don't remember the specific triangle formula.

Q: Why do we get a negative sign when we move $\vec{a}$ to the other side?

We are simply using the rules of vector algebra. If $\vec{a} + \vec{b} + \vec{c} = 0$, then subtracting $\vec{a}$ from both sides gives $\vec{b} + \vec{c} = -\vec{a}$. This is mathematically sound and allows us to isolate the vectors we are interested in. When we square both sides, the negative sign disappears because $(-\vec{a}) \cdot (-\vec{a}) = \vec{a} \cdot \vec{a} = |\vec{a}|^2$.

Question

If $\vec{a}, \vec{b}, \vec{c}$ are vectors show that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\vec{a}|=7, |\vec{b}|=5, |\vec{c}|=3$ then angle between vector $\vec{b}$ and $\vec{c}$ is

Options:

(A) $60^\circ$

(B) $30^\circ$

(C) $45^\circ$

(D) $90^\circ$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty space. You have three arrows—vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$—floating in front of you. You are told that their sum is zero:

$$\vec{a} + \vec{b} + \vec{c} = 0$$

This means that if you walk along $\vec{a}$, then turn and walk along $\vec{b}$, and finally walk along $\vec{c}$, you end up exactly where you started. You have traced a closed triangle.

We know the magnitudes: $|\vec{a}| = 7$, $|\vec{b}| = 5$, and $|\vec{c}| = 3$. These are the lengths of the sides of our triangle.

### The Angle Trap

Many students, when faced with this, immediately look at the interior angles of the triangle. They think the angle between $\vec{b}$ and $\vec{c}$ is just the angle inside the triangle. Stop!

Remember the definition of the angle between two vectors. It is the angle formed when the vectors are placed tail-to-tail. In our triangle, $\vec{b}$ ends where $\vec{c}$ begins, which is a head-to-tail arrangement.

To find the true angle $	heta$, we must extend the line of $\vec{b}$ forward. The angle $	heta$ we seek is the exterior angle at that vertex. This distinction is the difference between a correct answer and a common mistake.

### The Algebraic Alchemy

Now, we bridge the gap between this geometric visualization and a numerical answer. We need to isolate the vectors we care about.

Starting with $\vec{a} + \vec{b} + \vec{c} = 0$, we subtract $\vec{a}$ from both sides to get:

$$\vec{b} + \vec{c} = -\vec{a}$$

Now, we perform a piece of algebraic alchemy. We take the dot product of this equation with itself. Because the dot product is the bridge between vectors and scalars, it allows us to turn vector magnitudes and angles into simple numbers.

$$(\vec{b} + \vec{c}) \cdot (\vec{b} + \vec{c}) = (-\vec{a}) \cdot (-\vec{a})$$

Expanding this is just like expanding $(x+y)^2$. We get:

$$|\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c}) = |\vec{a}|^2$$

We know that the dot product $\vec{b} \cdot \vec{c}$ is defined as $|\vec{b}| |\vec{c}| \cos	heta$. Substituting this in, we obtain:

$$|\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos	heta = |\vec{a}|^2$$

### The Final Calculation

Now, the physics is done; the rest is arithmetic. We plug in our values:

$$5^2 + 3^2 + 2(5)(3)\cos	heta = 7^2$$

This simplifies to:

$$25 + 9 + 30\cos	heta = 49$$

Combining the constants, we have:

$$34 + 30\cos	heta = 49$$

Subtracting $34$ from $49$ gives us $15$. So:

$$30\cos	heta = 15$$

Dividing by $30$, we find:

$$\cos	heta = \frac{15}{30} = \frac{1}{2}$$

We know from our trigonometry tables that the angle whose cosine is $\frac{1}{2}$ is **$60^\circ$**. You have navigated the geometry, avoided the trap, performed the algebraic transformation, and arrived at the solution.

Animated Solution for Mathematics - Vector Algebra: If $a, b, c$ are vectors show that $a + b + c = 0$ and $∣ a ∣ = 7, ∣ b ∣ = 5, ∣ c ∣ = 3$ then angle between vector $b$ and $c$ is

Select Answer:

Visualized Solution

Visualizing the Vector Triangle

Defining the Angle $θ$

Isolating the Target Vectors

Squaring the Vector Equation

Expanding the Algebraic Terms

Applying the Dot Product Formula

Substituting the Given Values

Simplifying the Arithmetic

Isolating $cos θ$

Determining the Angle $θ$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Angle Trap

The Algebraic Alchemy

The Final Calculation

Similar Questions

If $(a + 3 b)$ is perpendicular to $(7 a - 5 b)$ and $(a - 4 b)$ is perpendicular to $(7 a - 2 b)$ , then the angle between $a$ and $b$ (in degrees) is ___

If $a$ and $b$ are two unit vectors such that $a + 2 b$ and $5 a - 4 b$ are perpendicular to each other then the angle between $a$ and $b$ is

Let $a$ and $b$ be two unit vectors. If the vectors $c = a + 2 b$ and $d = 5 a - 4 b$ are perpendicular to each other, then the angle between $a$ and $b$ is:

Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :

Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\overset{a}{^}$ and $\hat{b}$ be $sin^{- 1} (\frac{65}{9})$ . If the vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})$ , then the value of $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$ is

$a, b, c$ are 3 vectors, such that $a + b + c = 0$ , $∣ a ∣ = 1, ∣ b ∣ = 2, ∣ c ∣ = 3$ , then $a \cdot b + b \cdot c + c \cdot a$ is equal to

If $∣ a ∣ = 5, ∣ b ∣ = 4, ∣ c ∣ = 3$ thus what will be the value of $∣ a \cdot b + b \cdot c + c \cdot a ∣$ , given that $a + b + c = 0$

Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are vectors show that a+b+c=0 and ∣a∣=7,∣b∣=5,∣c∣=3 then angle between vector b and c is