If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, $|\vec{a}| = \sqrt{37}$, $|\vec{b}| = 3$ and $|\vec{c}| = 4$, then angle between $\vec{b}$ and $\vec{c}$ is (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{2}$

If veca + vecb + vecc = vec0, |veca| = √(37), |vecb| = 3 and |vecc| = 4, then angle between vecb and vecc is (

Question

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

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Vector Addition Triangle Law

Triangle Law of Vector Addition

How do you combine two vectors into a single one? If you make two journeys one after the other, the net journey is a single vector from where you started to where you finished. That is exactly the triangle law.

The law

Important

If two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order (the tip of the first joined to the tail of the second), then their sum is represented by the third side taken in the reverse order — from the tail of the first to the tip of the second.

Place $\vec{a}$ , then start $\vec{b}$ where $\vec{a}$ ends. The arrow that closes the triangle, drawn from the start of $\vec{a}$ to the end of $\vec{b}$ , is the resultant $\vec{a}+\vec{b}$ .

\vec{AB}+\vec{BC}=\vec{AC}

Why it works

Read the vectors as directed displacements: going from $A$ to $B$ and then $B$ to $C$ lands you at $C$ , and the single displacement that achieves the same is $A$ to $C$ . The intermediate point $B$ cancels — only the overall start and finish survive.

Consequences

Commutative: $\vec{a}+\vec{b}=\vec{b}+\vec{a}$ . Completing the triangle the other way gives the same closing side — which is why the parallelogram law agrees with the triangle law.
Closed triangle = zero: if three vectors form a triangle taken in order, $\vec{AB}+\vec{BC}+\vec{CA}=\vec{0}$ , since you return to the start.
To subtract, add the negative: $\vec{a}-\vec{b}=\vec{a}+(-\vec{b})$ , reversing $\vec{b}$ before joining it.

Tip

Triangle law (tail-to-tip) and parallelogram law (both vectors from a common tail) are two pictures of the same addition — use whichever fits the diagram.

Why it matters

This is the foundation of all vector addition: resolving and combining forces, velocities, and displacements in physics, and adding position vectors in geometry, all rest on the triangle law.

The triangle law of vector addition is one of the earliest and most tested ideas in the NCERT Class 12 Vector Algebra chapter, appearing in CBSE board diagram-based questions and forming the geometric basis for the parallelogram law. "Triangle law of vector addition proof" is a frequently searched query among students preparing for both boards and JEE Main.

Using the triangle law of vector addition, the three vectors form a closed triangle. Applying the cosine rule to the triangle formed by $\vec{b}$ and $\vec{c}$ (with $\vec{a}$ as their resultant) gives $\cos\theta = \frac{1}{2}$ , so the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$ .

The key insight here is that when three vectors add to zero, they form the sides of a triangle taken head-to-tail. This is the Triangle Law of Vector Addition in reverse: if $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then $\vec{a} + \vec{b} = -\vec{c}$ , meaning the sum of any two gives the negative of the third. Geometrically, the three vectors can be arranged as three sides of a triangle, with each side representing one vector's magnitude and direction.

So we have a triangle whose sides have lengths $|\vec{a}| = \sqrt{37}$ , $|\vec{b}| = 3$ , and $|\vec{c}| = 4$ . The angle between $\vec{b}$ and $\vec{c}$ is the interior angle of this triangle at the vertex where $\vec{b}$ and $\vec{c}$ meet. In the triangle, the side opposite this angle is $\vec{a}$ (since $\vec{a}$ connects the tail of $\vec{b}$ to the head of $\vec{c}$ when arranged head-to-tail).

Now we apply the cosine rule from trigonometry: in any triangle with sides $p$ , $q$ , $r$ , where $r$ is opposite the angle $\theta$ between $p$ and $q$ , we have $r^2 = p^2 + q^2 - 2pq\cos\theta$ .

Identify the sides: Let the angle between $\vec{b}$ and $\vec{c}$ be $\theta$ . Then the side opposite $\theta$ is $|\vec{a}| = \sqrt{37}$ . The two sides forming the angle are $|\vec{b}| = 3$ and $|\vec{c}| = 4$ .
Write the cosine rule:

|\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2|\vec{b}||\vec{c}|\cos\theta

Substitute the given magnitudes:

(\sqrt{37})^2 = 3^2 + 4^2 - 2(3)(4)\cos\theta

37 = 9 + 16 - 24\cos\theta

37 = 25 - 24\cos\theta

Solve for $\cos\theta$ :

37 - 25 = -24\cos\theta

12 = -24\cos\theta

\cos\theta = -\frac{12}{24} = -\frac{1}{2}

Watch out

A common mistake is to forget the minus sign in the cosine rule. The formula is $r^2 = p^2 + q^2 - 2pq\cos\theta$ , not $+2pq\cos\theta$ . Also, note that $\cos\theta$ came out negative here — that's fine; it just means the angle is obtuse. But wait — let's check: $\cos\theta = -\frac{1}{2}$ gives $\theta = \frac{2\pi}{3}$ , which is not among the options. Something is off.

Let's re-examine the geometry. The angle between $\vec{b}$ and $\vec{c}$ in the vector equation is not the interior angle of the triangle where they meet head-to-tail. When vectors are placed head-to-tail, the angle between $\vec{b}$ and $\vec{c}$ is actually the exterior angle at that vertex, because $\vec{c}$ starts at the head of $\vec{b}$ , so the direction of $\vec{c}$ is away from $\vec{b}$ 's head. The interior angle of the triangle is the supplement of the angle between the vectors.

So if $\phi$ is the interior angle (the one we used in the cosine rule), then the angle between $\vec{b}$ and $\vec{c}$ is $\pi - \phi$ . We found $\cos\phi = -\frac{1}{2}$ , so $\phi = \frac{2\pi}{3}$ . Then the angle between $\vec{b}$ and $\vec{c}$ is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ .

Alternatively, we can avoid this confusion by using the vector relation directly: from $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , we have $\vec{a} = -(\vec{b} + \vec{c})$ . Then:

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

where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$ (the vectors themselves, not the triangle sides). Substituting:

37 = 9 + 16 + 2(3)(4)\cos\theta

37 = 25 + 24\cos\theta

12 = 24\cos\theta

\cos\theta = \frac{1}{2}

\theta = \frac{\pi}{3}

Tip

Using $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos\theta$ directly from the vector equation avoids the geometric confusion about interior vs. exterior angles. Always prefer the algebraic vector approach when the angle between the vectors themselves is asked.

✓Final answer

\vec{b}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$
(a) $-4\hat{j}-\hat{k}$
(b) $4\hat{i}-\hat{k}$
(c) $4\hat{j}+5\hat{k}$
(d) $\hat{i}+4\hat{j}-\hat{k}$
›Reveal solutionSolution

Add the $\hat i,\hat j,\hat k$ components of the three vectors separately.

$\vec a+\vec b+\vec c$ :

$\hat i$ : $1-2+1=0$

$\hat j$ : $-2+4-6=-4$

$\hat k$ : $1+5-7=-1$

Sum $= 0\hat i - 4\hat j - \hat k = -4\hat j - \hat k$ .

✓Final answer
$-4\hat j-\hat k$
CBSE 20251 markMCQ
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
›Reveal solutionSolution
Using the triangle law of vector addition, the three vectors form a closed triangle. Applying the cosine rule to the triangle formed by $\vec{b}$ and $\vec{c}$ (with $\vec{a}$ as their resultant) gives $\cos\theta = \frac{1}{2}$ , so the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$ .

The key insight here is that when three vectors add to zero, they form the sides of a triangle taken head-to-tail. This is the Triangle Law of Vector Addition in reverse: if $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then $\vec{a} + \vec{b} = -\vec{c}$ , meaning the sum of any two gives the negative of the third. Geometrically, the three vectors can be arranged as three sides of a triangle, with each side representing one vector's magnitude and direction.

So we have a triangle whose sides have lengths $|\vec{a}| = \sqrt{37}$ , $|\vec{b}| = 3$ , and $|\vec{c}| = 4$ . The angle between $\vec{b}$ and $\vec{c}$ is the interior angle of this triangle at the vertex where $\vec{b}$ and $\vec{c}$ meet. In the triangle, the side opposite this angle is $\vec{a}$ (since $\vec{a}$ connects the tail of $\vec{b}$ to the head of $\vec{c}$ when arranged head-to-tail).

Now we apply the cosine rule from trigonometry: in any triangle with sides $p$ , $q$ , $r$ , where $r$ is opposite the angle $\theta$ between $p$ and $q$ , we have $r^2 = p^2 + q^2 - 2pq\cos\theta$ .

Identify the sides: Let the angle between $\vec{b}$ and $\vec{c}$ be $\theta$ . Then the side opposite $\theta$ is $|\vec{a}| = \sqrt{37}$ . The two sides forming the angle are $|\vec{b}| = 3$ and $|\vec{c}| = 4$ .

Write the cosine rule:

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

Substitute the given magnitudes:

$(\sqrt{37})^2 = 3^2 + 4^2 - 2(3)(4)\cos\theta$ $37 = 9 + 16 - 24\cos\theta$ $37 = 25 - 24\cos\theta$

Solve for $\cos\theta$ :

$37 - 25 = -24\cos\theta$ $12 = -24\cos\theta$ $\cos\theta = -\frac{12}{24} = -\frac{1}{2}$
Watch out

A common mistake is to forget the minus sign in the cosine rule. The formula is $r^2 = p^2 + q^2 - 2pq\cos\theta$ , not $+2pq\cos\theta$ . Also, note that $\cos\theta$ came out negative here — that's fine; it just means the angle is obtuse. But wait — let's check: $\cos\theta = -\frac{1}{2}$ gives $\theta = \frac{2\pi}{3}$ , which is not among the options. Something is off.

Let's re-examine the geometry. The angle between $\vec{b}$ and $\vec{c}$ in the vector equation is not the interior angle of the triangle where they meet head-to-tail. When vectors are placed head-to-tail, the angle between $\vec{b}$ and $\vec{c}$ is actually the exterior angle at that vertex, because $\vec{c}$ starts at the head of $\vec{b}$ , so the direction of $\vec{c}$ is away from $\vec{b}$ 's head. The interior angle of the triangle is the supplement of the angle between the vectors.

So if $\phi$ is the interior angle (the one we used in the cosine rule), then the angle between $\vec{b}$ and $\vec{c}$ is $\pi - \phi$ . We found $\cos\phi = -\frac{1}{2}$ , so $\phi = \frac{2\pi}{3}$ . Then the angle between $\vec{b}$ and $\vec{c}$ is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ .

Alternatively, we can avoid this confusion by using the vector relation directly: from $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , we have $\vec{a} = -(\vec{b} + \vec{c})$ . Then:
$|\vec{a}|^2 = |\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos\theta$
where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$ (the vectors themselves, not the triangle sides). Substituting:
$37 = 9 + 16 + 2(3)(4)\cos\theta$ $37 = 25 + 24\cos\theta$ $12 = 24\cos\theta$ $\cos\theta = \frac{1}{2}$ $\theta = \frac{\pi}{3}$
Tip

Using $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos\theta$ directly from the vector equation avoids the geometric confusion about interior vs. exterior angles. Always prefer the algebraic vector approach when the angle between the vectors themselves is asked.

✓Final answer
$\vec{b}$
CBSE 2025Set A1 mark
$Q. The vector sum of the three sides of a triangle taken in order is ______.$
›Reveal solutionSolution

Walking along all three sides of a triangle in order brings you back to the starting point, so the net vector displacement is zero.

Let the triangle have vertices $A$ , $B$ , $C$ . Taking the sides in order as vectors: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA}$ .

By the triangle law of vector addition, $\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$ . Adding $\overrightarrow{CA}$ (which is $-\overrightarrow{AC}$ ):

$\overrightarrow{AC} + \overrightarrow{CA} = \overrightarrow{AC} - \overrightarrow{AC} = \vec0$

✓Final answer
$\vec 0$
CBSE 2025Set ANNUAL1 mark
$\vec{a}, \vec{b}$
›Reveal solutionSolution

Vector addition is associative — the way three vectors are grouped while adding does not affect the result.

For any three vectors $\vec{a},\vec{b},\vec{c}$ , the associative property of vector addition states:

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

This means whether we first add $\vec{a}$ and $\vec{b}$ and then $\vec{c}$ , or first add $\vec{b}$ and $\vec{c}$ and then $\vec{a}$ , the resultant vector is the same.

✓Final answer
$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
CBSE 2023Set A1 mark
$Q. The vector sum of the three sides of a triangle taken in order is ______.$
›Reveal solutionSolution

Traversing a triangle's three sides in order returns you to the starting point, so the vector sum is zero.

Let the triangle have vertices $A,B,C$ . Taking the sides in order as $\vec{AB},\vec{BC},\vec{CA}$ :

$\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AC}-\vec{AC}=\vec{0}.$

Geometrically, walking along all three sides in sequence brings you back to where you started — net displacement zero.

✓Final answer
$\vec{0}$
CBSE 2022Set ANNUAL1 markMCQ
$\vec a = 2\hat i+3\hat k$
(a) $2\hat i-4\hat j+8\hat k$
(b) $\hat i-4\hat j-8\hat k$
(c) $\hat i+4\hat j-8\hat k$
(d) $\hat i+4\hat j+8\hat k$
›Reveal solutionSolution

Add/subtract the vectors component-by-component ( $\hat i,\hat j,\hat k$ separately).

Given $\vec a=2\hat i+0\hat j+3\hat k$ , $\vec b=\hat i+2\hat j-3\hat k$ , $\vec c=0\hat i+3\hat j-4\hat k$ .

$\vec a-\vec b=(2-1)\hat i+(0-2)\hat j+(3-(-3))\hat k=\hat i-2\hat j+6\hat k$ .

$2\vec c=6\hat j-8\hat k$ .

$\vec a-\vec b+2\vec c=(1+0)\hat i+(-2+6)\hat j+(6-8)\hat k=\hat i+4\hat j-2\hat k$ .

Checking against the given options — (a) $2\hat i-4\hat j+8\hat k$ , (b) $\hat i-4\hat j-8\hat k$ , (c) $\hat i+4\hat j-8\hat k$ , (d) $\hat i+4\hat j+8\hat k$ — none reproduces the correctly-computed $\hat k$ coefficient of $-2$ , though (c) and (d) match the $\hat i,\hat j$ parts. This is most likely an error introduced when the options were extracted/typeset, not a genuine ambiguity in the vector algebra. The mathematically correct result is $\hat i+4\hat j-2\hat k$ .

✓Final answer
$\vec a-\vec b+2\vec c=\hat i+4\hat j-2\hat k$
CBSE 2020Set HE8231 mark
$\vec{0}$
›Reveal solutionSolution

The statement is True — this is a direct consequence of the triangle law of vector addition.

Let the triangle have vertices $A,B,C$ . Taking the sides in order means $\overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CA}$ (each side's vector starts where the previous one ended).

By the triangle law, $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ .

Adding $\overrightarrow{CA}$ to both sides:

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{AC}+\overrightarrow{CA}=\vec0$

(since $\overrightarrow{CA}=-\overrightarrow{AC}$ ).

✓Final answer
$True.$
CBSE 2018Set ANNUAL1 markMCQ
$Q. If two forces of 3 units and 4 units are acting at an angle 90°, then its resultant force will be:$
(a) 3 units
(b) 4 units
(c) 5 units
(d) 0 unit
›Reveal solutionSolution

Two forces at right angles combine like the legs of a right triangle — the resultant is the hypotenuse, found with the Pythagorean form of the parallelogram law.

When two forces (or vectors) $P$ and $Q$ act at a point with an angle $\theta$ between them, the magnitude of their resultant is:

$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$

Here $P=3$ , $Q=4$ , $\theta = 90^\circ$ , so $\cos\theta = 0$ :

$R = \sqrt{3^2 + 4^2 + 0} = \sqrt{9+16} = \sqrt{25} = 5$

✓Final answer
$= 5$
CBSE 2016Set ANNUAL1 mark
$Q. State the triangle law of vectors.$
›Reveal solutionSolution

definition recall

Triangle law of vectors: If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant (sum) is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Symbolically, if $\overrightarrow{AB}=\vec a$ and $\overrightarrow{BC}=\vec b$ , then $\overrightarrow{AC}=\vec a+\vec b$ .

✓Final answer
$\vec a=\overrightarrow{AB}$

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