For two unit vectors $\vec{a}$ and $\vec{b}$, if $|\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}|$, then find the angle between $\vec{a}$ and $\vec{b}$.

For two unit vectors veca and vecb, if |veca + 2vecb| = — CBSE Class 12 Mathematics 2026 PYQ

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\vec{a}

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

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Concept understanding — Dot Product Angle

Finding the Angle Between Vectors

Suppose you have two arrows drawn from the same point. One question is unavoidable in geometry, physics, and mechanics: what is the angle between them? You could measure it with a protractor on paper, but that fails the moment the vectors live in 3D. The dot product gives you the angle by pure calculation.

The Core Idea

The scalar (dot) product of two vectors has two faces that describe the same number:

\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad \text{(components)}

\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta \qquad \text{(geometry)}

The first is easy to compute from coordinates; the second hides the angle $\theta$ (with $0 \le \theta \le \pi$ ) between the vectors. Setting them equal and solving for $\cos\theta$ gives the master formula.

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}, \qquad \theta = \cos^{-1}\!\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)

Why It Works

Both vectors have a fixed length, so the only thing the dot product can "vary" with is how aligned they are. When they point the same way, $\cos\theta = 1$ and the dot product is as large as possible, $|\vec{a}||\vec{b}|$ . When they are perpendicular, $\cos\theta = 0$ and the dot product vanishes. When they point opposite ways, $\cos\theta = -1$ . Dividing $\vec{a}\cdot\vec{b}$ by the two lengths simply strips away the size information and leaves behind a pure measure of alignment — exactly $\cos\theta$ .

Note

The sign of the dot product tells you the type of angle at a glance: positive $\Rightarrow$ acute, zero $\Rightarrow$ right angle, negative $\Rightarrow$ obtuse.

Using the Formula

For $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 0\hat{j} + 0\hat{k}$ :

\vec{a}\cdot\vec{b} = 1, \quad |\vec{a}| = 3, \quad |\vec{b}| = 1

$\cos\theta = \frac{1}{3\cdot 1} = \frac{1}{3} \;\Rightarrow\; \theta = \cos^{-1}\tfrac{1}{3} \approx 70.5^\circ$ …

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To find the angle between two unit vectors given an equality of magnitudes, we square both sides to convert magnitudes into dot products. This simplifies to $\vec{a} \cdot \vec{b} = 0$ , implying the angle between them is $\frac{\pi}{2}$ .

When you're given an equation involving the magnitudes (lengths) of vector sums or differences, a common and effective strategy is to square both sides of the equation. This is because the square of a vector's magnitude, $|\vec{v}|^2$ , is equal to the dot product of the vector with itself, $\vec{v} \cdot \vec{v}$ . This property allows us to expand the expressions using the distributive property of the dot product, which then naturally introduces the dot product of the individual vectors, $\vec{a} \cdot \vec{b}$ . The dot product, in turn, is directly related to the cosine of the angle between the vectors, which is what we need to find.

Let's break down the solution step-by-step.

Understand the given information.

We are given two unit vectors, $\vec{a}$ and $\vec{b}$ . This means their magnitudes are 1:

$|\vec{a}| = 1$

$|\vec{b}| = 1$

We are also given the equality:

$|\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}|$
Square both sides of the equation.

As discussed, squaring both sides is the key step to convert magnitudes into dot products, which are easier to work with algebraically.

$|\vec{a} + 2\vec{b}|^2 = |2\vec{a} - \vec{b}|^2$
$\vec{v}$
Expand the squared magnitudes using the dot product property.

Applying the formula $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + 2(\vec{u} \cdot \vec{v}) + |\vec{v}|^2$ to the left side:

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

Using the property $k(\vec{u} \cdot \vec{v}) = (k\vec{u}) \cdot \vec{v} = \vec{u} \cdot (k\vec{v})$ and $|k\vec{v}|^2 = k^2|\vec{v}|^2$ :

$|\vec{a} + 2\vec{b}|^2 = |\vec{a}|^2 + 4(\vec{a} \cdot \vec{b}) + 4|\vec{b}|^2$

Similarly, for the right side, using $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 - 2(\vec{u} \cdot \vec{v}) + |\vec{v}|^2$ :

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

$|2\vec{a} - \vec{b}|^2 = 4|\vec{a}|^2 - 4(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$
Substitute the known magnitudes and equate the expanded expressions.

We know $|\vec{a}| = 1$ and $|\vec{b}| = 1$ . Substitute these values into the expanded expressions:

Left side:

$|\vec{a} + 2\vec{b}|^2 = (1)^2 + 4(\vec{a} \cdot \vec{b}) + 4(1)^2$

$|\vec{a} + 2\vec{b}|^2 = 1 + 4(\vec{a} \cdot \vec{b}) + 4$ …

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

Showing the 12 most recent of 66 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$|\vec{a}| = 8$
›Reveal solutionSolution
The key is the identity $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$ . Substituting the given magnitudes gives $144 + (\vec{a} \cdot \vec{b})^2 = 576$ , so $|\vec{a} \cdot \vec{b}| = \sqrt{432} = 12\sqrt{3}$ . The correct option is (C).

The problem gives you the magnitudes of two vectors and the magnitude of their cross product, and asks for the magnitude of their dot product. This is a classic setup — it tests a single, powerful relationship that ties the dot product and cross product together.

The core idea: The dot product depends on $\cos\theta$ , and the cross product depends on $\sin\theta$ , where $\theta$ is the angle between the vectors. Since $|\vec{a}|$ and $|\vec{b}|$ are known, you can use the identity $\sin^2\theta + \cos^2\theta = 1$ to eliminate $\theta$ and directly connect the two products.

Write the definitions:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$

Here $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ , with $0 \le \theta \le \pi$ .

Square both equations:

$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2\theta$

$(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2\theta$

Add them together:

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 (\sin^2\theta + \cos^2\theta) = |\vec{a}|^2 |\vec{b}|^2$
This is the identity you need. It holds for any two vectors in 3D space.
$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$

Plug in the given numbers:

$|\vec{a}| = 8$ , $|\vec{b}| = 3$ , so $|\vec{a}|^2 |\vec{b}|^2 = 64 \times 9 = 576$

$|\vec{a} \times \vec{b}| = 12$ , so $|\vec{a} \times \vec{b}|^2 = 144$

Therefore: …
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CBSE 2026Set 65/2/11 markMCQ
$\vec{a}$
›Reveal solutionSolution
Assertion (A) is a fundamental identity relating the magnitudes of the cross product and dot product, which is true. Reason (R) is also a true relationship between the magnitudes of the cross product and dot product, but it does not explain Assertion (A). The correct option is (B).

To evaluate this assertion-reason question, we need to understand the definitions of the dot product and cross product of two vectors and the geometric meaning of the angle between them. Both the dot product and the cross product are fundamental operations in vector algebra, and their properties are frequently tested.

The dot product (or scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$ respectively, and $\theta$ is the angle between them ( $0 \le \theta \le \pi$ ). The result is a scalar.

The cross product (or vector product) of two vectors $\vec{a}$ and $\vec{b}$ results in a vector perpendicular to both $\vec{a}$ and $\vec{b}$ . Its magnitude is defined as:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

where $\theta$ is again the angle between $\vec{a}$ and $\vec{b}$ . The direction of $\vec{a} \times \vec{b}$ is given by the right-hand rule.

Now, let's evaluate the Assertion and Reason.

Evaluate Assertion (A):

The assertion states: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$ .

Let's substitute the definitions of $|\vec{a} \times \vec{b}|$ and $(\vec{a} \cdot \vec{b})$ into the left-hand side (LHS) of the equation.

LHS $= (|\vec{a}| |\vec{b}| \sin \theta)^2 + (|\vec{a}| |\vec{b}| \cos \theta)^2$

LHS $= |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$

We can factor out $|\vec{a}|^2 |\vec{b}|^2$ :

LHS $= |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta)$

Important

Recall the fundamental trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$ .

Using this identity:

LHS $= |\vec{a}|^2 |\vec{b}|^2 (1)$

LHS $= |\vec{a}|^2 |\vec{b}|^2$

This matches the right-hand side (RHS) of the assertion.

Therefore, Assertion (A) is True. This identity is often known as Lagrange's Identity for vectors.

Evaluate Reason (R):

The reason states: $|\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta$ , $\left(\theta \neq \frac{\pi}{2}\right)$ .

Let's substitute the definitions of $|\vec{a} \times \vec{b}|$ and $(\vec{a} \cdot \vec{b})$ into this equation.

LHS: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

RHS: $(\vec{a} \cdot \vec{b}) \tan \theta = (|\vec{a}| |\vec{b}| \cos \theta) \tan \theta$
$\tan \theta$
Substitute this into the RHS:

RHS $= (|\vec{a}| |\vec{b}| \cos \theta) \left(\frac{\sin \theta}{\cos \theta}\right)$

Since $\theta \neq \frac{\pi}{2}$ , $\cos \theta \neq 0$ , so we can cancel $\cos \theta$ : …
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CBSE 2026Set ANNUAL1 mark
$\vec{a}$
›Reveal solutionSolution

Use $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ to solve for $\theta$ .

$\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \dfrac{\sqrt6}{\sqrt3\cdot2} = \dfrac{\sqrt6}{2\sqrt3}=\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\hat{i} + 3\hat{j} + 3\hat{k}$
(a) $0°$
(b) $45°$
(c) $60°$
(d) $90°$
›Reveal solutionSolution

The two vectors have zero dot product, so they are perpendicular.

Let $\vec{a}=\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+\hat{k}$ .

$\vec{a}\cdot\vec{b}=1(3)+3(-2)+3(1)=3-6+3=0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If vector a . vector b = \sqrt3 |vector a \times vector b| then angle between vector a and vector b is:$
(a) π/2
(b) π/6
(c) π/4
(d) π/3
›Reveal solutionSolution

Writing the dot and cross product magnitudes in terms of $\cos\theta$ and $\sin\theta$ turns the given condition into $\tan\theta = \frac{1}{\sqrt3}$ .

We know:

$\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta, \qquad |\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$

Given $\vec a \cdot \vec b = \sqrt3\, |\vec a \times \vec b|$ :

$|\vec a||\vec b|\cos\theta = \sqrt3\,|\vec a||\vec b|\sin\theta$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.If |\vec a| = 1, |\vec b| = 2 and \vec a \cdot \vec b = 1, then angle between \vec a and \vec b is:$
(a) \pi/2
(b) \pi/6
(c) \pi/3
(d) \pi/4
›Reveal solutionSolution

Use $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ to solve for the angle $\theta$ .

Working:

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Dot Product of two vectors is defined as$
(a) a·b = |a||b| sin θ
(b) a·b = |a||b| cos θ
(c) a·b = |a||b| sin θ n̂
(d) None of the above
›Reveal solutionSolution

The dot (scalar) product uses cosine of the angle between the vectors; the cross (vector) product uses sine.

For two vectors $\vec a$ and $\vec b$ with angle $\theta$ between them, the dot product is defined as

$\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta,$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the scalar product of two vectors is zero, then the angle between them is:$
(a) 45°
(b) 0°
(c) 180°
(d) 90°
›Reveal solutionSolution

A.B = AB*cos(theta) is zero only when cos(theta) = 0, i.e., when the vectors are perpendicular, theta = 90 degrees.

The scalar product (dot product) of two vectors A and B is defined as

A.B = |A||B|*cos(theta)

where theta is the angle between the two vectors.

For the scalar product to be zero (assuming neither vector is itself the zero vector), we need

cos(theta) = 0

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If P = 4 î - 5 ĵ and Q = 5 î + 4 ĵ, then the angle between the two vectors is$
(a) 0
(b) π
(c) π/2
(d) π/4
›Reveal solutionSolution

P.Q = 0, so the angle between them is 90 degrees = pi/2. Answer (C).

P = 4 i - 5 j and Q = 5 i + 4 j.

Dot product: P.Q = (4)(5) + (-5)(4) = 20 - 20 = 0.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the angle between two vectors is 90°, then their dot product will be$
(a) 0
(b) 1
(c) − 1
(d) ∞
›Reveal solutionSolution

A.B = |A||B|cos(theta); at 90 degrees this is 0. Answer (A).

The scalar (dot) product of two vectors is A.B = |A||B| cos(theta), where theta is the angle between them.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\vec{a}$
(a) $\dfrac{\pi}{2}$
(b) $\dfrac{\pi}{6}$
(c) $\dfrac{\pi}{3}$
(d) $\dfrac{\pi}{4}$
›Reveal solutionSolution

$|\vec a\times\vec b|=7$ , so $\sin\theta=\tfrac{7}{2\cdot 7}=\tfrac12\Rightarrow\theta=\tfrac{\pi}{6}$ .

Step 1: $|\vec a\times\vec b|=\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7.$

…

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CBSE 20251 markMCQ
$\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$
›Reveal solutionSolution
The dot product of $\vec{a}$ and $\vec{b}$ is zero, which means the vectors are perpendicular. The correct option is (B).

The key to deciding whether two vectors are parallel, perpendicular, or neither lies in their dot product. The dot product (also called the scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$
where $\theta$ is the angle between them. This single formula tells us everything:

If $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|$ , then $\cos\theta = 1$ , so $\theta = 0^\circ$ — the vectors are parallel (and pointing in the same direction).

If $\vec{a} \cdot \vec{b} = -|\vec{a}| |\vec{b}|$ , then $\cos\theta = -1$ , so $\theta = 180^\circ$ — the vectors are anti-parallel.

If $\vec{a} \cdot \vec{b} = 0$ , then $\cos\theta = 0$ , so $\theta = 90^\circ$ — the vectors are perpendicular (orthogonal).

So instead of guessing, we just compute the dot product and see which case holds.

Write the vectors in component form.

$\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$

$\vec{b} = \hat{i} - \hat{j} + \hat{k}$

Compute the dot product.

The dot product is the sum of the products of corresponding components:

$\vec{a} \cdot \vec{b} = (3)(1) + (2)(-1) + (-1)(1)$ $= 3 - 2 - 1 = 0$

Interpret the result.

Since $\vec{a} \cdot \vec{b} = 0$ , the angle between them is $90^\circ$ . Therefore, $\vec{a}$ is perpendicular to $\vec{b}$ .

Check the other options for completeness.

Option (A) says they are parallel — but that would require the dot product to be $\pm |\vec{a}||\vec{b}|$ , not zero. So (A) is false.

Option (C) says $|\vec{b}| > |\vec{a}|$ . Let's check magnitudes: …
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Q.For two unit vectors a⃗\vec{a}a and b⃗\vec{b}b, if ∣a⃗+2b⃗∣=∣2a⃗−b⃗∣|\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}|∣a+2b∣=∣2a−b∣, then find the angle between a⃗\vec{a}a and b⃗\vec{b}b.

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Q.For two unit vectors $\vec{a}$ and $\vec{b}$ , if $|\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}|$ , then find the angle between $\vec{a}$ and $\vec{b}$ .

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