The position vectors of vertices of $\triangle ABC$ are $A(2\hat{i} - \hat{j} + \hat{k})$, $B(\hat{i} - 3\hat{j} - 5\hat{k})$ and $C(3\hat{i} - 4\hat{j} - 4\hat{k})$. Find all the angles of $\triangle ABC$.

The position vectors of vertices of triangle ABC are A( — CBSE Class 12 Mathematics 2024 PYQ

Question

\triangle ABC

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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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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The key idea is to use the dot product of vectors to find the angle between them. For any two vectors $\vec{u}$ and $\vec{v}$ , the angle $\theta$ between them is given by $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$ .

First, we find the vectors representing the sides of the triangle:

$\vec{AB} = \vec{B} - \vec{A} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2\hat{j} - 6\hat{k}$

$\vec{BC} = \vec{C} - \vec{B} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) = 2\hat{i} - \hat{j} + \hat{k}$

$\vec{CA} = \vec{A} - \vec{C} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) = -\hat{i} + 3\hat{j} + 5\hat{k}$
Next, we calculate the magnitudes of these vectors:

$|\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$

$|\vec{BC}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

$|\vec{CA}| = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$
Now, we find each angle using the dot product formula:
- For angle $A$ , we use $\vec{AB}$ and $\vec{AC} = \vec{C} - \vec{A} = \hat{i} - 3\hat{j} - 5\hat{k}$ . $|\vec{AC}| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$ $\cos A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|} = \frac{(-\hat{i} - 2\hat{j} - 6\hat{k}) \cdot (\hat{i} - 3\hat{j} - 5\hat{k})}{\sqrt{41}\sqrt{35}} = \frac{(-1)(1) + (-2)(-3) + (-6)(-5)}{\sqrt{41}\sqrt{35}} = \frac{-1 + 6 + 30}{\sqrt{41}\sqrt{35}} = \frac{35}{\sqrt{41}\sqrt{35}} = \frac{\sqrt{35}}{\sqrt{41}}$ $A = \cos^{-1}\left(\frac{\sqrt{35}}{\sqrt{41}}\right)$
- For angle $B$ , we use $\vec{BA} = -\vec{AB} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{BC}$ . …

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To find the angles of a triangle given its vertices, we first determine the vectors representing its sides. Then, we use the dot product formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$ to find the angle between the side vectors originating from each vertex. The angles of $\triangle ABC$ are $\arccos\left(\frac{\sqrt{35}}{\sqrt{41}}\right)$ , $\arccos\left(\frac{\sqrt{6}}{\sqrt{41}}\right)$ , and $90^\circ$ .

The core idea behind finding the angles of a triangle when its vertices are given as position vectors is to leverage the geometric interpretation of the dot product. A position vector simply tells us the coordinates of a point relative to the origin. To find the vector representing a side of the triangle, say from point A to point B, we subtract the position vector of A from the position vector of B.

Once we have the vectors representing the sides, we can use the dot product formula. For any two vectors $\vec{a}$ and $\vec{b}$ , their dot product is defined as:

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

where $\theta$ is the angle between the two vectors. Rearranging this formula, we can find the cosine of the angle:

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

Watch out

When calculating an internal angle of a triangle, ensure the two vectors you use for the dot product either both originate from the vertex (e.g., $\vec{AB}$ and $\vec{AC}$ for angle A) or both terminate at the vertex (e.g., $\vec{BA}$ and $\vec{CA}$ for angle A). Using one originating and one terminating vector (e.g., $\vec{AB}$ and $\vec{CA}$ ) will yield the exterior angle or its supplement, not the internal angle.

Let's apply this method step-by-step.

Determine the side vectors of the triangle.

Given the position vectors of the vertices:

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

$\vec{B} = \hat{i} - 3\hat{j} - 5\hat{k}$

$\vec{C} = 3\hat{i} - 4\hat{j} - 4\hat{k}$

We find the vectors representing the sides. For angle A, we need $\vec{AB}$ and $\vec{AC}$ . For angle B, we need $\vec{BA}$ and $\vec{BC}$ . For angle C, we need $\vec{CA}$ and $\vec{CB}$ .
- $\vec{AB} = \vec{B} - \vec{A} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
- $\vec{AC} = \vec{C} - \vec{A} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = (3-2)\hat{i} + (-4-(-1))\hat{j} + (-4-1)\hat{k} = \hat{i} - 3\hat{j} - 5\hat{k}$
- $\vec{BC} = \vec{C} - \vec{B} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) = (3-1)\hat{i} + (-4-(-3))\hat{j} + (-4-(-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$
We also need the reverse vectors for other angles:
- $\vec{BA} = -\vec{AB} = \hat{i} + 2\hat{j} + 6\hat{k}$
- $\vec{CA} = -\vec{AC} = -\hat{i} + 3\hat{j} + 5\hat{k}$
- $\vec{CB} = -\vec{BC} = -2\hat{i} + \hat{j} - \hat{k}$
Calculate the magnitudes of the side vectors.

The magnitude of a vector $x\hat{i} + y\hat{j} + z\hat{k}$ is $\sqrt{x^2 + y^2 + z^2}$ .
- $|\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$
- $|\vec{AC}| = \sqrt{(1)^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$
- $|\vec{BC}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
Note that $|\vec{BA}| = |\vec{AB}| = \sqrt{41}$ , $|\vec{CA}| = |\vec{AC}| = \sqrt{35}$ , and $|\vec{CB}| = |\vec{BC}| = \sqrt{6}$ .
Calculate angle A.

We use vectors $\vec{AB}$ and $\vec{AC}$ . …

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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 68 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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