Prove that (vec a×vec b)²+(vec a·vec b)²=|vec a|²| – Class 12 Mathematics | ConceptFirst

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(\vec a\times\vec b)^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2

Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 3mImportance★★★★★

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Concept understanding — Vector Magnitude Properties

Vector Magnitude Properties

An arrow has a direction and a length. That length — the straight-line distance from tail to tip — is the magnitude of the vector, written $|\vec{v}|$ or $\|\vec{v}\|$ . It is always non-negative and tells you how much of something there is, ignoring direction.

Definition

Magnitude is the distance from the origin to the point the vector reaches — the Pythagorean theorem in $n$ dimensions:

|\vec{v}| = \sqrt{x^2 + y^2} \quad (\text{2D}), \qquad |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \quad (\text{3D}).

The four key properties

1. Non-negativity.

|\vec{v}| \geq 0, \qquad |\vec{v}| = 0 \iff \vec{v} = \vec{0}.

A length is never negative, and only the zero vector has zero length.

2. Scaling.

|k\vec{v}| = |k|\,|\vec{v}|.

Stretching a vector by $k$ multiplies its length by $|k|$ — the absolute value appears because a negative $k$ flips direction but the length still grows by $|k|$ . E.g. if $|\vec{v}| = 3$ , then $|-2\vec{v}| = 2\times 3 = 6$ .

3. Triangle inequality.

|\vec{u} + \vec{v}| \leq |\vec{u}| + |\vec{v}|.

The direct path is never longer than going the long way: the straight line from $A$ to $C$ is at most the distance $A \to B \to C$ . Equality holds only when $\vec{u}$ and $\vec{v}$ point in exactly the same direction.

4. Dot-product relation.

|\vec{v}|^2 = \vec{v} \cdot \vec{v}.

The squared length equals the vector's dot product with itself, since $\vec{v}\cdot\vec{v} = x^2 + y^2 + z^2$ . This is the workhorse in proofs and in physics (kinetic energy $\tfrac{1}{2}m|\vec{v}|^2$ ). …

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Express both the cross-product-squared and dot-product-squared terms in terms of $|\vec a|,|\vec b|,\theta$ , then use the Pythagorean identity to combine them.

Let $\theta$ be the angle between $\vec a$ and $\vec b$ . Recall:

$|\vec a\times\vec b|=|\vec a|\,|\vec b|\sin\theta \qquad\Rightarrow\qquad (\vec a\times\vec b)^2=|\vec a|^2|\vec b|^2\sin^2\theta$

(here $(\vec a\times\vec b)^2$ means the square of the magnitude of the cross-product vector), and

$\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta \qquad\Rightarrow\qquad (\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2\cos^2\theta.$

Adding these: …

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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 29 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$|\vec{a}| = 5$
›Reveal solutionSolution
Scalar multiplication scales a vector's magnitude by the absolute value of the scalar; since $|\lambda \vec{a}| = |\lambda| \cdot |\vec{a}|$ , we find the extreme values of $|\lambda|$ over $[-2, 1]$ and multiply by $5$ .

When you multiply a vector by a scalar, the magnitude of the resulting vector depends only on the absolute value of that scalar. This is because direction reversals (negative scalars) don't affect length—they just flip the arrow. The fundamental property is:
$|\lambda \vec{a}| = |\lambda| \cdot |\vec{a}|$
Given $|\vec{a}| = 5$ , we have $|\lambda \vec{a}| = 5|\lambda|$ . The problem now reduces to finding the maximum and minimum values of $|\lambda|$ as $\lambda$ ranges over $[-2, 1]$ .

Finding the extreme values of $|\lambda|$ :

Understand the absolute value function on the interval. For $\lambda \in [-2, 1]$ , the function $|\lambda|$ equals $-\lambda$ when $\lambda < 0$ and equals $\lambda$ when $\lambda \ge 0$ . The graph is V-shaped with its vertex at $\lambda = 0$ .

Identify the minimum. The absolute value $|\lambda|$ is smallest at $\lambda = 0$ , where $|\lambda| = 0$ . Therefore, the minimum value of $|\lambda \vec{a}|$ is $5 \cdot 0 = 0$ .

Identify the maximum. Since $|\lambda|$ increases as we move away from zero in either direction, we check the endpoints of the interval: …
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CBSE 2026Set 65/3/11 markMCQ
$\vec{a}$
›Reveal solutionSolution

The dot product satisfies $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$ , and since $\cos\theta \leq 1$ , we always have $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$ . The correct option is (A).

The key here is to recall the geometric definitions of the dot product and cross product, and the triangle inequality for vector addition. Each option claims an inequality or equality that must hold for any two vectors — so we test each against the general formulas.

1. Option (A): $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$

The dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$ , where $\theta$ is the angle between the vectors. Since $\cos\theta$ ranges from $-1$ to $1$ , the product $|\vec{a}| |\vec{b}| \cos\theta$ can be as large as $|\vec{a}| |\vec{b}|$ (when $\cos\theta = 1$ ) and as small as $-|\vec{a}| |\vec{b}|$ (when $\cos\theta = -1$ ). Therefore, $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$ is always true — the dot product never exceeds the product of magnitudes.

Tip

The inequality $\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}|$ is actually a form of the Cauchy–Schwarz inequality, which holds for any inner product space.

2. Option (B): $|\vec{a} + \vec{b}| \geq |\vec{a}| + |\vec{b}|$

This is the reverse of the triangle inequality. The actual triangle inequality states $|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$ , with equality only when the vectors point in the same direction. So the given statement is false — for example, take $\vec{a} = (1,0)$ and $\vec{b} = (-1,0)$ ; then $|\vec{a} + \vec{b}| = 0$ , which is not $\geq 2$ .

3. Option (C): $|\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}|$

This would require the vectors to be parallel and pointing in the same direction, with $|\vec{a}| \geq |\vec{b}|$ . In general, $|\vec{a} - \vec{b}|$ depends on the angle between them. For instance, if $\vec{a} = (1,0)$ and $\vec{b} = (0,1)$ , then $|\vec{a} - \vec{b}| = \sqrt{2}$ , but $|\vec{a}| - |\vec{b}| = 0$ . So this is not always true. …
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CBSE 2026Set CX1 mark
$\vec{a}$
›Reveal solutionSolution

The dot product expands to $|\vec x|^{2}-|\vec a|^{2}=12$ ; with a unit $\vec a$ this gives $|\vec x|=\sqrt{13}$ .

Concept: For any vectors, $(\vec p-\vec q)\cdot(\vec p+\vec q)=\vec p\cdot\vec p-\vec q\cdot\vec q=|\vec p|^{2}-|\vec q|^{2}$ (the cross terms cancel).

$(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=|\vec{x}|^{2}-|\vec{a}|^{2}=12.$ …

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CBSE 2026Set A1 markMCQ
$|3\vec{i} + 4\vec{j} + 7\vec{k}| =$
(a) $\sqrt{14}$
(b) $\sqrt{74}$
(c) $\sqrt{61}$
(d) $\sqrt{94}$
›Reveal solutionSolution

Magnitude of $a\vec{i}+b\vec{j}+c\vec{k}$ is $\sqrt{a^2+b^2+c^2}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{1}{\sqrt{3}}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}+\dfrac{1}{\sqrt{3}}\hat{k}$
(a) 0
(b) 3
(c) 1
(d) -1
›Reveal solutionSolution

Use the magnitude formula $|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}$ on the given vector.

Given $\vec v = \dfrac{1}{\sqrt3}\hat{i}+\dfrac{1}{\sqrt3}\hat{j}+\dfrac{1}{\sqrt3}\hat{k}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A(2\hat i - \hat j + \hat k)$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Compute the three squared side-lengths using position vectors and check the Pythagoras relation.

$A(2,-1,1)$ , $B(1,-3,-5)$ , $C(3,-4,-4)$ .

$\overrightarrow{AB}=B-A=(-1,-2,-6)$ , so $|\overrightarrow{AB}|^2 = 1+4+36 = 41$ .

$\overrightarrow{BC}=C-B=(2,-1,1)$ , so $|\overrightarrow{BC}|^2 = 4+1+1 = 6$ .

$\overrightarrow{AC}=C-A=(1,-3,-5)$ , so $|\overrightarrow{AC}|^2 = 1+9+25 = 35$ .

Check: $|\overrightarrow{BC}|^2+|\overrightarrow{AC}|^2 = 6+35 = 41 = |\overrightarrow{AB}|^2$ .

…

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CBSE 2025Set E1 markMCQ
$|\vec{i} - \vec{j} - 3\vec{k}| =$
(a) $11$
(b) $\sqrt{11}$
(c) $\sqrt{7}$
(d) $\sqrt{10}$
›Reveal solutionSolution

The magnitude of $\vec{i}-\vec{j}-3\vec{k}$ is $\sqrt{11}$ .

For a vector $a\vec{i}+b\vec{j}+c\vec{k}$ the magnitude is $\sqrt{a^2+b^2+c^2}$ .

Here $a=1,\ b=-1,\ c=-3$ , so …

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CBSE 2025Set E1 markMCQ
$(4\vec{i} + 3\vec{j})^2 =$
(a) $7$
(b) $19$
(c) $25$
(d) $49$
›Reveal solutionSolution

The square of a vector means its dot product with itself, i.e. $|\vec{a}|^2$ , which is $25$ .

For any vector $\vec{a}$ , $\vec{a}^2=\vec{a}\cdot\vec{a}=|\vec{a}|^2$ .

…

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CBSE 2024Set 65/3/11 markMCQ
$|\vec{a}| = 2$
›Reveal solutionSolution
The magnitude of a scalar multiple is the absolute value of the scalar times the magnitude of the vector; since $|k|$ ranges from $0$ to $3$ when $-3 \le k \le 2$ , we have $|k\vec{a}| \in [0,\,6]$ .

The key concept here is how scalar multiplication affects vector magnitude. When you multiply a vector by a scalar, the magnitude scales by the absolute value of that scalar—the direction may flip (if the scalar is negative), but magnitude is always non-negative.

The fundamental property is:
$|k\vec{a}| = |k| \cdot |\vec{a}|$
This tells us that to find the range of $|k\vec{a}|$ , we need to find the range of $|k|$ and multiply by the fixed magnitude $|\vec{a}| = 2$ .

Now let's trace through the reasoning:

Identify the range of the scalar $k$ .

We're given $-3 \le k \le 2$ .

Find the range of $|k|$ .

The absolute value function $|k|$ measures distance from zero. On the interval $[-3,\,2]$ :

At $k = 0$ , we have $|k| = 0$ (the minimum).

At $k = -3$ , we have $|k| = 3$ .

At $k = 2$ , we have $|k| = 2$ .

The maximum value of $|k|$ occurs at the endpoint farthest from zero, which is $k = -3$ , giving $|k| = 3$ .

Therefore, $|k| \in [0,\,3]$ . …
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CBSE 2024Set D1 markMCQ
$|\vec{i} - \vec{j} - \vec{k}| =$
(a) $\sqrt{3}$
(b) $3$
(c) $\sqrt{2}$
(d) $2$
›Reveal solutionSolution

Magnitude = square root of sum of squares of components.

…

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CBSE 2023Set 65/1/11 markMCQ
$\vec{a} + \vec{b} = \hat{i}$
›Reveal solutionSolution

$\vec{b} = \hat{i} - \vec{a} = -\hat{i} + 2\hat{j} - 2\hat{k}$ , so $|\vec{b}| = 3$ — option (B).

We are given $\vec{a} + \vec{b} = \hat{i}$ and $\vec{a} = 2\hat{i} - 2\hat{j} + 2\hat{k}$ .

Isolate $\vec{b}$ :

$\vec{b} = \hat{i} - \vec{a} = \hat{i} - (2\hat{i} - 2\hat{j} + 2\hat{k}) = -\hat{i} + 2\hat{j} - 2\hat{k}.$ …
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CBSE 2023Set E1 markMCQ
$|\vec{i}-2\vec{j}+2\vec{k}| =$
(a) $3$
(b) $6$
(c) $7$
(d) $5$
›Reveal solutionSolution

The magnitude is $\sqrt{1^2+(-2)^2+2^2}=3$ .

The magnitude of a vector $a\vec{i}+b\vec{j}+c\vec{k}$ is $\sqrt{a^{2}+b^{2}+c^{2}}$ .

Here $a=1,\ b=-2,\ c=2$ , so …

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