If $|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=144$ and $|\vec{a}|=4$, then $|\vec{b}|$ is equal to ________.

> [!TLDR] > The key idea is to use the identity $|\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2$. Substituting the given values gives $|\vec{b}| = 3$. This problem is a beautiful example of how vector algebra has a built-in Pythagorean relationship. The cross product magnitude depends on $\sin\theta$, and the dot product magnitude depends on $\cos\theta$. When you square and add them, the angle dependence vanishes — leaving only the product of the squared magnitudes. Let’s walk through it. 1. **Recall the fundamental formulas.** For any two vectors $\vec{a}$ and $\vec{b}$ with angle $\theta$ between them: $$ |\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\,\sin\theta $$ $$ |\vec{a}\cdot\vec{b}| = |\vec{a}|\,|\vec{b}|\,\cos\theta $$ 2. **Square both and add.** $$ |\vec{a}\times\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2\theta $$ $$ |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2\theta $$ Adding them: $$ |\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 (\sin^2\theta + \cos^2\theta) $$ 3. **Use the identity $\sin^2\theta + \cos^2\theta = 1$.** This is the crucial simplification — the angle disappears entirely: $$ |\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 $$ > [!FORMULA] > $$ > |\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 > $$ 4. **Plug in the given values.** We are told: $$ |\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = 144 $$ and $|\vec{a}| = 4$. So: …

If |veca×vecb|²+|veca·vecb|²=144 and |veca|=4, then |vecb| is equal to ____.

NCERT Exemplar · Q23

|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=144

Yanam CbseShort· 1mImportance★★★★★

Appeared in past exams:COMEDK 2024· Set 2024-M· 1mexactKCET 2023· Set A-2· 1mexactMHT-CET 2021· Set pcm-2021-09-21-E· 2mexactKCET 2018· Set A-1· 1mexact

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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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The key idea is to use the identity $|\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2$ . Substituting the given values gives $|\vec{b}| = 3$ .

This problem is a beautiful example of how vector algebra has a built-in Pythagorean relationship. The cross product magnitude depends on $\sin\theta$ , and the dot product magnitude depends on $\cos\theta$ . When you square and add them, the angle dependence vanishes — leaving only the product of the squared magnitudes.

Let’s walk through it.

Recall the fundamental formulas. For any two vectors $\vec{a}$ and $\vec{b}$ with angle $\theta$ between them:

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

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

Square both and add.

|\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

Adding them:

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

Use the identity $\sin^2\theta + \cos^2\theta = 1$ . This is the crucial simplification — the angle disappears entirely:

|\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2

|\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2

Plug in the given values. We are told:

|\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = 144

and $|\vec{a}| = 4$ . So: …

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Q.If $|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=144$ and $|\vec{a}|=4$ , then $|\vec{b}|$ is equal to ________.

Concept understanding — Vector Magnitude Properties

Vector Magnitude Properties

Definition

The four key properties

Method: Using the Lagrange Identity to Find an Unknown Magnitude

Steps

Common Mistakes

Q.If ∣a⃗×b⃗∣2+∣a⃗⋅b⃗∣2=144|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=144∣a×b∣2+∣a⋅b∣2=144 and ∣a⃗∣=4|\vec{a}|=4∣a∣=4, then ∣b⃗∣|\vec{b}|∣b∣ is equal to ________.

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Q.If $|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=144$ and $|\vec{a}|=4$ , then $|\vec{b}|$ is equal to ________.

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