Let $\hat{a}$ and $\hat{b}$ be two unit vectors and $\theta$ be the angle between them such that $\sin \theta = \frac{3}{5}$. Then $\hat{a} \cdot \hat{b}$ is equal to: (A) $\pm \frac{3}{5}$ (B) $\pm \frac{3}{4}$ (C) $\pm \frac{4}{5}$ (D) $\pm \frac{4}{3}$

Let hata and hatb be two unit vectors and θ be the angl — CBSE Class 12 Mathematics 2024 PYQ

Question

\hat{a}

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

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Concept understanding — Dot Product Sign

The Dot Product Sign: What Does It Tell You?

Imagine pushing a heavy box across the floor at an angle — partly forward, partly sideways. The dot product tells you how much of your effort actually moves the box forward.

Its sign — positive, negative, or zero — is the simplest possible answer to one question: Are these two vectors working together or against each other?

The Intuition First

Take two arrows (vectors) and place their tails together. The sign depends entirely on the angle between them:

Positive → the vectors point in roughly the same direction (angle less than $90^\circ$ ). They reinforce each other.
Zero → the vectors are perpendicular (exactly $90^\circ$ ). One has no effect in the direction of the other.
Negative → the vectors point in roughly opposite directions (angle greater than $90^\circ$ ). They oppose each other.

Note

This is not about "good" or "bad" — it's about alignment. A negative dot product is just as useful as a positive one.

The Precise Statement

For two vectors $\vec{a}$ and $\vec{b}$ in any dimension:

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

where $\theta$ is the angle between them (between $0^\circ$ and $180^\circ$ ).

Since $|\vec{a}|$ and $|\vec{b}|$ are always positive lengths, the sign is entirely determined by $\cos\theta$ :

Angle $\theta$	$\cos\theta$	Dot product sign
$0^\circ \leq \theta < 90^\circ$	Positive	Positive
$\theta = 90^\circ$	Zero	Zero
$90^\circ < \theta \leq 180^\circ$	Negative	Negative

\text{sign}(\vec{a} \cdot \vec{b}) = \text{sign}(\cos\theta)

Why This Matters in Exams

Check perpendicularity — If $\vec{a} \cdot \vec{b} = 0$ , the vectors are orthogonal. A common exam shortcut.
Determine the angle type — Positive means acute, negative means obtuse, zero means right angle. …

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The dot product of two unit vectors equals $\cos\theta$ , and $\cos\theta = \pm \frac{4}{5}$ when $\sin\theta = \frac{3}{5}$ . So the answer is $\pm \frac{4}{5}$ .

The dot product of two unit vectors $\hat{a}$ and $\hat{b}$ is defined as $\hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos\theta$ . Since both are unit vectors, their magnitudes are 1, so $\hat{a} \cdot \hat{b} = \cos\theta$ . The problem gives $\sin\theta = \frac{3}{5}$ , and asks for $\hat{a} \cdot \hat{b}$ , which is $\cos\theta$ .

The key insight: the sign of $\cos\theta$ is not fixed by $\sin\theta$ alone. The angle $\theta$ could be in the first quadrant (where both sine and cosine are positive) or in the second quadrant (where sine is positive but cosine is negative). So we must consider both possibilities.

Use the Pythagorean identity. For any angle $\theta$ , we have $\sin^2\theta + \cos^2\theta = 1$ . Substituting $\sin\theta = \frac{3}{5}$ :

\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1

\frac{9}{25} + \cos^2\theta = 1

\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}

Take the square root.

\cos\theta = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}

The $\pm$ is essential: $\cos\theta$ could be $+\frac{4}{5}$ or $-\frac{4}{5}$ , depending on which quadrant $\theta$ lies in.

Relate back to the dot product. …

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

CBSE 2024Set 65/1/11 markMCQ
$\hat{a}$
›Reveal solutionSolution
The dot product of two unit vectors equals $\cos\theta$ , and $\cos\theta = \pm \frac{4}{5}$ when $\sin\theta = \frac{3}{5}$ . So the answer is $\pm \frac{4}{5}$ .

The dot product of two unit vectors $\hat{a}$ and $\hat{b}$ is defined as $\hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos\theta$ . Since both are unit vectors, their magnitudes are 1, so $\hat{a} \cdot \hat{b} = \cos\theta$ . The problem gives $\sin\theta = \frac{3}{5}$ , and asks for $\hat{a} \cdot \hat{b}$ , which is $\cos\theta$ .

The key insight: the sign of $\cos\theta$ is not fixed by $\sin\theta$ alone. The angle $\theta$ could be in the first quadrant (where both sine and cosine are positive) or in the second quadrant (where sine is positive but cosine is negative). So we must consider both possibilities.

Use the Pythagorean identity. For any angle $\theta$ , we have $\sin^2\theta + \cos^2\theta = 1$ . Substituting $\sin\theta = \frac{3}{5}$ :

$\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1$ $\frac{9}{25} + \cos^2\theta = 1$ $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$

Take the square root.

$\cos\theta = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$
The $\pm$ is essential: $\cos\theta$ could be $+\frac{4}{5}$ or $-\frac{4}{5}$ , depending on which quadrant $\theta$ lies in.

Relate back to the dot product. …
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