Verified Solution • 2025–26 Session moderate

Q6.4: NCERT Physics - Part 1 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.4 Solution: Show that the area of the triangle contained between the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently write the triangle area as $\frac12|\vec a||\vec b|$ or forget that $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$, thus either omitting the sine or dividing by two twice. Remember: the cross‑product magnitude already incorporates the sine of the angle, so the triangle’s area is $\frac12|\vec a\times\vec b|$.

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a _ b.

Step-by-Step Verified Solution

**Answer:** Area =

\frac{1}{2}|\vec{a} \times \vec{b}|

**Step-by-Step Explanation:** Let the vectors

\vec{a}

and

\vec{b}

form two sides of a triangle. The area of the triangle is:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} |\vec{a}| (|\vec{b}| \sin\theta) = \frac{1}{2} |\vec{a} \times \vec{b}|

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All Questions in this Chapter

Q6.1 Q6.2 Q6.3 Q6.4 Q6.5 Q6.6 Q6.7 Q6.8 Q6.9 Q6.10 Q6.11 Q6.12 Q6.13 Q6.14 Q6.15 Q6.16 Q6.17

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