Question

If \(\vec{a}\) and \(\vec{b}\) are two non zero vectors such that \(|\vec{a}|\)=10, \(|\vec{b}|=2\) and \(\vec{a}.\vec{b}=12\), then value of \(|\vec{a}\times\vec{b}|\) is :

• A. 5
• B. 10
• C. 14
• D. 16

Answer 1

The Correct Option is D

To find the magnitude of the cross product \( |\vec{a}\times\vec{b}| \), we use the formula \( |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). We are given:

• \(|\vec{a}|=10\)
• \(|\vec{b}|=2\)
• \(\vec{a}\cdot\vec{b}=12\)

We can find \(\cos\theta\) using the dot product formula: \(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\).

Substituting the given values, we have:

\(12 = 10 \times 2 \times \cos\theta\)

\(\cos\theta = \frac{12}{20} = \frac{3}{5}\)

We know \(\sin^2\theta + \cos^2\theta = 1\), so:

\(\sin^2\theta = 1 - \cos^2\theta\)

\(\sin^2\theta = 1 - \left(\frac{3}{5}\right)^2\)

\(\sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\)

\(\sin\theta = \sqrt{\frac{16}{25}} = \frac{4}{5}\)

Now, calculate the cross product magnitude:

\(|\vec{a}\times\vec{b}| = 10 \times 2 \times \frac{4}{5}\)

\(|\vec{a}\times\vec{b}| = 20 \times \frac{4}{5}\)

\(|\vec{a}\times\vec{b}| = 16\)

Thus, the value of \(|\vec{a}\times\vec{b}|\) is 16.