KCET 2025 Math Paper

To find the value of $|\vec{a} \times \vec{b}|$, remember that the relationship with the dot product and magnitudes is given by:$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin\theta$where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$.We use the formula for the dot product to calculate $\cos\theta$ : $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$Given:$\;\left|\overset{\rightarrow}{a}\right|=10$$\;\left|\overset{\rightarrow}{b}\right|=2$$\;\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b}=12$ Substitute these values into the dot product formula:$12 = 10 \times 2 \times \cos\theta$Thus,$12 = 20 \cos\theta \Rightarrow \cos\theta = \frac{12}{20} = \frac{3}{5}$Since $\sin^2 \theta + \cos^2 \theta = 1$, we find $\sin\theta$ : $\;\sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$\;\sin\theta = \frac{4}{5}$Now, substitute back to find the magnitude of the cross product:$\left|\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}\right| = 10 \times 2 \times \frac{4}{5} = 20 \times \frac{4}{5} = 16$Therefore, the value of $\left|\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}\right|$ is 16 .