Question

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$ Then $(\vec{a} \cdot \vec{b})^2$ is equal to

Hint:

When solving problems involving vectors, use the identity \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \) to relate the dot and cross product magnitudes. This identity helps simplify many vector algebra problems.

Answer 1

Correct Answer: 36

The correct answer is 36.
$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}|\vec{a}\times \vec{b}|=\sqrt{48}$
$|\vec{a}\times \vec{b}{|}^2+|\vec{a}\cdot \vec{b}{|}^2=|\vec{a}{|}^2\times |\vec{b}{|}^2$
$\Rightarrow (\vec{a}\cdot \vec{b}{)}^2=84-48=36$

Answer 2

Step 1: Use the vector identity for the magnitude of the cross product and dot product: \[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2. \] This identity relates the cross product and dot product of two vectors. 
Step 2: Substitute the given values for the magnitudes of the vectors: \[ |\vec{a} \times \vec{b}|^2 = (\sqrt{48})^2 = 48, \quad |\vec{a}|^2 = (\sqrt{14})^2 = 14, \quad |\vec{b}|^2 = (\sqrt{6})^2 = 6. \] Here, we calculate the squared magnitudes of the vectors \( \vec{a} \) and \( \vec{b} \), as well as the squared magnitude of their cross product. Step 3: Substitute these values into the equation: \[ 48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6 = 84. \] Now, substitute the known values for the cross product and dot product magnitudes into the identity. 
Step 4: Solve for \( (\vec{a} \cdot \vec{b})^2 \): \[ (\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36. \] This gives us the squared value of the dot product of the vectors \( \vec{a} \) and \( \vec{b} \).