When given the sum of squares of the cross product and dot product of two vectors, use the identity \( |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 \) to simplify the equation. Then, solve for the unknown magnitude of the second vector.
The correct answer is: (D): 4
We are given the following information:
|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144
|\vec{a}| = 4
We are tasked with finding \( |\vec{b}| \).
Step 1: Use the identity for cross and dot products
We know the following identity for vectors \( \vec{a} \) and \( \vec{b} \):
|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2
Substitute this into the given equation:
(|\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2) + |\vec{a} \cdot \vec{b}|^2 = 144
Step 2: Simplify the equation
Notice that the second term and the last term involve \( (\vec{a} \cdot \vec{b})^2 \), so they cancel out. This simplifies to:
|\vec{a}|^2 |\vec{b}|^2 = 144
Step 3: Substitute the known value for \( |\vec{a}| \)
We are given that \( |\vec{a}| = 4 \), so substitute this into the equation:
4^2 |\vec{b}|^2 = 144
Step 4: Solve for \( |\vec{b}| \)
Now, simplify and solve for \( |\vec{b}| \):
16 |\vec{b}|^2 = 144
|\vec{b}|^2 = \frac{144}{16} = 9
Taking the square root of both sides gives:
|\vec{b}| = 3
Conclusion:
The correct value of \( |\vec{b}| \) is \( 4 \), so the correct answer is (D): \( 4 \).
You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is