The magnitude of scalar product of two vectors is 8 and the magnitude of their vector product is 8√3. If the angle between them is α, find 1/|cosα|.
Physics · Mixed Chapter
JEE Main Practice

Question

The magnitude of scalar product of two vectors is 8 and the magnitude of their vector product is 8√3. If the angle between them is α, find 1/|cosα|.

G1 Physics · Accepted Answer

Step 1: |A⃗·B⃗|=AB|cosα|=8; |A⃗×B⃗|=AB|sinα|=8√3. Step 2: |tanα|=(8√3)/8=√3, hence |cosα|=1/2. Step 3: The requested reciprocal is 1/(1/2)=2.