Question

If \(\rm |\vec{a}\times \vec{b}|^2 + |\vec{a}\cdot \vec{b}|^2=100\) and \(\rm |\vec{a}|=5\), then what is \(\rm |\vec{b}|\) equal to?
1. 3
2. 4
3. 2
4. None of these

Answer 1

Correct Answer - Option 3 : 2

 Concept:

Let the two vectors be \(\rm \overrightarrow{a}\)  and \(\rm \overrightarrow{b}\) then

The dot product of two vectors is given by: \(\rm \overrightarrow{a} . \overrightarrow{b} = |a||b| \cosθ\) 

The cross product of the two vectors is given by: \(\rm \overrightarrow{a} × \overrightarrow{b} = |a||b| \sinθ\)

Where \(\rm {|\vec a|}\)= Magnitude of vectors a and \(\rm {|\vec b|}\) = Magnitude of vectors b and θ is the angle between a and b.

Trigonometric identities: sin² θ + cos2 θ = 1

 

Calculation:

The given equation is,

\(\rm |\vec{a}× \vec{b}|^2 + |\vec{a}\cdot \vec{b}|^2=100\) this can be simplified as,

⇒ \(\rm {|\vec a|}^2 {|\vec b|}^2\)sin2 θ +  \(\rm {|\vec a|}^2 {|\vec b|}^2\)​cos2 θ = 100

⇒ \(\rm {|\vec a|}^2 {|\vec b|}^2\) [sin2 θ + cos2 θ] = 100

⇒ \(\rm {|\vec a|}^2 {|\vec b|}^2\) = 100

⇒ 25 × \(\rm {|\vec b|}^2\) = 100

⇒ \(\rm {|\vec b|}^2\) = 4

∴ \(\rm {|\vec b|}\) = ± 2

So, \(\rm {|\vec b|}\) = 2