Question

The condition vector (a.b)²=vector(a²×b²) is satisfied when:-

Answer 1

see attachment......hope it will help u

Answer 2

Concept:

We need to know what scalar and cross products are.

The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

a×b=|a| |b| sin∅

Scalar product:  It is equal to the product of the magnitudes of the two vectors and the cosine of the angle between them.

a.b=ab cos∅

∅ is the angle between a and b

Given:

The following equation is given:

$\overline{(a.b)^{2} }=\overline{a^{2} \times b^{2}} }$

To find:

we need to find which condition satisfies the above equation.

Solution:

Now, we know

$\overline{(a.b)^{2} }=(abCos\theta)^{2} \\ \overline{(a.b)^{2} }=a^{2} b^{2}cos^{2} \theta \ \ \ \ \ \ \------(1)$

Also,

$\overline{a^{2} \times b^{2}} }=(a \times b \times sin \theta)^{2}$

$\overline{a^{2} \times b^{2}} }=a ^{2} b ^{2} sin \theta^{2}$       --------------(2)

Substituting values equations (1) and (2)

$a^{2} b^{2} cos^{2}\theta =a^{2} b^{2} sin^{2}\theta$

⇒$cos^{2}\theta =sin^{2} \theta$

⇒$\frac{sin^{2}\theta}{cos^{2} \theta}=1$

⇒$tan^{2}\theta =1$⇒$tan\theta=$±1

Therefore, value of $\theta=\frac{\pi }{4}$ will be the best suitable answer.