Question

The magnitude of the vector product of two
vectors is √3 times then scalar product. The
angle between vectors is
(a) ???? (b) ????
4 6

(c) ???? (d) ????
3 2

Answer 1

Answer:

Let us consider the two vector be a and b and ¥ the angle between them

The scalar product of a and b = 2√3

a . b = 2√3

ab cos¥= 2√3 ………………………….(1)

The vector product of a and b = 2

a × b = 2

ab sin ¥ =2 ………………………………..(2)

Now dividing equation (2) with equation (1) we get

tan ¥ = 1/ √3

¥ = 30 ° is required angle

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

We know that ,

• Vector product = ABsin(x)
• Scalar product = ABcos(x)

By the given condition ,

$\sf \hookrightarrow ABSin(x) = \sqrt{3} \times ABCos(x) \\ \\ \sf \hookrightarrow</p><p> \frac{Sin(x) }{Cos(x)} = \sqrt{3} \\ \\ \sf \hookrightarrow</p><p>Cot(x) = \sqrt{3} \\ \\ \sf \hookrightarrow</p><p>Cot(x) = Cos(30) \\ \\ \sf \hookrightarrow</p><p>x = 30$

Hence , the angle between two vectors is 30° degrees

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