Question

If the velocity of a particle is 2i^+3j^-4^ and it's acceleration is -i^+2j^+k^ and angle btw them is nπ÷4 the value of n is

Answer 1

Answer:

Given:

2 vectors have been provided.

$\vec v = 2 \hat i + 3 \hat j - 4 \hat k$

$\vec a = -1 \hat i + 2 \hat j + 1 \hat k$

Angle between the vectors be θ

$\boxed{ \theta = \dfrac{n\pi}{4} }$

To find:

Value of n

Concept:

In this type of questions , we will try to use Dot Product or Scalar product to find out the angle between the vectors .

Calculation:

$\vec v. \vec a = | \vec v| \times | \vec a| \times \cos( \theta)$

$= > ( 2 \hat i + 3 \hat j - 4 \hat k).( - \hat i + 2 \hat j + \hat k)= \sqrt{29} \times \sqrt{6} \times \cos( \theta)$

$= > (-2 + 6 - 4) = \sqrt{29} \times \sqrt{6} \times \cos( \theta)$

$= > \cos( \theta) = 0$

$= > \cos( \frac{n\pi}{4} ) = 0$

$= > \dfrac{n\pi}{4} = \dfrac{\pi}{2}$

$= > n = \dfrac{4}{2}$

$= > n = 2$

So final answer :

$\boxed{\boxed{ \red{ \huge{ \bold{ \underline{n = 2}}}}}}$

Answer 2

$</p><p>\huge{\tt{\pink {Hello!!! }}}$

$</p><p>\tt{\red{vec\: v=2 \hat i + 3 \hat j - 4 \hat k }}$

$</p><p>\tt{\purple{\vec \:a = -1 \hat i + 2 \hat j + 1 \hat k }}$

$</p><p>\tt{\blue{Let \:Angle\: between \:the \:vectors\: be \: \phi }}$

$</p><p>\tt{\huge{\violet{\boxed{\boxed{ \phi= \frac{n\pi}{4} } }}}}$

$</p><p>\tt{\fbox{\fbox{\rightarrow {\mathfrak {\red{ Dot\: Product\: or\: Scalar \:product \: - }}}}}}$

$</p><p>\vec v. \vec a = | \vec v| \times | \vec a| \times \cos( \phi)$

$</p><p>= > ( 2 \hat i + 3 \hat j - 4 \hat k).( - \hat i + 2 \hat j + \hat k)= \sqrt{29} \times \sqrt{6} \times \cos( \theta)$

$</p><p>= > (-2 + 6 - 4) = \sqrt{29} \times \sqrt{6} \times\cos( \phi)$

$</p><p>= > \cos( \phi) = 0$

$= > \cos( \frac{n\pi}{4} ) = 0$

$</p><p>= > \dfrac{n\pi}{4} = \dfrac{\pi}{2}$

$</p><p>\purple{\boxed{\boxed{\huge{\mathfrak{= > n = \frac{4}{2} = 2 }}}}}$