**A, B and C are three non-collinear, non co-planar vectors. What can you say about direction of A * (B * C)?**

3 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

7 months ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the direction of (B $* C$ ) will be perpendicular the plane containing B and B by right hand rule. A $* (B * C)$ will lie in the plane of B and C and is perpendicular to vector A.

Upvote

Similar Questions for you

A ball is projected from ground at angle q with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s^–2]

Alok Kumar Singh

Please find the solution below:

An electron in a research apparatus follows a circle path. On the electron’s first circuit of the apparatus, its speed is v₀ and the radius of its circular path is R. Each time it makes on circuit, it passes a short region where it receives a “kick” and gains an additional speed of $\frac{v_{0}}{100} .$ The electron follows a circular path such that the magnitude of its acceleration is always the same. What is the radius of the circular path after the electron has received 10 kicks?

Vishal Baghel

$a = \frac{v_{0}^{2}}{R}$ $\frac{_{}^{}}{}$

after 10 kicks, $v = v_{0} + \frac{10 v_{0}}{100} = 1.1 v_{0}$ $_{} \frac{_{}}{}_{}$

$\Rightarrow R_{1} = 1.21 R$

Two particles start moving from the same point in two different directions. The first particle moves along the positive x-axis with a constant acceleration of 3 m/s². The second particle moves in a straight line making an angle of 60° with the positive x-axis with a constant speed of 24 m/s. Find the time after which the relative velocity of the particles is minimum.

Vishal Baghel

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:


Vishal Baghel

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

A flying disc of radius r is moving with constant speed v? while rotating anticlockwise with angular speed v?/r along a curve PQ as shown below. Radius of curvature of curve at the instant is 4r. The magnitude of acceleration of point A at the instant is (3v₀²)/(nr). Find n.

Vishal Baghel

a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r