4.7 Given a + b + c + d = 0, which of the following statements are correct:

(a) a, b, c, and d must each be a null vector

(b) The magnitude of (a + c) equals the magnitude of (b + d)

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

191 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

4.7

(a) This statement is In order to make a + b+ c + d = 0, it is not necessary to have all four given vectors to be null vectors.

(b) a + b + c +d = 0, ( a + c ) = - (b + d)

Taking the magnitude of both LHS and RHS, | a + c| = | -(b + d)| = | b + d|

So the statement is True.

(c) Given a + b + c + d = 0, a = - (b + c + d)

Taking magnitude of both LHS and RHS, |a| = | -( b + c+ d)| = |b + c +d|

|a| $\leq |b| + |c| + | d |$ ……….(1)

Equation (1) shows that the magnitude of a can never be greater than the sum of magnitudes of b,c and d. So the statement is True.

(d) In the given expression a + b + c + d = 0, the sum of 3 vectors

...more

4.7

(a) This statement is In order to make a + b+ c + d = 0, it is not necessary to have all four given vectors to be null vectors.

(b) a + b + c +d = 0, ( a + c ) = - (b + d)

Taking the magnitude of both LHS and RHS, | a + c| = | -(b + d)| = | b + d|

So the statement is True.

(c) Given a + b + c + d = 0, a = - (b + c + d)

Taking magnitude of both LHS and RHS, |a| = | -( b + c+ d)| = |b + c +d|

|a| $\leq |b| + |c| + | d |$ ……….(1)

Equation (1) shows that the magnitude of a can never be greater than the sum of magnitudes of b,c and d. So the statement is True.

(d) In the given expression a + b + c + d = 0, the sum of 3 vectors a, (b+c), d can be zero only if (b+c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle. If a and d are collinear , then it implies that the vector (b+c) is in the line of a and d. This statement holds good only then the sum of all three vectors will be zero. The statement is True.

less

0 Upvotes 0 Downvotes

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