Question

11.
The sum of the magnitudes of two vectors is 18.
The magnitude of their resultant is 12. If the
resultant is perpendicular to one of the vectors,
then the magnitudes of the two vectors are
c
​

Answer 1

Answer:

13 , 5 are the magnitudes of vectors,.

Explanation:

Given :

The sum of the magnitudes of two vectors is 18.

The magnitude of their resultant is 12.

If the resultant is perpendicular to one of the vectors,

then the magnitudes of the two vectors are,

Solution :

Let the vectors be X & Y,

Then,

The sum of the magnitudes of two vectors is 18.

⇒ X + Y = 18  ..(i)

_

The magnitude of their resultant is 12.

⇒ $\sqrt{X^2 + Y^2 + 2XYcos\theta} = 12$

⇒ $X^2 + Y^2 + 2XYcos\theta= 12^2$

⇒ $X^2 + Y^2 + 2XYcos\theta= 144$ ..(ii)

_

The resultant is perpendicular to one of the vectors,

Perpendicular ↔ 90°

⇒ $\frac{Ysin\theta}{X+Ycos\theta} = tan 90 \degree = \infty$

⇒ $X + Ycos\theta = 0$

⇒ $X = -Ycos\theta$

Substituting value of X in (ii),

We get,

⇒ $X^2 + Y^2 + 2XYcos\theta= 144$

⇒ $(-Ycos\theta)^2 + Y^2 + 2(-Ycos\theta)Ycos\theta= 144$

⇒ $Y^2cos^2\theta + Y^2 - 2Y^2cos^2\theta= 144$

⇒ $Y^2 - Y^2cos^2\theta= 144$

⇒ $Y^2(1-cos^2\theta)= 144$

⇒ $Y^2(sin^2\theta)= 144$

⇒ $Ysin\theta= 12$ ..(iii)

⇒ $Y^2(sin^2\theta)= 144$

⇒ $sin^2\theta = \frac{144}{Y^2}$

⇒ $1-sin^2\theta = 1-\frac{144}{Y^2}=\frac{Y^2-144}{Y^2}$

⇒ $cos^2\theta = \frac{Y^2-144}{Y^2}$

⇒ $Y^2cos^2\theta =Y^2-144}$

⇒ $X^2 =Y^2-144}$

⇒ $Y^2 - X^2 =144}$

⇒ $(Y + X)(Y-X) =144}$  (by (i) )

⇒ $18(Y - X) = 144$

⇒ $(Y - X) = \frac{144}{18} = 8$ ...(iv)

By adding (i) & (iv)

⇒ $(X+Y)+(Y-X) =18 + 8$

⇒ $2Y = 26$

⇒ Y = 13

_

By substituting value of Y in (i),

We get,

⇒ $X+Y=18$

⇒ $X+13=18$

⇒ $X=18-13$

⇒ X = 5

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The magnitudes of 2 vectors are , 13 & 5,.

Answer 2

Answer:

A=5N and B=13N

Explanation:

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