Question

v/ 45.
The sum of magnitudes of two forces acting at a
point is 16N. If the resultant force is 8N and its
direction is perpendicular to smaller force, then the
forces are
(1) 6N and 10N (2) 8N and 8N
(3) 4N and 12N (4) 2N and 14N
46.The vector P = ai + aj+3ſ and Q = aî - 29-ki
are perpendicular to each other. The positive value
of a is
713 (2) 2 (3) 1 (4) zero
46.
33

pls pls solve its urgent​

Answer 1

45)

Let the two vectors be $\sf \vec{A}\ and\ \vec{B}$, where $\sf \vec A > \vec B$

Given, their sum is 16. This means that the sum of their Magnitude is 16.

⇒ |A| + |B| = 16

Also, given that the resultant is 8 and the resultant is perpendicular to the smaller force.

We know that the angle between resultant with respect to a vector is given as

$\sf tan\alpha = \frac{Asin\theta}{B+Acos\theta}$

where, α is the angle between resultant and vector B and θ is the angle between vector A and vector B.

Here, given α = 90° (since, Resultant is ⊥ vector B)

tan(90°) = undefined.

But, tan(90°) = sin(90)°/cos(90°)

⇒ tan(90°) = 1/0

So, put the values and we get

$\sf \frac{1}{0} = \frac{Asin\theta}{B + Acos\theta}\\\\\implies B + Acos\theta = 0\\\\\implies Acos\theta = -B \\\\\implies 2ABcos\theta = -2B^2$

(by multiplying 2B on both sides)

Now, Resultant = $\sf \sqrt{A^2+B^2+2ABcos\theta}$

So,

$\sf 8 = \sqrt{A^2 + B^2 + 2ABcos\theta}\\\\\implies 64 = A^2 + B^2 + 2ABcos\theta\\\\\implies 64 = A^2 + B^2- 2B^2\\\\(since, 2ABcos\theta = -2B^2)\\\\\implies 64 = A^2 - B^2 \\\\\implies 64 = (A + B)(A-B) \\\\\implies 64 = (16)(A-B)\\\\(since, A + B =16)\\\\\implies A - B = \frac{64}{16}\\\\\implies A - B = 4$

Now, A + B = 16

A - B = 4

adding the two equations, we get

2A = 20

⇒ A = 10

Substituing the value of A, we get B = 6

Hence, the answer is option 1) 6N and 10N

$\rule{200}2$

46)

Given that P and Q vectors are perpendicular to each other. Hence, their dot product would be 0.

⇒ P.Q = 0

⇒ 0 = (ai + aj + 3k)(ai - 2j - k)

⇒ 0 = (ai)(ai) - (2j)(ai) - (k)(ai) + (aj)(ai) - (aj)(2j) - (aj)(k) + (3k)(ai) - (3k)(2j) - (3k)(k)

⇒ 0 = a² - 2a- 3

(since, i.i = j.j = k.k = 1 and i.j = j.i = i.k = k.i = j.k = k.k = 0)

⇒ a² - 3a + a - 3 = 0

⇒ a(a - 3) + 1(a - 3) = 0

⇒ (a + 1)(a - 3) = 0

So, either a + 1 = 0 or a - 3 = 0

so, a = -1 or a = 3

Now, positive value is asked, hence answer is a = 3.

option 1) a = 3

Answer 2

Explanation:

45) Let one force be X and another be Y.

X = magnitude of larger force

Y = magnitude of smaller force

R = resultant

It's clear from above X = Hypoteneous

Y = Base

R = Perpendicular distance between X and Y

From Pythagoras Theorem

X² = R² + Y²

R² = Y² - X² __(A)

R² = (8)²

R² = 64 ..(1)

X+Y = 16 (Given)

Y = 16-X ..(2)

64 = (16-X)² - X² (From A)

64 = (16)² + X² -2×16X - X²

64 = 256 - 32X

32X = 256 - 64

32X = 192

X = 6

Y = 16 - 6

Y = 10

Option 1) 6N and 10N

46) (ai + aj + 3k).(ai + 2j - 1k)

(a×a) - (a×2) + (3×(-1))

a² - 2a - 3 = 0

a² - 3a + a - 3 = 0

a(a-3) +1(a-3) = 0

(a+1) (a-3) = 0

a + 1 = 0

a = -1

a - 3 = 0

a = +3

Option 1) 3