Question

Class 11th,
Two vectors p and q are the sum of 18 and resultant is 12 the resultant is perpendicular to smaller of two vectors find the value of P and Q and angle between them.

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Answer 1

given that,

Two vectors p and q are the sum of 18 and resultant is 12

here,

we have,

sum of magnitude of vector = 18

so,

p + q = 18 ... (1)

and sum of both vectors I. e. resultant = 12

so,

p^-> + q^-> = 12 = R

Now,

ACCORDING TO THE FIGURE,

R² = p² + q² + 2pq cosθ

given R = 12

so,

12² = p² + q² + 2pq cosθ .... (2)

now,

here, since the resultant is perpendicular

so,

value of α = 90°

so,

tanα =

$\large{ \frac{ \large{q \sin \theta }}{ \large{p + qcos \theta}}}$

so,

tan90.=

$\large{ \frac{ \large{q \sin \theta }}{ \large{p+ qcos \theta}}} = \infty$

so,

p + qcosθ = 0

qcosθ = -p

now,

putting the value of qcosθ on (2)

12² = p² + q² + 2pqcosθ

144 = p² + q² + 2p(-p)

144 = p² + q² - 2p²

q² - p² = = 144

using algebraic identity

a² - b² = (a + b)(a - b)

so,

(q + p)(q - p) = 144

now

from eqn (1)

p + q = 18

putting the values,

18(q - p) = 144

q - p = 144/18

q - p = 8 .....(3)

now,

we have,

q + p = 18 ..(1)

q - p = 8 ...(3)

adding of both equations

q + p + q - p = 18 + 8

2q = 26

q = 26/2

q = 13

now,

putting the value of q in (1)

p + q = 18

p + 13 = 18

p = 18 - 13

p = 5

q = 13

now,

putting the value of p and q on (2)

p² + q² + 2pqcosθ = 144

(5)² + (13)³ + 2(5)(13)cosθ = 144

25 + 169 + 130cosθ = 144

194 + 130cosθ = 144

130cosθ = 144 - 194

130cosθ = -50

cosθ = -50/130

cosθ = -5/13

Answer 2

Explanation:

refer to attachment

hope it will help u