Question

the vector Q which has a magnitude of 8 is added to the vector P which lies along the x-axis. the resultant of these vectors is third vector R,which lies along the y-axis and has magnitude twice that of P. the magnitude of P is?

Answer 1

Let the magnitude of the Vector P be a.

∴ In Vector Form,

Vector P = a$\hat{i}$

Let the magnitude of Vector R be r.

∴ r = 2a

In Vector Form,

$\hat{R}$ = 2a $\hat{j}$

Now, We know,

$\hat{R}$ = $\hat{P}$ + $\hat{Q}$

⇒ |R²| = |P|² + |Q|² + 2 |PQ| Cosθ

Cos θ = 90° [Angle between the x-y axis is 90°]

⇒ |R²| = |P|² + |Q|² + 2 |PQ|

∴ (2a)² = (a)² + (8)² + 2(a)(8)

⇒ 4a² = a² + 64 + 16a

⇒ 3a² - 16a - 64 = 0

∴ 3a² - 24a + 8a - 64 = 0

⇒ 3a(a - 8) + 8(a - 8) = 0

∴ (a - 8)(3a + 8) = 0

⇒ a = 8 and a = -8/3

a ≠ -8/3.

Hence, the Magnitude of the Vector P is 8.

Hope it helps.

Answer 2

Answer: $8/\sqrt{5$

Explanation:  As the resultant is on the y-axis therefore it makes a 90 degree with vector P

Therefore tan90° =QsinФ/P+QcosФ

                  0/1 = 8sinФ/P+8cosФ

                    P+8cosФ  =  0

                     8cosФ =  -P

                          cosФ =-P/cosФ

Therefore

                          R² = P² + Q² + 2PQ(-P/8)

                          (2P)² = P² + (8)² + 2P × 8 (-P/8)

                          4P² = P² + 64 + 2P × -P

                         4P² = P² + 64 - 2P²

                                        4P² + P² = 64

                                         5P² = 64

                                           P² = 64/5

                                           P = 8/√5