A vector Q has a magnitude of 8 is added to the vector p which ties along the X-axis. The resultant of these two vectors is a third vector R which lies along the Y-axis and has a magnitude twice that of P The magnitude of P is

Question

A vector Q has a magnitude of 8 is added to the vector p which ties along the X-axis. The resultant of these two vectors is a third vector R which lies along the Y-axis and has a magnitude twice that of P The magnitude of P is (A) 6/ √5 (B) 8/ √5 (C) 12/ √5 (D) 16/ √5

Tardigrade Admin · Accepted Answer

Given Q = 8 units R = 2 P Since. R is along Y-axis and P is along x-axis. Therefore. P and R are perpendicular vectors. Hence, Q2 = R2 + P2 Putting the given values in Eq. (i), we get (8)2 = (2P)2 + P2 = 4P2 + P2 = 5 P2 or 5 P2 = 64 P2 = (64/5) ∴ P = (8/√5)

Q. A vector $Q$ has a magnitude of 8 is added to the vector $p$ which ties along the X-axis. The resultant of these two vectors is a third vector $R$ which lies along the Y-axis and has a magnitude twice that of $P$ The magnitude of $P$ is

$6/ 5$

$8/ 5$

$12/ 5$

$16/ 5$

Q. A vector Q has a magnitude of 8 is added to the vector p which ties along the X-axis. The resultant of these two vectors is a third vector R which lies along the Y-axis and has a magnitude twice that of P The magnitude of P is

6/5​

8/5​

12/5​

16/5​

Given Q = 8 units R=2P Since. R is along Y-axis and P is along x-axis. Therefore. P and R are perpendicular vectors. Hence, Q2=R2+P2 Putting the given values in Eq. (i), we get (8)2=(2P)2+P2=4P2+P2=5P2or5P2=64 P2=564​ ∴P=5​8​

Q. A vector $Q$ has a magnitude of 8 is added to the vector $p$ which ties along the X-axis. The resultant of these two vectors is a third vector $R$ which lies along the Y-axis and has a magnitude twice that of $P$ The magnitude of $P$ is

$6/ 5$

$8/ 5$

$12/ 5$

$16/ 5$