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. $6/ \sqrt{5}$
• B. $8/ \sqrt{5}$
• C. $12/ \sqrt{5}$
• D. $16/ \sqrt{5}$

Answer 1

Answer: (B)

$8/ \sqrt{5}$

Answer 2

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, $Q^2 = R^2 + P^2$
Putting the given values in E (i), we get
$(8)^2 = (2P)^2 + P^2 = 4P^2 + P^2 = 5 P^2 \, \, \, or \, \, \, 5 P^2 = 64$
$P^2 = \frac{64}{5}$
$\therefore P = \frac{8}{\sqrt5}$