Question

If p is an unit vector perpendicular to q, and (7p-q).(4p+3q) =1 then find |q| (magnitude of q)(p and q are vectors)

Answer 1

Therefore the magnitude of $\vec q$ is 3 units.

Step-by-step explanation:

Given that, $\vec p$ is unit vector that is $|\veac p|=1$.

$\vec p$ is perpendicular to $\vec q$.

$\therefore \vec p.\vec q=0$

∴$(7\vec p-\vec q).(4\vec p+3\vec q)=1$

$\Rightarrow 7\vec p.4\vec q+7\vec p.3\vec q-\vec q.4\vec p-\vec q. 3\vec q=1$

$\Rightarrow 28 (\vec p)^2+21 \vec p.\vec q-4 \vec p. \vec q-3 (\vec q)^2=1$   [ since dot product commutative ]

$\Rightarrow 28 |\vec p|^2+17 \vec p.\vec q-3 |\vec q|^2=1$               [$(\vec a)^2=|a|^2$ ]

$\Rightarrow 28. 1^2+17 .0-3 |\vec q|^2=1$                [  $\vec p.\vec q=0$ and  $|\veac p|=1$ ]

$\Rightarrow 28 -3 |\vec q|^2=1$

$\Rightarrow3 |\vec q|^2=27$

$\Rightarrow |\vec q|^2=9$

$\Rightarrow |\vec q|= 3$

Therefore the magnitude of $\vec q$ is 3 units.