Question

if a is a unit vector and (x+a).(x-a)=15, then find |x|. please consider x as a vector. |x|means magnitude of vector.

Answer 1

it will help u..... I hope

Answer 2

$|\vec{x}|=4$

Step-by-step explanation:

Given:

$\vec{a}$ is a unit vector

So $\vec{a}$ is a magnitude of 1.

⇒ $|\vec{a}| = 1$

We have,

$(\vec{x}+\vec{a}) \cdot (\vec{x}-\vec{a}) = 15$

⇒ $\vec{x}\cdot\vec{x}-\vec{x}\cdot\vec{a}+\vec{a}\cdot\vec{x}-\vec{a}\cdot\vec{a}=15$

⇒ $\vec{x}\cdot\vec{x}-\vec{x}\cdot\vec{a}+\vec{x}\cdot\vec{a}-\vec{a}\cdot\vec{a}=15$      [property, $\vec{x}\cdot\vec{a} = \vec{a}\cdot\vec{x}$ ]

⇒ $\vec{x}\cdot\vec{x}-\vec{a}\cdot\vec{a}=15$  

⇒ $|\vec{x}|^2- |\vec{a}|^2=15$           [property,  $\vec{x}\cdot\vec{x} =|\vec{x}|^2$ ]

⇒ $|\vec{x}|^2- (1)^2=15$      [given $|\vec{a}| = 1$]

⇒ $|\vec{x}|^2- 1=15$

⇒ $|\vec{x}|^2=15+1$

⇒ $|\vec{x}|^2=16$

⇒ $|\vec{x}|=\pm\sqrt{16}$

⇒ $|\vec{x}|=\pm4$

We know that magnitude is not negative,

Therefore, $|\vec{x}|=4$