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# Find $| \overrightarrow x |$, if for a unit vector $\overrightarrow a , (\overrightarrow x − \overrightarrow a) ⋅ (\overrightarrow x + \overrightarrow a) =12$ .

$\begin{array}{1 1} (A) |\overrightarrow x|=\sqrt{11} \\ (B) |\overrightarrow x|=\sqrt{12} \\ (C) |\overrightarrow x|=\sqrt{13} \\ (D) |\overrightarrow x|=\sqrt{14} \end{array}$

Can you answer this question?

Toolbox:
• $\overrightarrow a.\overrightarrow a=\mid\overrightarrow a\mid^2$
• $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
• $(\overrightarrow x-\overrightarrow a).(\overrightarrow x+\overrightarrow a)=\mid \overrightarrow x\mid^2-\mid\overrightarrow a\mid^2$
Step 1:
$(\overrightarrow x-\overrightarrow a).(\overrightarrow x+\overrightarrow a)=12$
Magnitude of $\mid \overrightarrow a\mid=1$
$(\overrightarrow x-\overrightarrow a).(\overrightarrow x+\overrightarrow a)=\overrightarrow x(\overrightarrow x+\overrightarrow a)-\overrightarrow a(\overrightarrow x+\overrightarrow a)$
$\quad\qquad\qquad\qquad\;\;=\mid \overrightarrow x \mid^2+\overrightarrow x.\overrightarrow a-\overrightarrow a.\overrightarrow x-\mid\overrightarrow a\mid^2$
Step 2:
But $\overrightarrow x.\overrightarrow a=\overrightarrow a.\overrightarrow x$
Therefore $(\overrightarrow x-\overrightarrow a)(\overrightarrow x+\overrightarrow a)=\mid\overrightarrow x\mid^2-\mid\overrightarrow a\mid^2$------(1)
But the product is 12 and $|\overrightarrow a|=1$
$(\overrightarrow x-\overrightarrow a).(\overrightarrow x+\overrightarrow a)=12$
Step 3:
Substitute these values in eq(1) we get
$12=\mid\overrightarrow x\mid^2-1$
$\mid\overrightarrow x\mid^2=12+1$
$\mid\overrightarrow x\mid^2=13$
$\mid\overrightarrow x\mid=\sqrt {13}$
answered May 21, 2013