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Let a and b be vectors such that \(\|\mathbf{a}\| = 5,\) \(\|\mathbf{a} + \mathbf{b}\| = 7\), and \(\|\mathbf{a} - \mathbf{b}\| = 9\) find \(\|\mathbf{b}\|.\)

 Apr 2, 2022
 #1
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By Cosine Law, |b| = 24/5.

 Apr 2, 2022
 #2
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thats wrong

Guest Apr 2, 2022
 #3
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Hello,

Here is an algebraic approach; although the quickiest way is to sketch a diagram and use some geometry.

\(\text{Let}\) \(a=(x_a,y_a)\) and \(b=(x_b,y_b)\)

 

(*)........\(x^2_a+y^2_a=25\)   (Given: \(\left | \left | a \right | \right |=5 \iff x^2_a+y^2_a=25\) )

(**)......\((x_a+x_b)^2+(y_a+y_b)^2=49\) (Similarly, find the vector a+b, then find its magnitude)

(***).....\((x_a-x_b)^2+(y_a-y_b)^2=81\) (Similarly)

We want: 

\(\sqrt{x^2_b+y^2_b}\) (Which is the magnitude of vector b)

Expanding (**) and (***) yields, respectively:

\(x^2_a+2x_ax_b+x^2_b+(y^2_a+2y_ay_b+y^2_b)=49\)        Eq.(1)

\(x^2_a-2x_ax_b+x^2_b+(y^2_a-2y_ay_b+y^2_b)=81\)        Eq.(2)

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Add Eq.(1) and Eq.(2) to get:

 

\(2x^2_a+2x^2_b+2y^2_a+2y^2_b=130\)  (Divide by 2 or factor as below, and make \(x^2_b+y^2_b\) the subject).

\(2(x^2_a+y^2_a)+2(x^2_b+y^2_b)=130\)   From (*), \(x^2_a+y^2_a=25\)

\(2(25)+2(x^2_b+y^2_b)=130\)    

Thus,

\(x^2_b+y^2_b=40\)
Then, taking the squareroot of both sides gives:
\(\left |\left |b \right | \right |=x^2_b+y^2_b=2\sqrt{10}\)

 Apr 3, 2022

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