Let a and b be vectors such that \(\|\mathbf{a}\| = 5,\) \(\|\mathbf{a} + \mathbf{b}\| = 7\), and \(\|\mathbf{a} - \mathbf{b}\| = 9\) find \(\|\mathbf{b}\|.\)
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)
----------------------------------------------------------------
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:
\(\left |\left |b \right | \right |=x^2_b+y^2_b=2\sqrt{10}\)
