+0

Let a and b be vectors such that , and find ​

0
95
3

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
0

By Cosine Law, |b| = 24/5.

Apr 2, 2022
#2
0

thats wrong

Guest Apr 2, 2022
#3
0

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:
$$\left |\left |b \right | \right |=x^2_b+y^2_b=2\sqrt{10}$$

Apr 3, 2022