Let a and b be 3-D Vectors and the norm of a is 20, and the norm of b is 36. What is the biggest possible value of the norm (a+b) and what is the smallest possible value of the norm (a+b). Also, the question says to use computational and geometrical arguments.
The way I approached the problem was I let v= a+b, and then the norm of a+b is equal to the norm of V squared. Then, I found out the a+b is 56. Any help would be appreciated!
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First, you should recall that the norm is used to measure the size of a vector. This can help in thinking about the problem geometrically.
One way to think about this problem computationally is to use the triangle inequality. The triangle inequality states that given two vectors $x$ and $y$, $\|x + y\| \leq (\|x\| + \|y\|)$.
Let $a, b \in \mathbb{R}^3$, then we have that \begin{align} ||a+b||^2 =& ||a||^2+||b||^2 +2a \cdot b = 20^2+36^2+2(20)(36)\cos\theta\\ \leq& 20^2+36^2+2(20)(36) = 3136. \end{align} which means $||a+b|| \leq \sqrt{3136} = 56$. Thus the biggest value $||a+b||$ can be is $56$ which is attained when $a$ and $b$ are in the same direction. With a similar argument, you could show the minimum value $||a+b|| = 16$, which occurs when $a$ and $b$ are directly opposite vectors.