Question

Show that if a and b are
(a) in the same direction then |a + b| = |a| + |b|,
(b) in the opposite direction then |a − b| = |a| + |b|

Answer 1

a) In case, when we are speaking about the vectors, then the given notation means the following: the length of the vector, which is a sum of the other two vectors is directly the sum of their lengths (in case, when both are parallel and co-directional).

This can be proved schematically with using of a so-called triangle rule. On the other hand, this can be proved in the other way:

$\mid{\vec{a}+\vec{b}}\mid^2=\vec{a}^2+\vec{b}^2+2\vec{a}\vec{b}=\mid{\vec{a}}\mid^2+\mid{\vec{b}}\mid^2+2\mid{\vec{a}}\mid\mid{\vec{b}}\mid cos(0^0)=(\mid{\vec{a}}\mid+\mid{\vec{b}}\mid)^2$

Taking the square root of both sides, this is equal to $\mid\vec{a}+\vec{b}\mid=\mid{\vec{a}}\mid+\mid{\vec{b}}\mid$

(angle between vectors is equal to 0⁰ and cos(0⁰) = 1).

b) This case is practically the same as the previous one (just the vectors are parallel and the opposite-directional):

$\mid\vec{a}-\vec{b}\mid^2=\vec{a}^2+\vec{b}^2-2\vec{a}\vec{b}=\mid{\vec{a}}\mid^2+\mid{\vec{b}}\mid^2-2\mid{\vec{a}}\mid\mid{\vec{b}}\mid cos(180^0)=\mid{\vec{a}}\mid^2+\mid{\vec{b}}\mid^2+2\mid{\vec{a}}\mid\mid{\vec{b}}\mid=(\mid{\vec{a}}\mid+\mid{\vec{b}}\mid)^2$

Taking the square root of both sides, this is equal to $\mid{\vec{a}}-\vec{b}\mid=\mid{\vec{a}}\mid+\mid{\vec{b}}\mid$

(angle between vectors is equal to 180⁰ and cos(180⁰) = -1).