Answer on Question #71180 – Math – Linear Algebra
Question
Can the magnitude of the resultant of the two vectors be greater than the sum of magnitude of individual vector?
Solution
Let a = ∣ a ⃗ ∣ , b = ∣ b ⃗ ∣ a = |\vec{a}|, b = |\vec{b}| a = ∣ a ∣ , b = ∣ b ∣ , then
∣ a ⃗ + b ⃗ ∣ = ∣ a ⃗ ∣ 2 + ∣ b ⃗ ∣ 2 + 2 ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ = a 2 + b 2 + 2 a b cos θ , \left| \vec {a} + \vec {b} \right| = \sqrt {\left| \vec {a} \right| ^ {2} + \left| \vec {b} \right| ^ {2} + 2 \left| \vec {a} \right| \left| \vec {b} \right| \cos \theta} = \sqrt {a ^ {2} + b ^ {2} + 2 a b \cos \theta}, ∣ ∣ a + b ∣ ∣ = ∣ a ∣ 2 + ∣ ∣ b ∣ ∣ 2 + 2 ∣ a ∣ ∣ ∣ b ∣ ∣ cos θ = a 2 + b 2 + 2 ab cos θ ,
where θ \theta θ is angle between a ⃗ \vec{a} a and b ⃗ \vec{b} b .
The maximum value of ∣ a ⃗ + b ⃗ ∣ |\vec{a} + \vec{b}| ∣ a + b ∣ is reached when cos θ = 1 \cos \theta = 1 cos θ = 1 :
∣ a ⃗ + b ⃗ ∣ m a x = a 2 + b 2 + 2 a b = a + b , \left| \vec {a} + \vec {b} \right| _ {m a x} = \sqrt {a ^ {2} + b ^ {2} + 2 a b} = a + b, ∣ ∣ a + b ∣ ∣ ma x = a 2 + b 2 + 2 ab = a + b ,
which is actually the sum of magnitudes of individual vectors.
Answer: no.
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