Is |$\vec{a}+\lambda\vec{b}$|>$|\vec{a}|$ true or false?

Question

|$\vec{a}+\lambda\vec{b}|\geq|\vec{a}|$ for all real $\lambda$ if $\vec{a}$ and $\vec{b}$ are perpendicular.

I need to verify the truth or falsity of this statement.What would be the best method?

P.S:Now I realize it was a really silly question which I asked.I guess my book's answer key is wrong as it claims the statement is false.

Thanks a lot to all those who answered!

Answer 1

$|\vec{a}+\lambda \vec{b}|^2=(\vec{a}+\lambda \vec{b})\cdot (\vec{a}+\lambda \vec{b})=|\vec{a}|^2+\lambda^2|\vec{b}|^2+2\lambda\vec{a}\cdot\vec{b}=|\vec{a}|^2+\lambda^2|\vec{b}|^2\geq |\vec{a}|^2$

Thus the statement is true.

Another approach may be a visual one seeing that $\vec{a}$ is a side in a right triangle with hypotenuse $\vec{a}+\lambda\vec{b}$

Answer 2

$$|\vec a+\lambda\vec b|=\sqrt{|\vec a|^2+\lambda^2|\vec b|^2+2|\vec a|\cdot|\vec b|\cos\frac{\pi}2}\ge|\vec a|$$ since $\vec a$ and $\vec b$ are perpendicular.

So, the inequality is true.

Answer 3

Let $\lambda\in\mathbb{R}$, and $a,b\in V$ be vectors in a real inner product space $V$. As the quantities $|a+\lambda b|$ and $|a|$ are positive, the inequality in question holds exactly when it holds for the squares: $$ |a+\lambda b|^2 \geq |a|^2 $$ Apply the fact that $|v|^2 = v\cdot v$ for all $v\in V$: $$ |a+\lambda b|^2 = (a+\lambda b)\cdot (a+\lambda b) = a\cdot a + 2\lambda a\cdot b + b\cdot b $$ As $a,b$ are perpendicular, $a\cdot b = 0$, and also $b\cdot b \geq 0$. This establishes the desired result.

Answer 4

Here is a picture displaying (a small part) of vectors having the form $u+\lambda \vec{v}$.

One can observe values of $\lambda$ for which $\|u+\lambda \vec{v}\|<\|u\|$.