CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Chandragupta Maurya

For any two v...

Question

For any two vectors $\to a$ and $\to b$ , we always have $∣ ∣ \begin{matrix} \to a + \to b \end{matrix} ∣ ∣ \leq ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ + ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣$ (triangle inequality).

Open in App

Solution

Let us assume $\to a \neq \to 0$ and $\to b \neq \to 0$
we know, ${∣ ∣ \begin{matrix} \to a + \to b \end{matrix} ∣ ∣}^{2}$
$= {∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣}^{2} + 2 ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣ cos θ + {∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣}^{2}$ (i)
We know that $cos θ \leq 1$
Multiplying $2 ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣$ on both sides then we get,
$2 ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣ cos θ \leq 2 ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣$
Adding ${∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣}^{2} + {∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣}^{2}$ on both sides then we get,
${∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣}^{2} + {∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣}^{2} + 2 ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ + ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ cos θ \leq {∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣}^{2}$
${∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣}^{2} + 2 ∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣$
${∣ ∣ \begin{matrix} \to a + \to b \end{matrix} ∣ ∣}^{2} \leq (∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ + {∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣}^{2})$ Form (i)
Taking square root on both sides then we grt,
$∣ ∣ \begin{matrix} \to a + \to b \end{matrix} ∣ ∣ \leq (∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ + {∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣}^{2})$
Hence, proved.

Suggest Corrections

thumbs-up

8

Similar questions

Q. For any two vectors

\vec{a} and \vec{b}, find (\vec{a} \times \vec{b}) . \vec{b} .

Q. For any two vectors

\vec{a}

\vec{b}

\vec{a} . (\vec{b} \times \vec{a}) .

Q. For any two vectors

[\vec{a} \vec{b} \vec{a} \times \vec{b}] + (\vec{a} . \vec{b})

= ________________.

Q. For any two vectors

\vec{a} and \vec{b}

(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \Leftrightarrow |\vec{a}| = |\vec{b}|

Q. For any two vectors

\vec{a} and \vec{b}

|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Chandragupta Maurya

Standard IX History

Join BYJU'S Learning Program

CrossIcon