For any two vectors a and b- we always have a - b-a - b triangle inequality-

Name: NCERT - Grade 12 - Mathematics - Vector Algebra - Q44
Uploaded: 2022-07-04T10:36:42+05:30
Description: Let us assume a≠0 and b≠0 we know, a + b^2 = nbsp;a^2 + 2 nbsp; nbsp;a nbsp;bcos θ + nbsp;b^2 (i) We know that cos θ≤ 1 Multiplying 2 nbsp;a nbsp; ...

Let us assume a≠0 and b≠0 we know, a + b^2 = nbsp;a^2 + 2 nbsp; nbsp;a nbsp;bcos θ + nbsp;b^2 (i) We know that cos θ≤ 1 Multiplying 2 nbsp;a nbsp; ...

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Byju's Answer

Standard IX

History

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).

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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.

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For any two vectors →a and →b, we always have ∣∣→a+→b∣∣≤∣∣→a∣∣+∣∣→b∣∣ (triangle inequality).

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).