Question

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Find the dot product of A = 2i + 3j + k and B = I - j + 2k​

Answer 1

Given :

• $\sf{ \vec A \: = \: 2 \hat{i} \: + \: 3 \hat{j} \: + \: \hat{k}}$
• $\sf{\vec B \: = \: \hat{i} \: - \: \hat {j} \: + \: 2 \hat{k}}$

To Find :

• Dot Product

Solution :

We are given, $\sf{\vec A \: = \: 2 \hat{i} \: + \: 3 \hat{j} \: + \: \hat{k}}$ and $\sf{\vec B \: = \: \hat{i} \: - \: \hat {j} \: + \: 2 \hat{k}}$. We know that in dot product,

$\implies \sf{\vec A. \vec B \: = \: \big[ \: 2 \hat{i} \: + \: 3 \hat{j} \: + \: \hat{k} \big] . \big[ \hat{i} \: - \: \hat {j} \: + \: 2 \hat{k} \big] } \\ \\ \implies \sf{\vec A. \vec B \: = \: \big[ (2 \: \times \: 1) \: + \: (3 \: \times \: -1) \: + \: (1 \: \times \: 2)\big]} \\ \\ \implies \sf{\vec A. \vec B \: = \: 2 \: + \: (-3 ) \: + \: 2 } \\ \\ \implies \sf{\vec A. \vec B \: = \: 2 \: - \: 3 \: + \: 2} \\ \\ \implies \sf{\vec A . \vec B \: = \: 4 \: - \: 3} \\ \\ \implies \sf{\vec A. \vec B \: = \: 1}$

$\therefore$ Dot Product is 1

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Now, we will find angle between vectors :

$\implies \sf{|A| \: = \: \sqrt{(2)^2 \: + \: (3)^2 \: + \: (1)^2}} \\ \\ \implies \sf{|A| \: = \: \sqrt{4 \: + \: 9 \: + \: 1}} \\ \\ \implies \sf{|A| \: = \: \sqrt{14}}$

$\implies \sf{|B| \: = \: \sqrt{(1)^2 \: + \: (-1)^2 \: + \: (2)^2}} \\ \\ \implies \sf{|B| \: = \: \sqrt{1 \: + \: 1 \: 4}} \\ \\ \implies \sf{|B| \: = \: \sqrt{6}}$

Now, angle is :

$\implies \sf{\cos \theta \: = \: \dfrac{\vec A. \vec B}{|A| . |B|}} \\ \\ \implies \sf{\cos \theta \: = \: \dfrac{1}{\sqrt{14} \: \times \: \sqrt{6}}} \\ \\ \implies \sf{\cos \theta \: = \: \dfrac{1}{\sqrt{84}}} \\ \\ \implies \sf{\cos \theta \: = \: \dfrac{1}{9.16}} \\ \\ \implies \sf{\theta \: = \: \cos^{-1} \bigg( \dfrac{1}{9.16} \bigg) } \\ \\ \implies \sf{\theta \: = \: 83.73}$

Answer 2

Given :

To Find :

Dot Product

Solution :

We are given,  and . We know that in dot product,

Dot Product is 1

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Now, we will find angle between vectors :

Now, angle is :