Kinematics Question 14

Question: The three vectors $ \overrightarrow{A}=3\hat{i}-2\hat{j}+\hat{k},\overrightarrow{B}=\hat{i}-3\hat{j}+5\hat{k} $ and $ \overrightarrow{C}=2\hat{i}+\hat{j}-4\hat{k} $ form

Options:

A) An equilateral triangle

B) isosceles triangle

C) A right angled triangle

D) No triangle

Show Answer

Answer:

Correct Answer: C

Solution:

$ \vec{A}=3\hat{i}-2\hat{j}+\hat{k} $ , $ \vec{B}=\hat{i}-3\hat{j}+5\hat{k} $ , $ \vec{C}=2\hat{i}-\hat{j}+4\hat{k} $

$ |\vec{A}|=\sqrt{3^{2}+{{(-2)}^{2}}+1^{2}}=\sqrt{9+4+1}=\sqrt{14} $

$ |\vec{B}|=\sqrt{1^{2}+{{(-3)}^{2}}+5^{2}}=\sqrt{1+9+25}=\sqrt{35} $

$ |\vec{A}|=\sqrt{2^{2}+1^{2}+{{(-4)}^{2}}}=\sqrt{4+1+16}=\sqrt{21} $ As $ B=\sqrt{A^{2}+C^{2}} $

therefore ABC will be right angled triangle.



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