SATHEE: Kinematics Question 14

Kinematics Question 14

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