The three vectors overrightarrow A = 3hat i - | vedclass

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Question

(c) $\vec A = 3\hat i - 2\hat j + \hat k$, $\vec B = \hat i - 3\hat j + 5\hat k$, $\vec C = 2\hat i - \hat j + 4\hat k$

$|\vec A| = \sqrt {{3^2} +

Vedclass · Accepted Answer

A right angled triangle

Vedclass · Answer

(c) $\vec A = 3\hat i - 2\hat j + \hat k$, $\vec B = \hat i - 3\hat j + 5\hat k$, $\vec C = 2\hat i - \hat j + 4\hat k$

$|\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.

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

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