Question

Given that
Vector A = i-cap+ 2j-cap
Vector B = -j-cap + 2k-cap than find
1. |Vector A + Vector B|
2. |Vector A - Vector B|​

Answer 1

vector A+vector B= i-cap + 1j-cap + 2k- cap

vector A- vector B = i-cap +3j-cap -2k-cap

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Answer 2

Answer:

$\huge \boxed{1. |\vec{A} + \vec{B}| = \sqrt{6} }$

$\huge \boxed{2. |\vec{A} - \vec{B}| = \sqrt{14} }$

Explanation:

Given

$\vec{A} = \hat{i} + 2\hat{j} \\ \vec{B} = -\hat{ j} + 2\hat{k}$

To find

$1.| \vec{A} + \vec{B}| \\ 2.| \vec{A} - \vec{B}|$

Solutions;

$\bold{1.| \vec{A} + \vec{B}| } \\ | \vec{A} + \vec{B}| = ( \hat{i} + 2\hat{j}) + ( - \hat{j} + 2\hat{k}) \\ |\vec{A} + \vec{B}| = \hat{i} + 2\hat{j} - \hat{j} + 2\hat{k}) \\ | \vec{A} + \vec{B}| = \hat{i} + \hat{j} + 2\hat{k} \\ | \vec{A} + \vec{B}| = \sqrt{ {1}^{2} + {1}^{2} + {2}^{2} } \\ | \vec{A} + \vec{B}| = \sqrt{1 + 4 + 1} \\ \huge \boxed{| \vec{A} + \vec{B}| = \sqrt{6} }$

$\bold{2.| \vec{A} - \vec{B}| } \\|\vec{A} - \vec{B}| = ( \hat{i} + 2\hat{j} ) - ( - \hat{j} + 2\hat{k}) \\ |\vec{A} - \vec{B}| = \hat{i} + 2\hat{j} + \hat{j} - 2\hat{k} \\ |\vec{A} - \vec{B}| = \hat{i} + 3\hat{j} - 2 \hat{k} \\ |\vec{A} - \vec{B}| = \sqrt{ {1}^{2} + {3}^{2} + {2}^{2} } \\ |\vec{A} - \vec{B}| = \sqrt{1 + 9 + 4} \\ \huge \boxed{|\vec{A} - \vec{B}| = \sqrt{14} }$