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Question: Find a unit vector parallel to the vector $2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarr...

Find a unit vector parallel to the vector 2a→−b→+3c→2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}

A

−222i^−322j^+322k^-\frac{2}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{3}{\sqrt{22}}\hat{k}

B

322i^−322j^+222k^\frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}

C

−322i^+322j^+222k^-\frac{3}{\sqrt{22}}\hat{i} + \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}

D

−222i^+322j^+322k^-\frac{2}{\sqrt{22}}\hat{i} + \frac{3}{\sqrt{22}}\hat{j} + \frac{3}{\sqrt{22}}\hat{k}

Answer

322i^−322j^+222k^\frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}

Explanation

Solution

To find a unit vector parallel to the vector 2a→−b→+3c→2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}, we follow these steps:

  1. Calculate the resultant vector:
    Let V→=2a→−b→+3c→\overrightarrow{V} = 2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}.
    Given:
    a→=i^+j^+k^\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}
    b→=2i^−j^+3k^\overrightarrow{b} = 2\hat{i} - \hat{j} + 3\hat{k}
    c→=i^−2j^+k^\overrightarrow{c} = \hat{i} - 2\hat{j} + \hat{k}

    First, calculate the scalar multiples:
    2a→=2(i^+j^+k^)=2i^+2j^+2k^2\overrightarrow{a} = 2(\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + 2\hat{k}
    3c→=3(i^−2j^+k^)=3i^−6j^+3k^3\overrightarrow{c} = 3(\hat{i} - 2\hat{j} + \hat{k}) = 3\hat{i} - 6\hat{j} + 3\hat{k}

    Now, substitute these into the expression for V→\overrightarrow{V}:
    V→=(2i^+2j^+2k^)−(2i^−j^+3k^)+(3i^−6j^+3k^)\overrightarrow{V} = (2\hat{i} + 2\hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k})

    Combine the components:
    For i^\hat{i}: (2−2+3)i^=3i^(2 - 2 + 3)\hat{i} = 3\hat{i}
    For j^\hat{j}: (2−(−1)−6)j^=(2+1−6)j^=−3j^(2 - (-1) - 6)\hat{j} = (2 + 1 - 6)\hat{j} = -3\hat{j}
    For k^\hat{k}: (2−3+3)k^=2k^(2 - 3 + 3)\hat{k} = 2\hat{k}

    So, the resultant vector is V→=3i^−3j^+2k^\overrightarrow{V} = 3\hat{i} - 3\hat{j} + 2\hat{k}.

  2. Calculate the magnitude of the resultant vector:
    The magnitude of V→\overrightarrow{V} is ∣V→∣=(3)2+(−3)2+(2)2|\overrightarrow{V}| = \sqrt{(3)^2 + (-3)^2 + (2)^2}.
    ∣V→∣=9+9+4|\overrightarrow{V}| = \sqrt{9 + 9 + 4}
    ∣V→∣=22|\overrightarrow{V}| = \sqrt{22}

  3. Find the unit vector:
    A unit vector parallel to V→\overrightarrow{V} is given by V^=V→∣V→∣\hat{V} = \frac{\overrightarrow{V}}{|\overrightarrow{V}|}.
    V^=3i^−3j^+2k^22\hat{V} = \frac{3\hat{i} - 3\hat{j} + 2\hat{k}}{\sqrt{22}}
    V^=322i^−322j^+222k^\hat{V} = \frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}