If vectors A→ and B→ are 3i^−4j^+5k^ and 2i^+3j^−4k^ respectively, then find the unit vector parallel to A→+B→.

Let C^=C¯|C¯|;C=A+B=5i^−j^+k^|C¯|=(5)2+(1)2+(1)2=25+1+1=27C=C¯|C¯|=5i^−j+k27

If vectors A→ and B→ are 3i^−4j^+5k^ and 2i^+3j^−4k^ respectively, then find the unit vector parallel to A→+B→.

Let C^=C¯|C¯|;C=A+B=5i^−j^+k^|C¯|=(5)2+(1)2+(1)2=25+1+1=27C=C¯|C¯|=5i^−j+k27

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If vectors $\vec{A}$ and $\vec{B}$ are $3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ respectively, then find the unit vector parallel to $\vec{A} + \vec{B}$ .

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$\frac{(5 \hat{i} - \hat{j} + \hat{k})}{\sqrt{27}}$

$\frac{(5 \hat{i} + \hat{j} + \hat{k})}{\sqrt{27}}$

$\frac{(5 \hat{i} + \hat{j} + \hat{k})}{27}$

$\frac{(5 \hat{i} - \hat{j} - \hat{k})}{27}$

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Detailed Solution

$Let \hat{C} = \frac{\bar{C}}{| \bar{C} |}; C = A + B = 5 \hat{i} - \hat{j} + \hat{k}$

$| \bar{C} | = \sqrt{(5)^{2} + (1)^{2} + (1)^{2}} = \sqrt{25 + 1 + 1} = \sqrt{27}$

$C = \frac{\bar{C}}{| \bar{C} |} = \frac{5 \hat{i} - j + k}{\sqrt{27}}$

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If vectors A→ and B→ are 3i^−4j^+5k^ and 2i^+3j^−4k^ respectively, then find the unit vector parallel to A→+B→.

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If vectors $\vec{A}$ and $\vec{B}$ are $3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ respectively, then find the unit vector parallel to $\vec{A} + \vec{B}$ .