Find the projection of $b = 2 i + 3 j - 2 k$ in the direction of vector $a = i + 2 j + 3 k .$ What is the vector determined by the projection?

\frac{2}{\sqrt{14}}, \frac{1}{7} (- i - 2 j - 3 k) .

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\frac{2}{\sqrt{14}}, \frac{1}{7} (i + 2 j + 3 k) .

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\frac{2}{\sqrt{17}}, \frac{2}{17} (2 i + 3 j - 2 k) .

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\frac{2}{\sqrt{17}}, \frac{2}{17} (- 2 i - 3 j + 2 k) .

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Solution

The correct option is B $\frac{2}{\sqrt{14}}, \frac{1}{7} (i + 2 j + 3 k) .$
The projection of vector $2 i + 3 j - 2 k$ along vector, $i + 2 j + 3 k$ is

$= \frac{(2 i + 3 j - 2 k) \cdot (i + 2 j + 3 k)}{| i + 2 j + 3 k |} = \frac{2}{\sqrt{14}}$

Hence component of vector $2 i + 3 j - 2 k$ along $i + 2 j + 3 k$ is

$= \frac{2}{\sqrt{14}} . \frac{i + 2 j + 3 k}{| 1 + 2 j + 3 k |} = \frac{1}{7} (i + 2 j + 3 k)$

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\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} + 2 \hat{j} + 7 \hat{k}, \vec{c} = 5 \hat{i} + 6 \hat{j} + 5 \hat{k}

\vec{a} = - 4 \hat{i} - 6 \hat{j} - 2 \hat{k}, \vec{b} = - \hat{i} + 4 \hat{j} + 3 \hat{k}, \vec{c} = - 8 \hat{i} - \hat{j} + 3 \hat{k}

\hat{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \hat{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \hat{c} = \hat{i} - 3 \hat{j} + 5 \hat{k}

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