The direction cosines of a vector A→ are cos⁡α=452,cos⁡β=12 and cos⁡γ=352, then the vector A→ is

cos⁡α=AxA=452cos⁡β=AyA=12=552cos⁡γ=AzA=352∴A¯=Axi^+Ayj^+Azk^=4i^+5j^+3k^

The direction cosines of a vector A→ are cos⁡α=452,cos⁡β=12 and cos⁡γ=352, then the vector A→ is

cos⁡α=AxA=452cos⁡β=AyA=12=552cos⁡γ=AzA=352∴A¯=Axi^+Ayj^+Azk^=4i^+5j^+3k^

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The direction cosines of a vector $\vec{A}$ are $\cos α = \frac{4}{5 \sqrt{2}}, \cos β = \frac{1}{\sqrt{2}} and \cos γ = \frac{3}{5 \sqrt{2}},$ then the vector $\vec{A}$ is

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$4 \hat{i} + \hat{j} + 3 \hat{k}$

$4 \hat{i} + 5 \hat{j} + 3 \hat{k}$

$4 \hat{i} - 5 \hat{j} - 3 \hat{k}$

$\hat{i} + \hat{j} - \hat{k}$

answer is B.

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Detailed Solution

$\begin{array}{l} \cos α = \frac{A_{x}}{A} = \frac{4}{5 \sqrt{2}} \\ \cos β = \frac{A_{y}}{A} = \frac{1}{\sqrt{2}} = \frac{5}{5 \sqrt{2}} \\ \cos γ = \frac{A_{z}}{A} = \frac{3}{5 \sqrt{2}} \\ ∴ \bar{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k} \end{array}$