A particle moves 21, m along vector 6hat{i} + 2hat{j} + 3hat{k} , then 14, m along vector 3hat

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3-2.Motion in Plane

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A particle moves $21\, m$ along the vector $6\hat i + 2\hat j + 3\hat k$ , then $14\, m$ along the vector $3\hat i - 2\hat j + 6\hat k$ . Its total displacement (in meters) is

$15\hat i + 2\hat j + 12\hat k$

$9\hat i + 12\hat k$

$9\hat i + 6\hat j$

$24\hat i + 2\hat j + 21\hat k$

Solution

Total displacement

$=21\left(\frac{6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}}{7}\right)+14\left(\frac{3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}}{7}\right)$

$=18 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}+6 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}$

$=(24 \hat{i}+2 \hat{j}+21 \hat{k}) m$

Standard 11

Physics

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Solution

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If the position vector of a particle is

$\vec r = – \cos \,t\hat i + \sin \,t\hat j – 18\,t\hat k$

then what is the magnitude of its acceleration ?