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(D) The displacement vector $\vec{d}$ is given by the sum of individual displacements $\vec{d_1}$ and $\vec{d_2}$.First,find the unit vectors for the

Vedclass · Accepted Answer

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

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(D) The displacement vector $\vec{d}$ is given by the sum of individual displacements $\vec{d_1}$ and $\vec{d_2}$.First,find the unit vectors for the given directions:For the first vector $\vec{v_1} = 6\hat{i} + 2\hat{j} + 3\hat{k}$,the magnitude is $|\vec{v_1}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.The unit vector is $\hat{u_1} = \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7}$.So,$\vec{d_1} = 21 	imes \hat{u_1} = 21 	imes \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = 3(6\hat{i} + 2\hat{j} + 3\hat{k}) = 18\hat{i} + 6\hat{j} + 9\hat{k}$.For the second vector $\vec{v_2} = 3\hat{i} - 2\hat{j} + 6\hat{k}$,the magnitude is $|\vec{v_2}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.The unit vector is $\hat{u_2} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7}$.So,$\vec{d_2} = 14 	imes \hat{u_2} = 14 	imes \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} = 2(3\hat{i} - 2\hat{j} + 6\hat{k}) = 6\hat{i} - 4\hat{j} + 12\hat{k}$.Total displacement $\vec{d} = \vec{d_1} + \vec{d_2} = (18\hat{i} + 6\hat{j} + 9\hat{k}) + (6\hat{i} - 4\hat{j} + 12\hat{k}) = 24\hat{i} + 2\hat{j} + 21\hat{k}$.

$A$ particle moves $21 \ m$ along the direction of the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$,then it moves $14 \ m$ along the direction of the vector $3\hat{i} - 2\hat{j} + 6\hat{k}$. What is its total displacement (in meters)?

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