(A) Let the vector be $\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.The magnitude of the vector is given by $|\vec{A}| = \sqrt{2^2 + 3^2 + 4^2}$.$|\vec{A

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$\frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}}$

(A) Let the vector be $\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.The magnitude of the vector is given by $|\vec{A}| = \sqrt{2^2 + 3^2 + 4^2}$.$|\vec{A}| = \sqrt{4 + 9 + 16} = \sqrt{29}$.The unit vector $\hat{A}$ in the direction of $\vec{A}$ is defined as $\hat{A} = \frac{\vec{A}}{|\vec{A}|}$.Therefore,$\hat{A} = \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}}$.

Class 11 Physics 3-1.Vectors Fundamentals of Vectors

Easy

Find the unit vector in the direction of $2\hat{i} + 3\hat{j} + 4\hat{k}$.

A
$\frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}}$
B
$\frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{29}$
C
$\frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{9}}$
D
$\frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{9}$

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Find the unit vector in the direction of $2\hat{i} + 3\hat{j} + 4\hat{k}$.

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