1. Key Concepts and Formulas

• Vector Components (Projections) and Direction Ratios: If a vector $\vec{v}$ makes angles $\alpha, \beta, \gamma$ with the positive x, y, z axes respectively, its components (or projections) along these axes are $a = |\vec{v}|\cos\alpha$, $b = |\vec{v}|\cos\beta$, and $c = |\vec{v}|\cos\gamma$. These components $(a, b, c)$ are also known as the direction ratios of the vector. The vector can be written as $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$.
• Magnitude of a Vector: The length or magnitude of a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula: $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$
• Direction Cosines: The direction cosines $(l, m, n)$ of a vector are the cosines of the angles it makes with the positive coordinate axes, i.e., $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$. They are obtained by dividing each direction ratio by the magnitude of the vector: $l = \frac{a}{|\vec{v}|}, \quad m = \frac{b}{|\vec{v}|}, \quad n = \frac{c}{|\vec{v}|}$ An important property is that the sum of the squares of the direction cosines is always 1: $l^2 + m^2 + n^2 = 1$.

2. Step-by-Step Solution

Step 1: Identify the Vector and its Direction Ratios. The problem states that the projections of a vector on the three coordinate axes are $6, -3, 2$.

• What we are doing: We are using the given projections to define the vector in component form and identify its direction ratios.
• Why: The projections of a vector on the coordinate axes are precisely its components along those axes, which serve as its direction ratios. Let the vector be $\vec{v}$. Based on the given projections, we can write the vector in component form: $\vec{v} = 6\hat{i} - 3\hat{j} + 2\hat{k}$ From this, the direction ratios $(a, b, c)$ of the vector are $(6, -3, 2)$.

Step 2: Calculate the Magnitude of the Vector.

• What we are doing: We are calculating the length of the vector using its direction ratios.
• Why: To find the direction cosines, we need to normalize the direction ratios. This means dividing each direction ratio by the magnitude of the vector, effectively scaling the vector to a unit length while preserving its direction. Using the formula for the magnitude of a vector: $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$ Substitute the direction ratios $a=6, b=-3, c=2$: $|\vec{v}| = \sqrt{(6)^2 + (-3)^2 + (2)^2}$ $|\vec{v}| = \sqrt{36 + 9 + 4}$ $|\vec{v}| = \sqrt{49}$ $|\vec{v}| = 7$ The magnitude of the vector is 7 units.

Step 3: Determine the Direction Cosines.

• What we are doing: We are calculating the direction cosines by dividing each direction ratio by the magnitude of the vector.
• Why: This process normalizes the direction ratios, yielding the components of a unit vector in the same direction, which are the direction cosines. Using the formulas for direction cosines $l = \frac{a}{|\vec{v}|}, m = \frac{b}{|\vec{v}|}, n = \frac{c}{|\vec{v}|}$:
• For the x-axis: $l = \frac{6}{7}$
• For the y-axis: $m = \frac{-3}{7}$
• For the z-axis: $n = \frac{2}{7}$ Thus, the direction cosines of the vector are $\left({6 \over 7}, {{ - 3} \over 7}, {2 \over 7}\right)$.

3. Common Mistakes & Tips

• Confusing Projections with Direction Cosines: Remember that projections are the components of the vector (direction ratios), not the direction cosines themselves. The direction cosines are obtained from the projections by dividing by the magnitude.
• Sign Errors: Always retain the correct signs of the components when calculating the magnitude and especially when determining the direction cosines. A negative component leads to a negative direction cosine.
• Verification: After calculating the direction cosines, always perform a quick check: $l^2 + m^2 + n^2 = 1$. This helps confirm the accuracy of your magnitude calculation and subsequent divisions. For our result: $\left(\frac{6}{7}\right)^2 + \left(\frac{-3}{7}\right)^2 + \left(\frac{2}{7}\right)^2 = \frac{36}{49} + \frac{9}{49} + \frac{4}{49} = \frac{49}{49} = 1$. This confirms our answer is correct.

4. Summary

To find the direction cosines of a vector given its projections on the coordinate axes, first, recognize that these projections are the direction ratios of the vector. Next, calculate the magnitude (length) of the vector using the Pythagorean theorem in three dimensions. Finally, divide each direction ratio by the magnitude to obtain the respective direction cosines. The calculated direction cosines for the given vector are $\left({6 \over 7}, {{ - 3} \over 7}, {2 \over 7}\right)$, which can be verified by ensuring the sum of their squares equals 1.

5. Final Answer

The direction cosines of the vector are $\left({6 \over 7}, {{ - 3} \over 7}, {2 \over 7}\right)$. This corresponds to option (B).

The final answer is $\boxed{\text{(B)}}$.