Vector Algebra
Direction Cosines
Grade 12
Question:
<p>The projections of a vector on the three coordinate axes are 6, \(-3\), 2, respectively. The direction cosines of the vector are</p>
<p>\(6, -3, 2\)</p>
<p>\(\dfrac{6}{5}, -\dfrac{3}{5}, \dfrac{2}{5}\)</p>
<p>\(\dfrac{6}{7}, -\dfrac{3}{7}, \dfrac{2}{7}\)</p>
<p>\(-\dfrac{6}{7}, -\dfrac{3}{7}, \dfrac{2}{7}\)</p>
Step-by-Step Solution
Key Concept: Direction cosines are found by dividing each component of a vector by its magnitude. The magnitude is calculated using the Pythagorean theorem in 3D: √(a² + b² + c²).
Step 1: Identify the vector components (projections on axes). The vector is v = 6 i − 3 j + 2 k Step 2: Calculate the magnitude of the vector. | v | = √(6^2 + (−3)^2 + 2^2) = √(36 + 9 + 4) = √49 = 7 Step 3: Find direction cosines by dividing each component by magnitude. Direction cosines are: (l, m, n) = (6/7, −3/7, 2/7) ∴ Answer: C
Correct Answer: C