Vector Algebra
Direction Cosines
Grade 12
Question:
<p>The projections of a vector on coordinate axes are \(x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1\). Also, \(x_2 - x_1 = 6,\ y_2 - y_1 = -3,\ z_2 - z_1 = 2\). Find the direction cosines of the vector.</p>
<p>\(\dfrac{6}{7},\ \dfrac{-3}{7},\ \dfrac{2}{7}\)</p>
<p>\(\dfrac{6}{7},\ \dfrac{3}{7},\ \dfrac{2}{7}\)</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 obtained by dividing each component of a vector by its magnitude. The magnitude is found using √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²], and each direction cosine equals the corresponding component divided by this magnitude.
Step 1: Write the vector components from projections on coordinate axes: Vector = (6, -3, 2) 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: l = 6/7, m = -3/7, n = 2/7 Step 4: Verify: l^2 + m^2 + n^2 = (6/7)^2 + (-3/7)^2 + (2/7)^2 = 36/49 + 9/49 + 4/49 = 49/49 = 1 ✓ ∴ Direction cosines = (6/7, -3/7, 2/7)
Correct Answer: A