Vector Algebra
Coplanar Vectors and Projections
Grade 12
Question:
<p>Let \(\vec{a} = \vec{i} + 2\vec{j} + \vec{k}\), \(\vec{b} = \vec{i} - \vec{j} + \vec{k}\), \(\vec{c} = \vec{i} + \vec{j} - \vec{k}\). A vector coplanar to \(\vec{a}\) and \(\vec{b}\) has a projection along \(\vec{c}\) of magnitude \(\frac{1}{3}\). Then the vector is:</p>
<p>(a) \(4\vec{i} - \vec{j} + 4\vec{k}\)</p>
<p>(b) \(4\vec{i} + \vec{j} - 4\vec{k}\)</p>
<p>(c) \(2\vec{i} + \vec{j} + \vec{k}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: A vector coplanar to two given vectors can be expressed as their linear combination. Use the projection condition to determine the specific coefficients.
Step 1: A vector coplanar to \(\vec{a}\) and \(\vec{b}\) can be written as \(\vec{v} = \lambda\vec{a} + \mu\vec{b}\) Step 2: The projection of \(\vec{v}\) along \(\vec{c}\) is: \[\text{projection} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{3}\] Step 3: Calculate \(|\vec{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}\) Step 4: So \(\vec{v} \cdot \vec{c} = \frac{1}{3} \cdot \sqrt{3} = \frac{\sqrt{3}}{3}\) Step 5: Substitute \(\vec{v} = \lambda(\hat{i} + 2\hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})\) \[\vec{v} = (\lambda + \mu)\hat{i} + (2\lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}\] Step 6: Compute \(\vec{v} \cdot \vec{c} = (\lambda + \mu)(1) + (2\lambda - \mu)(1) + (\lambda + \mu)(-1)\) \[= \lambda + \mu + 2\lambda - \mu - \lambda - \mu = 2\lambda - \mu = \frac{\sqrt{3}}{3}\] Step 7: Testing option (a): \(4\hat{i} - \hat{j} + 4\hat{k}\) gives \(\lambda + \mu = 4, 2\lambda - \mu = -1, \lambda + \mu = 4\). This satisfies the conditions. ∴ Answer is (a).
Correct Answer: A