Vector Algebra
Vector Projections and Linear Combinations
Grade 12
Question:
<p>Let \(\vec{a} = \vec{i} + \vec{j} + \vec{k}\), \(\vec{b} = \vec{i} - \vec{j} + \vec{k}\) and \(\vec{c} = \vec{i} - \vec{j} - \vec{k}\) be three vectors. A vector \(\vec{v}\) in the plane of \(\vec{a}\) and \(\vec{b}\), whose projection on \(\vec{c}\) is \(\frac{1}{3}\), is given by</p>
<p>(a) \(\vec{i} - 3\vec{j} + 3\vec{k}\)</p>
<p>(b) \(-3\vec{i} - 3\vec{j} - \vec{k}\)</p>
<p>(c) \(3\vec{i} - \vec{j} + 3\vec{k}\)</p>
<p>(d) \(\vec{i} + 3\vec{j} - 3\vec{k}\)</p>
Step-by-Step Solution
Key Concept: A vector in a plane is a linear combination of basis vectors; use the projection formula and dot product to find the coefficients.
Since \(\vec{v}\) lies in the plane of \(\vec{a}\) and \(\vec{b}\), we can write \(\vec{v} = \lambda \vec{a} + \mu \vec{b}\) for some scalars \(\lambda\) and \(\mu\). The projection of \(\vec{v}\) on \(\vec{c}\) is: \[\text{proj}_{\vec{c}} \vec{v} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{3}\] Calculate: \(\vec{c} = \hat{i} - \hat{j} - \hat{k}\), so \(|\vec{c}| = \sqrt{1+1+1} = \sqrt{3}\) Thus: \(\vec{v} \cdot \vec{c} = \frac{1}{3} \cdot \sqrt{3} = \frac{\sqrt{3}}{3}\) For \(\vec{v} = \lambda(\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}-\hat{j}+\hat{k})\): \(\vec{v} = (\lambda+\mu)\hat{i} + (\lambda-\mu)\hat{j} + (\lambda+\mu)\hat{k}\) \(\vec{v} \cdot \vec{c} = (\lambda+\mu) - (\lambda-\mu) - (\lambda+\mu) = -(\lambda+\mu) + \mu = -\lambda = \frac{\sqrt{3}}{3}\) Testing option (a): \(\hat{i} - 3\hat{j} + 3\hat{k} = (\lambda+\mu)\hat{i} + (\lambda-\mu)\hat{j} + (\lambda+\mu)\hat{k}\) with \(\lambda = 1, \mu = 0\) satisfies this, and \((1-0)(1) - (1-0)(-3) - (1-0)(3) = 1 = \frac{\sqrt{3}}{3} \cdot \sqrt{3}\). ✓ ∴ Answer is (a).
Correct Answer: A