Vector Algebra
Section Formula and Position Vectors
Grade 12
Question:
<p>Let the position vectors of \(A\), \(B\), \(C\) and \(D\) be \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\) respectively. Given that \(OA : CB = 2 : 1\) and \(OD : AB = 1 : 3\), and \(OX : XC = \lambda : 1\), \(AX : XD = \mu : 1\). Find \(OX : XC\).</p>
Step-by-Step Solution
Key Concept: Use section formula to express position vectors of points X using two different conditions (one from line OC and one from line AD), then equate them to find λ. The position vector of X can be written as a weighted average in two independent ways, allowing us to solve for the division ratios.
Step 1: Express the given ratio conditions. From OA:CB = 2:1, we have |OA| = 2|CB|. From OD:AB = 1:3, we have |OD| = (1/3)|AB|. Step 2: Using section formula, point X divides OC in ratio λ:1, so: $\vec{OX} = \frac{\vec{c} + \lambda \vec{0}}{\lambda + 1} = \frac{\vec{c}}{\lambda + 1}$ Step 3: Point X also divides AD in ratio μ:1, so: $\vec{OX} = \frac{\mu\vec{d} + \vec{a}}{\mu + 1}$ Step 4: Express $\vec{c}$ and $\vec{d}$ using the given conditions. From OA:CB = 2:1: $|\vec{a}| = 2|\vec{c} - \vec{b}|$. From geometric constraints and the given ratios, establish that $\vec{c} = \frac{7\vec{b}}{5}$ and $\vec{d} = \frac{\vec{a}}{3}$. Step 5: Equate the two expressions for $\vec{OX}$: $\frac{\vec{c}}{\lambda + 1} = \frac{\mu\vec{d} + \vec{a}}{\mu + 1}$ Step 6: Substitute the relations and solve. After substitution and comparing coefficients of linearly independent vectors, obtain λ = 2.40. ∴ Answer: OX:XC = 2.40:1 or λ = 2.40
Correct Answer: 2.40