Question

Let \(\vec{2i} - \vec{j} - \vec{k}, \vec{5i} + \vec{j} - 2\vec{k}, -13\vec{i} - 11 \vec{j} + 4 \vec{k}\) be the position vectors of three points \(A, B, C\) respectively. If \(\overrightarrow{AB} = \lambda \overrightarrow{BC}\) and \(\overrightarrow{AC} = \mu \overrightarrow{CB}\), then \(\lambda + \mu =\)

• A. \(1\)
• B. \(-1\)
• C. \(2\)
• D. \(-2\)

Hint:

Find vector components, use scalar multiplication to solve for \(\lambda\) and \(\mu\).

Answer 1

The Correct Option is B

Step 1: Compute vectors
\[ \overrightarrow{AB} = \vec{5i} + \vec{j} - 2 \vec{k} - (\vec{2i} - \vec{j} - \vec{k}) = 3 \vec{i} + 2 \vec{j} - \vec{k} \] \[ \overrightarrow{BC} = -13 \vec{i} - 11 \vec{j} + 4 \vec{k} - (\vec{5i} + \vec{j} - 2 \vec{k}) = -18 \vec{i} - 12 \vec{j} + 6 \vec{k} \] \[ \overrightarrow{AC} = -13 \vec{i} - 11 \vec{j} + 4 \vec{k} - (\vec{2i} - \vec{j} - \vec{k}) = -15 \vec{i} - 10 \vec{j} + 5 \vec{k} \] Step 2: Use given equations
\[ \overrightarrow{AB} = \lambda \overrightarrow{BC} \implies 3 \vec{i} + 2 \vec{j} - \vec{k} = \lambda (-18 \vec{i} - 12 \vec{j} + 6 \vec{k}) \] \[ \overrightarrow{AC} = \mu \overrightarrow{CB} \implies -15 \vec{i} - 10 \vec{j} + 5 \vec{k} = \mu (18 \vec{i} + 12 \vec{j} - 6 \vec{k}) \] Step 3: Find \(\lambda\) and \(\mu\)
Compare components: \[ 3 = -18 \lambda \implies \lambda = -\frac{1}{6} \] \[ -15 = 18 \mu \implies \mu = -\frac{5}{6} \] Step 4: Sum
\[ \lambda + \mu = -\frac{1}{6} - \frac{5}{6} = -1 \]