Question

\(\vec a,\vec b,\vec c\) are the position vectors of three points \(A,B,C\) respectively. If \[ \angle ABC=\frac{\pi}{2}, \] and \[ \overrightarrow{AB} =\hat{i}+4\hat{j}+(4-\lambda)\hat{k}, \qquad \overrightarrow{AC} =(\lambda-1)\hat{i}+6\hat{j}+(2-\lambda)\hat{k}, \] then \(\lambda=\)

• A. \(0\)
• B. \(-\dfrac13\)
• C. \(-\dfrac34\)
• D. \(\dfrac23\)

Hint:

If the angle between two vectors is \(90^\circ\), then \[ \boxed{\vec u\cdot\vec v=0.} \] Always form the vectors from the vertex of the given angle before taking the dot product.

Answer 1

The Correct Option is D

Step 1: Find \(\overrightarrow{BC}\). Since \[ \overrightarrow{BC} = \overrightarrow{AC}-\overrightarrow{AB}, \] we get \[ \overrightarrow{BC} = (\lambda-2)\hat{i} +2\hat{j} -2\hat{k}. \]

Step 2: Use the perpendicular condition. Given \[ \angle ABC=\frac{\pi}{2}, \] therefore, \[ \overrightarrow{BA}\cdot\overrightarrow{BC}=0. \] Now, \[ \overrightarrow{BA} = -\hat{i}-4\hat{j}-(4-\lambda)\hat{k}. \] Hence, \[ (-1)(\lambda-2) +(-4)(2) -(4-\lambda)(-2)=0. \]

Step 3: Solve for \(\lambda\). Simplifying, \[ -\lambda+2-8+8-2\lambda=0, \] \[ 2-3\lambda=0, \] \[ \lambda=\frac23. \] Therefore, \[ \boxed{\frac23}. \] Thus, \[ \boxed{(D)} \] is the correct answer.