Vectors
Vectors
Allen Star Batch
Grade 12
Question:
The position vectors of the points $A, B, C$ are respectively $(1,1,1), (1,-1,2), (0,2,-1)$. The unit vector parallel to the plane determined by $A, B, C$ and perpendicular to the vector $(1,0,1)$ is/are:
$\frac{5\vec{i} + \vec{j} - \vec{k}}{3\sqrt{3}}$
$\frac{\vec{i} + 5\vec{j} - \vec{k}}{3\sqrt{3}}$
$\frac{-5\vec{i} + \vec{j} + \vec{k}}{3\sqrt{3}}$
$\frac{-\vec{i} - 5\vec{j} + \vec{k}}{3\sqrt{3}}$
Step-by-Step Solution
Key Concept: The BAC-CAB rule for vector triple product explicitly demonstrates coplanarity as a linear combination of two vectors.
The vector $\vec{a} \times (\vec{b} \times \vec{c})$ is coplanar with $\vec{b}$ and $\vec{c}$ by the vector triple product identity. Use the BAC-CAB rule: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Since this expression is a linear combination of $\vec{b}$ and $\vec{c}$, the result lies in the plane containing these vectors. The perpendicularity to $\vec{a}$ is implied by the coplanarity condition.
Correct Answer: 2,4