Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Let $\vec{a}=a\vec{i}+b\vec{j}+c\vec{k}$, $\vec{\beta}=b\vec{i}+c\vec{j}+a\vec{k}$ and $\vec{\gamma}=c\vec{i}+a\vec{j}+b\vec{k}$ be three coplanar vectors, and $\vec{v}=\vec{i}+\vec{j}+\vec{k}$. Then $\vec{v}$ may be:
$\perp$ to $\vec{a}$, $\vec{\beta}$, $\vec{\gamma}$
$\parallel$ to $\vec{a}$, $\vec{\beta}$, $\vec{\gamma}$
anti $\parallel$ to $\vec{a}$, $\vec{\beta}$, $\vec{\gamma}$
None of these
Step-by-Step Solution
Key Concept: The coplanarity condition forces either $a+b+c=0$ or $a=b=c$, leading to three distinct geometric relationships between $\vec{v}$ and the given vectors.
Since $\vec{\alpha}$, $\vec{\beta}$, $\vec{\gamma}$ are coplanar, we have $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$. This determinant equals $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Therefore either $a+b+c=0$ or $a=b=c$. When $a+b+c=0$: $\vec{\alpha}+\vec{\beta}+\vec{\gamma}=\vec{0}$, so $\vec{v}$ is perpendicular to all three vectors. When $a=b=c$: $\vec{\alpha}=\vec{\beta}=\vec{\gamma}=a(\vec{i}+\vec{j}+\vec{k})=a\vec{v}$, making $\vec{v}$ parallel (or anti-parallel if $a<0$) to all three vectors. All three cases are valid.
Correct Answer: 1,2,3