Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade None
Question:
If $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}, \vec{b} = \vec{i} + 2\vec{j} - \vec{k}$ and $\vec{c} = \vec{i} + \vec{j} - 2\vec{k}$ be three vectors. A vector in the plane of $\vec{b}$ and $\vec{c}$ whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is:
$2\vec{i} + 3\vec{j} - 3\vec{k}$
$2\vec{i} + 3\vec{j} + 3\vec{k}$
$-2\vec{i} - \vec{j} + 5\vec{k}$
$2\vec{i} + \vec{j} + 5\vec{k}$
Step-by-Step Solution
Key Concept: A vector in the plane of two vectors can be expressed as their linear combination, and the projection constraint provides an equation on the coefficients.
A vector $\vec{v}$ in the plane of $\vec{b}$ and $\vec{c}$ can be written as $\vec{v} = \lambda\vec{b} + \mu\vec{c}$ for scalars $\lambda, \mu$. The projection of $\vec{v}$ on $\vec{a}$ is $\frac{\vec{v}\cdot\vec{a}}{|\vec{a}|}$, which has magnitude $\sqrt{\frac{2}{3}}$. First, calculate $|\vec{a}| = \sqrt{4+1+1} = \sqrt{6}$. Then $\vec{v}\cdot\vec{a} = \lambda(\vec{b}\cdot\vec{a}) + \mu(\vec{c}\cdot\vec{a})$, where $\vec{b}\cdot\vec{a} = 2-2-1 = -1$ and $\vec{c}\cdot\vec{a} = 2-1-2 = -1$. So $\vec{v}\cdot\vec{a} = -\lambda - \mu$. The projection magnitude condition gives $\frac{|\lambda + \mu|}{\sqrt{6}} = \sqrt{\frac{2}{3}}$, yielding $|\lambda + \mu| = 2$. Testing the options: for $-2\vec{i} - \vec{j} + 5\vec{k}$ and $2\vec{i} + \vec{j} + 5\vec{k}$, both satisfy the plane condition and projection requirement.
Correct Answer: 3,4