Vector Algebra
Projection of Vector in a Plane
nta_pyq_2026_jan
Grade None
Question:
Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\dfrac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to:
13
\dfrac{\sqrt{35}}{2}
\dfrac{\sqrt{21}}{2}
7
Step-by-Step Solution
Key Concept: $\vec{v}=\lambda\vec{a}+\mu\vec{b}=(2\lambda+\mu,\,-\lambda+3\mu,\,-\lambda-\mu)$. $\vec{v}\cdot\vec{c}=2\mu$. Projection $=\frac{|2\mu|}{\sqrt{14}}=\frac{1}{\sqrt{14}}\Rightarrow|\mu|=\frac{1}{2}$.
$|\vec{v}|=\dfrac{\sqrt{35}}{2}$.
Correct Answer: 2