Vector ProjectionJEE Main

Concept

The projection of $\vec a$ onto $\vec b$ is the 'shadow' $\vec a$ casts along $\vec b$. Its signed length is the scalar projection $(\vec a\cdot\vec b)/|\vec b|$; multiply by the unit vector $\hat b$ to get the vector projection.

Key formula

Derivation

By definition $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$, so $|\vec a|\cos\theta = (\vec a\cdot\vec b)/|\vec b|$ is the length of the shadow.

To turn that length into a vector along $\vec b$, multiply by $\hat b = \vec b/|\vec b|$, giving $\text{proj}_{\vec b}\vec a = \dfrac{\vec a\cdot\vec b}{|\vec b|^2}\vec b$.

The leftover $\vec a - \text{proj}_{\vec b}\vec a$ is perpendicular to $\vec b$ — the basis of orthogonal decomposition.

Scenarios to explore

• Vector Projection — Projection of one vector onto another.
• Perpendicular vectors — a · b = 0, so the projection is zero.
• Parallel vectors — The projection of a on b is a itself.

Real-world applications

• Resolving forces along and perpendicular to a surface.
• Work done $= \vec F\cdot\vec d$ is force projected on displacement.
• Gram–Schmidt orthogonalisation and least-squares fitting.

JEE exam tips

• If $\vec a\cdot\vec b = 0$ the vectors are perpendicular and the projection is zero.
• The component of $\vec a$ perpendicular to $\vec b$ has length $|\vec a|\sin\theta$.

Common mistakes

• Dividing by $|\vec b|$ once when the vector projection needs $|\vec b|^2$.
• Reporting the scalar projection as always positive — it carries the sign of $\cos\theta$.
• Projecting onto $\vec a$ when the question asks for projection onto $\vec b$.

Exam traps to avoid

• Scalar projection is a number (can be negative); vector projection is a vector.
• Projection onto $\vec b$ depends only on the direction of $\vec b$, not its magnitude.