Q.If with reference to the right handed system of mutually perpendicular unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$, $\vec{\alpha}=3\hat{i}-\hat{j}$ and $\vec{\beta}=2\hat{i}+\hat{j}-3\hat{k}$, then express $\vec{\beta}$ in the form $\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}$, where $\vec{\beta_1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta_2}$ is perpendicular to $\vec{\alpha}$.

Concept: Vector Projection — we decompose $\vec{\beta}$ into a component parallel to $\vec{\alpha}$ (the projection) and a component perpendicular to $\vec{\alpha}$.

Step 1: Find $\vec{\beta_1}$, the projection of $\vec{\beta}$ onto $\vec{\alpha}$.

The formula is

Step 2: Compute the dot product and magnitude.

$\vec{\beta} \cdot \vec{\alpha} = (2)(3) + (1)(-1) + (-3)(0) = 6 - 1 = 5$.

$|\vec{\alpha}|^2 = 3^2 + (-1)^2 = 9 + 1 = 10$.

Thus

Step 3: Find $\vec{\beta_2}$ as $\vec{\beta} - \vec{\beta_1}$. …

Given $\vec\alpha=3\hat i-\hat j$ and $\vec\beta=2\hat i+\hat j-3\hat k$, write $\vec\beta=\vec{\beta_1}+\vec{\beta_2}$ where $\vec{\beta_1}$ is the projection of $\vec\beta$ on $\vec\alpha$.

Parallel part (projection):

Perpendicular part: …

Method: Resolving a vector into parallel and perpendicular parts

To split $\vec{\beta}$ into a piece along $\vec{\alpha}$ and a piece perpendicular to $\vec{\alpha}$, take the projection for the parallel part and subtract it off for the perpendicular part.

Steps

Step 1: Parallel part = projection of $\vec{\beta}$ onto $\vec{\alpha}$

The denominator is $|\vec{\alpha}|^2$ (equivalently $\vec{\alpha}\cdot\vec{\alpha}$), not $|\vec{\alpha}|$ — this is what makes $\vec{\beta_1}$ come out parallel to $\vec{\alpha}$ with the correct length.

Step 2: Perpendicular part = what is left over

$\vec{\beta_2} = \vec{\beta} - \vec{\beta_1}.$ …

Common Mistakes

Mistake 1: Dividing by $|\vec{\alpha}|$ instead of $|\vec{\alpha}|^2$ in the projection.

Why it's wrong: $\dfrac{\vec{\beta}\cdot\vec{\alpha}}{|\vec{\alpha}|}\vec{\alpha}$ has the right direction but the wrong length, so the "parallel part" comes out scaled incorrectly. Correct approach: the vector projection uses $|\vec{\alpha}|^2$ in the denominator.

Mistake 2: Computing $\vec{\beta_2}$ as a separate projection instead of $\vec{\beta} - \vec{\beta_1}$. …