Q.Find the projection of the vector $\vec{a}=2\hat{i}+3\hat{j}+2\hat{k}$ on the vector $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$.

Concept: Vector Projection — the scalar projection of $\vec{a}$ onto $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

Step 1: Compute the dot product.

$\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10$.

Step 2: Find the magnitude of $\vec{b}$.

$|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$. …

Concept First: What Does "Projection" Mean?

When we talk about the projection of one vector onto another, we're asking: If you shine a light straight down onto $\vec{b}$, how long is the shadow that $\vec{a}$ casts along $\vec{b}$?

This "shadow length" is a scalar — it tells you the magnitude of $\vec{a}$'s component in the direction of $\vec{b}$. The formula comes directly from the dot product, which measures how much two vectors "align":

Why does this work? The dot product $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between them. Dividing by $|\vec{b}|$ leaves $|\vec{a}|\cos\theta$ — exactly the length of $\vec{a}$ along $\vec{b}$.

Step-by-Step Solution

1. Compute the dot product $\vec{a} \cdot \vec{b}$

The dot product is the sum of the products of corresponding components:

2. Find the magnitude of $\vec{b}$ …

Method: Scalar projection of one vector onto another

Use this for "projection of $\vec{a}$ on $\vec{b}$" — how much of $\vec{a}$ lies along $\vec{b}$.

Steps

Step 1: Use the scalar-projection formula.

The vector you project onto ($\vec{b}$) supplies the magnitude in the denominator — divide by $|\vec{b}|$, not $|\vec{a}|$.

Step 2: Compute the dot product and the correct magnitude.

$\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3$ and $|\vec{b}|=\sqrt{b_1^2+b_2^2+b_3^2}$. …

Common Mistakes

Mistake 1: Dividing by $|\vec{a}|$ instead of $|\vec{b}|$.

Why it's wrong: the projection of $\vec{a}$ on $\vec{b}$ uses the vector projected onto in the denominator: $\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$. Dividing by $|\vec{a}|$ answers a different question. Correct approach: use $|\vec{b}|=\sqrt6$, giving $\frac{10}{\sqrt6}$.

Mistake 2: Confusing the scalar projection with the vector projection (or using $|\vec{b}|^2$). …