What will be the projection of vector A = i + j + k on vector B = i + j ?
(a) √2(i + j + k)
(b) 2(i

Q: Why do we divide by $|\mathbf{B}|^2$ instead of just $|\mathbf{B}|$?

The scalar projection (which is just the length of the shadow) is $\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}$. To convert this length into a vector, we must multiply it by the unit vector pointing in the direction of $\mathbf{B}$, which is $\hat{\mathbf{B}} = \frac{\mathbf{B}}{|\mathbf{B}|}$. Multiplying these two terms gives $\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2} \mathbf{B}$.

Q: Is the projection of $\mathbf{A}$ on $\mathbf{B}$ always in the exact same direction as $\mathbf{B}$?

Yes, if the dot product is positive. However, if the dot product is negative (meaning the angle between them is obtuse), the projection will point in the exact opposite direction of $\mathbf{B}$ (i.e., $-\mathbf{B}$).

Q: Can the projection vector be longer than the original vector $\mathbf{A}$?

No. The length of the projection is $|\mathbf{A}| \cos\theta$. Since the value of $\cos\theta$ is always between -1 and 1, the projection's length can never exceed the magnitude of the original vector $\mathbf{A}$.

Q: Why did the answer turn out to be exactly vector $\mathbf{B}$?

Because the scalar multiplier $\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2}$ evaluated to exactly 1. Geometrically, dropping a perpendicular from the point $(1,1,1)$ down to the $xy$-plane lands exactly on the point $(1,1,0)$, which corresponds perfectly to vector $\mathbf{B}$.

Question

What will be the projection of vector $\mathbf{A} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$ on vector $\mathbf{B} = \hat{\mathbf{i}} + \hat{\mathbf{j}}$?

Options:

(A) $\sqrt{2}(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}})$

(B) $2(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}})$

(C) $\sqrt{2}(\hat{\mathbf{i}} + \hat{\mathbf{j}})$

(D) $(\hat{\mathbf{i}} + \hat{\mathbf{j}})$

Sigma Percentile · Accepted Answer

### Visualizing the Vectors in Space

Imagine you are standing in a three-dimensional coordinate system. Let's look at the two vectors given to us.

Vector $\mathbf{A} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$ has components in all three directions: $x$, $y$, and $z$. It points diagonally outwards into space.

On the other hand, vector $\mathbf{B} = \hat{\mathbf{i}} + \hat{\mathbf{j}}$ only has components in the $x$ and $y$ directions. This means it lies perfectly flat on the floor, which is the $xy$-plane.

### The Master Equation for Projection

Our goal is to find the **vector projection** of $\mathbf{A}$ onto $\mathbf{B}$. Think of this geometrically: if you were to shine a light straight down from the tip of vector $\mathbf{A}$ onto the line created by vector $\mathbf{B}$, the shadow it casts is our projection.

Mathematically, the formula for the vector projection of $\mathbf{A}$ on $\mathbf{B}$ is given by:

$$ 	ext{Proj}_{\mathbf{B}}\mathbf{A} = \left( \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2} ight) \mathbf{B} $$

This formula might look intimidating, but it is incredibly logical. The term $\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}$ gives us the *length* of the shadow (the scalar projection). We then divide by another $|\mathbf{B}|$ and multiply by the vector $\mathbf{B}$ itself to give that shadow the correct *direction*.

### Executing the Dot Product

Let's carefully substitute our known vectors into this formula. First, we need the dot product of the two vectors.

$$ \mathbf{A} \cdot \mathbf{B} = (\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) \cdot (\hat{\mathbf{i}} + \hat{\mathbf{j}}) $$

Remember, the dot product simply multiplies the corresponding components of each vector. 
- The $\hat{\mathbf{i}}$ components give $1 	imes 1 = 1$.
- The $\hat{\mathbf{j}}$ components give $1 	imes 1 = 1$.
- Vector $\mathbf{B}$ has no $\hat{\mathbf{k}}$ component (it is $0\hat{\mathbf{k}}$), so that gives $1 	imes 0 = 0$.

Adding these up:

$$ \mathbf{A} \cdot \mathbf{B} = 1 + 1 + 0 = 2 $$

### Finding the Magnitude Squared

Next, we evaluate the denominator, which is the magnitude squared of vector $\mathbf{B}$. The magnitude squared is simply the sum of the squares of its components.

Since vector $\mathbf{B} = \hat{\mathbf{i}} + \hat{\mathbf{j}}$, we square its coefficients:

$$ |\mathbf{B}|^2 = 1^2 + 1^2 = 2 $$

### The Final Calculation

We are at the final step. Let's plug our results back into the master projection formula. We have the dot product ($2$) divided by the magnitude squared ($2$).

$$ 	ext{Proj}_{\mathbf{B}}\mathbf{A} = \left( \frac{2}{2} ight) (\hat{\mathbf{i}} + \hat{\mathbf{j}}) $$

The fraction simplifies perfectly to $1$. This scalar multiplies our vector $\mathbf{B}$.

$$ 	ext{Proj}_{\mathbf{B}}\mathbf{A} = 1 	imes (\hat{\mathbf{i}} + \hat{\mathbf{j}}) = \hat{\mathbf{i}} + \hat{\mathbf{j}} $$

This is our final projected vector.

Take a moment to appreciate what just happened. The projection of vector $\mathbf{A}$ onto $\mathbf{B}$ turned out to be **exactly vector $\mathbf{B}$ itself**! Geometrically, because $\mathbf{A}$ is just $\mathbf{B}$ plus an extra $\hat{\mathbf{k}}$ component, dropping a perpendicular from $\mathbf{A}$ straight down to the $xy$-plane lands perfectly on the tip of $\mathbf{B}$. Beautiful, isn't it?

Animated Solution for Physics - Kinematics: What will be the projection of vector $A = \hat{i} + \hat{j} + \hat{k}$ on vector $B = \hat{i} + \hat{j}$ ?

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Visualized Solution

\text{Visualizing the Vectors}

\text{Vector Projection Formula}

\text{Setting up the Values}

\text{Calculating Dot Product}

\text{Calculating Magnitude Squared}

\text{Final Projection Vector}

\text{Geometric Insight}

The Sigma Insight: Dot and Cross Products

Visualizing the Vectors in Space

The Master Equation for Projection

Executing the Dot Product

Finding the Magnitude Squared

The Final Calculation

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Animated Solution for Physics - Kinematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What will be the projection of vector A=i^+j^​+k^ on vector B=i^+j^​?