If a is nonzero vector such that its projections on the vectors 2i - j + 2k, i + 2j - 2k and k are e

Q: Wait, why do we divide by the magnitude in the projection formula? Can't we just use the dot product?

That is a classic trap! The dot product $\vec{a} \cdot \vec{b}$ gives you the projection *scaled* by the magnitude of $\vec{b}$. If you only use the dot product, you are implicitly assuming that all the vectors you are projecting onto are unit vectors. Since your vectors have different magnitudes, you must divide by $|\vec{b}|$ to get the true geometric shadow length.

Q: I have three variables ($x, y, z$) but only two equations. How can I find a unique vector?

You are absolutely right that you cannot find a unique vector $\vec{a}$ with just two equations—but you don't need to! The question asks for a *unit vector* along $\vec{a}$. A unit vector depends only on the *direction* of $\vec{a}$, which is determined by the ratios of its components ($x:y:z$). Any vector proportional to the one you find will yield the same unit vector.

Q: Is there a trick to solving the system of equations quickly?

Yes! Since the equations are homogeneous (they equal zero), you can use the cross-multiplication method (or Cramer's rule) to find the ratios $x:y:z$ instantly. It is much faster than substitution and less prone to algebraic errors. Just be very careful with the signs during the cross-multiplication!

Question

If $\vec{a}$ is nonzero vector such that its projections on the vectors $2\hat{i} - \hat{j} + 2\hat{k}, \hat{i} + 2\hat{j} - 2\hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is:

Options:

(A) $\frac{1}{\sqrt{155}}(-7\hat{i} + 9\hat{j} + 5\hat{k})$

(B) $\frac{1}{\sqrt{155}}(-7\hat{i} + 9\hat{j} - 5\hat{k})$

(C) $\frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})$

(D) $\frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} - 5\hat{k})$

Sigma Percentile · Accepted Answer

### The Mystery of the Hidden Vector

Imagine you are standing in a vast, three-dimensional space. You have a mysterious vector, $\vec{a}$, whose direction and magnitude are unknown.

We represent this vector as $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$. Our goal is to uncover the identity of this vector using the provided clues: the projections of $\vec{a}$ onto three specific vectors are identical.

### The Geometry of Shadows

To solve this, we must first understand the concept of a projection. When we project a vector $\vec{a}$ onto a vector $\vec{b}$, we are finding the length of the shadow that $\vec{a}$ casts along the direction of $\vec{b}$.

The formula for this scalar projection is:

$$	ext{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

We are given three vectors: $\vec{b}_1 = 2\hat{i} - \hat{j} + 2\hat{k}$, $\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}$, and $\vec{b}_3 = \hat{k}$.

First, we calculate the magnitudes of these vectors:

$$|\vec{b}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$$

$$|\vec{b}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = 3$$

$$|\vec{b}_3| = \sqrt{0^2 + 0^2 + 1^2} = 1$$

### Setting the Stage

The problem states that the projections are equal. This gives us the following triple equality:

$$\frac{\vec{a} \cdot \vec{b}_1}{|\vec{b}_1|} = \frac{\vec{a} \cdot \vec{b}_2}{|\vec{b}_2|} = \frac{\vec{a} \cdot \vec{b}_3}{|\vec{b}_3|}$$

Substituting our known values, we obtain:

$$\frac{2x - y + 2z}{3} = \frac{x + 2y - 2z}{3} = \frac{z}{1}$$

By equating the first and third parts, we get $2x - y + 2z = 3z$, which simplifies to:

$$2x - y - z = 0$$

By equating the second and third parts, we get $x + 2y - 2z = 3z$, which simplifies to:

$$x + 2y - 5z = 0$$

### The Elegance of Ratios

We now have a system of two homogeneous linear equations: $2x - y - z = 0$ and $x + 2y - 5z = 0$. We solve for the ratios of $x, y,$ and $z$ using the cross-multiplication method.

We set up the ratios as follows:

$$\frac{x}{(-1)(-5) - (-1)(2)} = \frac{y}{(-1)(1) - (2)(-5)} = \frac{z}{(2)(2) - (-1)(1)}$$

Simplifying the denominators, we get:

$$\frac{x}{5 + 2} = \frac{y}{-1 + 10} = \frac{z}{4 + 1} \Rightarrow \frac{x}{7} = \frac{y}{9} = \frac{z}{5}$$

This indicates that our vector $\vec{a}$ is proportional to $7\hat{i} + 9\hat{j} + 5\hat{k}$.

### The Final Normalization

The question asks for a unit vector, which must have a magnitude of $1$. We take our proportional vector $\vec{v} = 7\hat{i} + 9\hat{j} + 5\hat{k}$ and divide it by its magnitude:

$$|\vec{v}| = \sqrt{7^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155}$$

Thus, the final unit vector is:

$$\mathbf{\hat{a} = \frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})}$$

Animated Solution for Mathematics - Vector Algebra: If $a$ is nonzero vector such that its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $a$ is:

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Visualized Solution

$Defining Vector a$

$The Projection Formula$

$Identifying the Target Vectors$

$Calculating Magnitudes$

$Equating the Projections$

$Extracting Equation 1$

$Extracting Equation 2$

$Solving by Cross-Multiplication$

$Finding the Direction Ratios$

$Magnitude of the Proportional Vector$

$The Final Unit Vector$

The Sigma Insight: Scalar (Dot) Product

The Mystery of the Hidden Vector

The Geometry of Shadows

Setting the Stage

The Elegance of Ratios

The Final Normalization

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining Vector a

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Projection Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying the Target Vectors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Magnitudes

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Projections

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting Equation 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting Equation 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving by Cross-Multiplication

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Direction Ratios

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitude of the Proportional Vector

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Final Unit Vector