Let a = 2i - j + k, b = i + 2j - k and c = i + j - 2k be three vectors. A vector in the plane of b a

Q: Wait, why did we use $\vec{u} = \vec{b} + \lambda\vec{c}$? Can't we use $\vec{u} = x\vec{b} + y\vec{c}$?

That is a brilliant question! You absolutely could use two variables, $x$ and $y$. However, notice that the problem asks for 'a' vector, and the scalar $\lambda$ effectively captures the ratio of the components. Using one variable simplifies the algebra significantly without losing generality. If you used two variables, you would find that one is dependent on the other, leading you back to the same result!

Q: Why do we have two answers? Did I make a calculation error?

Not at all! In fact, finding two answers is a sign that you are on the right track. The absolute value in the projection formula $|\frac{\vec{u} \cdot \vec{a}}{|\vec{a}|}| = \sqrt{\frac{2}{3}}$ implies that the dot product $\vec{u} \cdot \vec{a}$ can be either positive or negative. This corresponds to the vector $\vec{u}$ making either an acute or an obtuse angle with $\vec{a}$. Both are valid vectors in the plane!

Q: Is the projection formula $\frac{\vec{u} \cdot \vec{a}}{|\vec{a}|}$ or $\frac{\vec{u} \cdot \vec{a}}{|\vec{u}|}$?

Always remember: you project *onto* the vector you are dividing by. Since we are projecting $\vec{u}$ onto $\vec{a}$, we divide by the magnitude of $\vec{a}$. Think of it as 'normalizing' the direction of $\vec{a}$ to see how much of $\vec{u}$ lies along it.

Question

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$ be three vectors. A vector in the plane of $\vec{b}$ and $\vec{c}$, whose projection on $\vec{a}$ is of magnitude $\sqrt{2/3}$, is:

Options:

(A) $2\hat{i} + 3\hat{j} - 3\hat{k}$

(B) $2\hat{i} + 3\hat{j} + 3\hat{k}$

(C) $-2\hat{i} - \hat{j} + 5\hat{k}$

(D) $2\hat{i} + \hat{j} + 5\hat{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Vector $\vec{a}$ is like a spear pointing into the distance, while $\vec{b}$ and $\vec{c}$ define a flat, infinite sheet—a plane. Our goal is to find a specific vector $\vec{u}$ that lies perfectly flat on this sheet, such that its shadow, or projection, onto our spear $\vec{a}$ has a specific length of $\sqrt{\frac{2}{3}}$.

### The Linear Combination: Defining Our Vector

Since $\vec{u}$ must lie in the plane of $\vec{b}$ and $\vec{c}$, it must be a linear combination of them. We can write this as $\vec{u} = \vec{b} + \lambda\vec{c}$.

$\lambda$ is our degree of freedom. It allows us to slide along the plane, stretching and shrinking $\vec{c}$ to find the exact vector that satisfies our projection condition. Substituting the given components, we have:

$$\vec{u} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} + \hat{j} - 2\hat{k})$$

By grouping the components, we obtain:

$$\vec{u} = (1+\lambda)\hat{i} + (2+\lambda)\hat{j} + (-1-2\lambda)\hat{k}$$

This is our target vector, waiting for the right value of $\lambda$ to reveal itself.

### The Projection: The Shadow of the Vector

Now, we invoke the projection formula. The magnitude of the projection of $\vec{u}$ onto $\vec{a}$ is given by:

$$\left| \frac{\vec{u} \cdot \vec{a}}{|\vec{a}|} ight| = \sqrt{\frac{2}{3}}$$

First, let us find the magnitude of $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$. It is:

$$|\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$$

Next, the dot product $\vec{u} \cdot \vec{a}$ is calculated by multiplying corresponding components:

$$\vec{u} \cdot \vec{a} = (1+\lambda)(2) + (2+\lambda)(-1) + (-1-2\lambda)(1)$$

Expanding this, we get $2 + 2\lambda - 2 - \lambda - 1 - 2\lambda$, which simplifies beautifully to $-\lambda - 1$.

### The Algebraic Climax: Solving for Lambda

We now have our equation:

$$\left| \frac{-\lambda - 1}{\sqrt{6}} ight| = \sqrt{\frac{2}{3}}$$

Multiplying both sides by $\sqrt{6}$, we get:

$$|-\lambda - 1| = \sqrt{\frac{2}{3}} 	imes \sqrt{6} = \sqrt{4} = 2$$

The absolute value tells us that $-\lambda - 1$ can be either $2$ or $-2$.

**Case 1:** $-\lambda - 1 = 2$ leads to $\lambda = -3$. Substituting this back into our expression for $\vec{u}$, we get:

$$\vec{u} = -2\hat{i} - \hat{j} + 5\hat{k}$$

**Case 2:** $-\lambda - 1 = -2$ leads to $\lambda = 1$. Substituting this back, we get:

$$\vec{u} = 2\hat{i} + 3\hat{j} - 3\hat{k}$$

Both vectors are perfectly valid, existing in the plane and casting the exact shadow required. You have successfully navigated the geometry and the algebra.

Animated Solution for Mathematics - Vector Algebra: Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

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Visualized Solution

Defining the Vectors $a, b, c$

Vector in the Plane of $b$ and $c$

Expressing $u$ in Component Form

Grouping the Components of $u$

The Projection Formula

Calculating the Magnitude of $a$

Calculating the Dot Product $u \cdot a$

Setting up the Projection Equation

Simplifying the Equation

Solving for $λ$ : Case 1

Solving for $λ$ : Case 2

Final Answer

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Linear Combination

The Projection

The Algebraic Climax

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Visualized Solution

Defining the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

Vector in the Plane of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u in Component Form

Grouping the Components of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u