Let a = 4i + 3j and b = 3i - 4j + 5k and c is a vector such that c · (a × b) + 25 = 0, c · (i + j +

Q: Wait, why do we use the cross product $\vec{a} \times \vec{b}$? Can't we just solve for $\vec{c}$ directly?

Great question! We use the cross product because the problem gives us a condition involving $\vec{c} \cdot (\vec{a} \times \vec{b})$. This is the scalar triple product, which is a powerful tool in vector algebra. It provides a linear constraint on the components of $\vec{c}$ that we wouldn't be able to access otherwise. It's not just about solving for $\vec{c}$ directly; it's about using the geometry of the vectors to build the equations we need.

Q: Why is the projection formula $\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$? Is it always that?

Absolutely! The projection of any vector $\vec{c}$ onto another vector $\vec{a}$ is defined as the scalar component of $\vec{c}$ in the direction of $\vec{a}$. Mathematically, that is the dot product of $\vec{c}$ with the unit vector of $\vec{a}$, which is $\vec{c} \cdot \frac{\vec{a}}{|\vec{a}|}$. So, $\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|}$ is the standard and correct formula.

Q: Is there a faster way to solve the system of equations than substitution?

You could use Cramer's Rule or matrix inversion, but for a $3 \times 3$ system like this, substitution is often the most reliable way to avoid arithmetic errors. Since the third equation only involves $x$ and $y$, substitution is actually quite efficient here. Don't rush the algebra; take it one step at a time, and you'll find the solution with confidence.

Question

Let $\vec{a} = 4\hat{i} + 3\hat{j}$ and $\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot (\vec{a} 	imes \vec{b}) + 25 = 0, \vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ and projection of $\vec{c}$ on $\vec{a}$ is 1, then the projection of $\vec{c}$ on $\vec{b}$ equals:

Options:

(A) $\frac{5}{\sqrt{2}}$

(B) $\frac{1}{5}$

(C) $\frac{1}{\sqrt{2}}$

(D) $\frac{3}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system. You have two vectors, $\vec{a}$ and $\vec{b}$, firmly planted in this space.

Vector $\vec{a} = 4\hat{i} + 3\hat{j}$ is a flat, two-dimensional arrow, while $\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ reaches out into the third dimension. We are tasked with finding a third vector, $\vec{c}$, using the laws of vector algebra.

### The Foundation: The Cross Product

Our first clue is the condition $\vec{c} \cdot (\vec{a} 	imes \vec{b}) + 25 = 0$. Before we can use this, we need to calculate $\vec{a} 	imes \vec{b}$.

We set up our determinant:

$$
\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 3 & 0 \ 3 & -4 & 5 \end{vmatrix}
$$

Expanding this, we get $15\hat{i} - 20\hat{j} - 25\hat{k}$. By setting $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$, the condition becomes:

$$
(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (15\hat{i} - 20\hat{j} - 25\hat{k}) = -25
$$

Simplifying this, we get $15x - 20y - 25z = -25$, which reduces to our first solid piece of evidence:

$$
3x - 4y - 5z = -5
$$

### Building the System

The second clue is $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$. This implies that the sum of the components of $\vec{c}$ is exactly four:

$$
x + y + z = 4
$$

The third clue is the projection of $\vec{c}$ on $\vec{a}$, which is $1$. Using the formula $\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$, and knowing $|\vec{a}| = \sqrt{4^2 + 3^2} = 5$, we get:

$$
\frac{4x + 3y}{5} = 1 \Rightarrow 4x + 3y = 5
$$

### The Unmasking

We now have a system of three equations:
1) $3x - 4y - 5z = -5$
2) $x + y + z = 4$
3) $4x + 3y = 5$

From the second equation, we write $z = 4 - x - y$. Substituting this into the first equation:

$$
3x - 4y - 5(4 - x - y) = -5 \Rightarrow 8x + y = 15
$$

Now we solve the system of two equations: $4x + 3y = 5$ and $8x + y = 15$. Solving for $y$ in the second gives $y = 15 - 8x$.

Plugging this into the first:

$$
4x + 3(15 - 8x) = 5 \Rightarrow 4x + 45 - 24x = 5 \Rightarrow -20x = -40
$$

Thus, $x = 2$. Substituting back, we find $y = 15 - 16 = -1$, and $z = 4 - 2 - (-1) = 3$. Our mystery vector is **$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$**.

### The Final Victory

The problem asks for the projection of $\vec{c}$ on $\vec{b}$, given by the formula $\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|}$.

The dot product is:

$$
\vec{c} \cdot \vec{b} = (2)(3) + (-1)(-4) + (3)(5) = 6 + 4 + 15 = 25
$$

The magnitude $|\vec{b}|$ is:

$$
|\vec{b}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{50} = 5\sqrt{2}
$$

The final projection is:

$$
\frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}
$$

The final answer is **$\frac{5}{\sqrt{2}}$**.

Select Answer:

Visualized Solution

Identify Given Vectors $a$ and $b$

Calculate Cross Product $a \times b$

Define Vector $c$ and Equation 1

Formulate Equation 2

Formulate Equation 3 using Projection

Eliminate $z$ from Equations 1 and 2

Solve for $x$ and $y$

Find $z$ and Vector $c$

Calculate Projection of $c$ on $b$

Final Result

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Foundation

Building the System

The Unmasking

The Final Victory

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Select Answer:

Visualized Solution

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