Key Concepts and Formulas

• Scalar Triple Product (STP): For vectors $\overrightarrow u, \overrightarrow v, \overrightarrow w$, $\overrightarrow u \cdot (\overrightarrow v \times \overrightarrow w)$ can be computed as a determinant if components are known.
• Dot Product: $\overrightarrow u \cdot \overrightarrow v = u_1v_1 + u_2v_2 + u_3v_3$.
• Cross Product: $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$.
• Magnitude of a Vector: $|\overrightarrow u| = \sqrt{u_1^2 + u_2^2 + u_3^2}$.
• Scalar Projection: The projection of $\overrightarrow u$ on $\overrightarrow v$ is $\frac{\overrightarrow u \cdot \overrightarrow v}{|\overrightarrow v|}$.

Step-by-Step Solution

Step 1: Calculate the cross product $\overrightarrow a \times \overrightarrow b$. We are given $\overrightarrow a = 4\widehat i + 3\widehat j$ and $\overrightarrow b = 3\widehat i - 4\widehat j + 5\widehat k$. To use the first condition, we need to compute $\overrightarrow a \times \overrightarrow b$. $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{vmatrix} = \widehat i(3 \cdot 5 - 0 \cdot (-4)) - \widehat j(4 \cdot 5 - 0 \cdot 3) + \widehat k(4 \cdot (-4) - 3 \cdot 3)$ $\overrightarrow a \times \overrightarrow b = \widehat i(15 - 0) - \widehat j(20 - 0) + \widehat k(-16 - 9) = 15\widehat i - 20\widehat j - 25\widehat k$

Step 2: Use the first condition to form an equation involving $\overrightarrow c$. The first condition is $\overrightarrow c \cdot (\overrightarrow a \times \overrightarrow b) + 25 = 0$. Let $\overrightarrow c = x\widehat i + y\widehat j + z\widehat k$. Substituting the result from Step 1: $(x\widehat i + y\widehat j + z\widehat k) \cdot (15\widehat i - 20\widehat j - 25\widehat k) + 25 = 0$ Using the dot product: $15x - 20y - 25z + 25 = 0$ Dividing by 5, we get our first equation: $3x - 4y - 5z + 5 = 0 \quad \text{(Equation 1)}$

Step 3: Use the second condition to form another equation involving $\overrightarrow c$. The second condition is $\overrightarrow c \cdot (\widehat i + \widehat j + \widehat k) = 4$. $(x\widehat i + y\widehat j + z\widehat k) \cdot (\widehat i + \widehat j + \widehat k) = 4$ Using the dot product: $x(1) + y(1) + z(1) = 4$ $x + y + z = 4 \quad \text{(Equation 2)}$

Step 4: Use the third condition to form a third equation involving $\overrightarrow c$. The projection of $\overrightarrow c$ on $\overrightarrow a$ is 1. The formula for projection is $\frac{\overrightarrow c \cdot \overrightarrow a}{|\overrightarrow a|}$. First, find the magnitude of $\overrightarrow a$: $|\overrightarrow a| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ Next, find the dot product $\overrightarrow c \cdot \overrightarrow a$: $\overrightarrow c \cdot \overrightarrow a = (x\widehat i + y\widehat j + z\widehat k) \cdot (4\widehat i + 3\widehat j) = 4x + 3y$ Now, apply the projection condition: $\frac{4x + 3y}{5} = 1$ $4x + 3y = 5 \quad \text{(Equation 3)}$

Step 5: Solve the system of linear equations for $x, y, z$. We have the system:

1. $3x - 4y - 5z = -5$
2. $x + y + z = 4$
3. $4x + 3y = 5$

From Equation 2, we can express $z$ as $z = 4 - x - y$. Substitute this into Equation 1: $3x - 4y - 5(4 - x - y) = -5$ $3x - 4y - 20 + 5x + 5y = -5$ $8x + y - 20 = -5$ $8x + y = 15 \quad \text{(Equation 4)}$ Now we have a system of two equations with two variables: 3. $4x + 3y = 5$ 4. $8x + y = 15$

From Equation 4, $y = 15 - 8x$. Substitute this into Equation 3: $4x + 3(15 - 8x) = 5$ $4x + 45 - 24x = 5$ $-20x = 5 - 45$ $-20x = -40$ $x = 2$ Substitute $x=2$ into $y = 15 - 8x$: $y = 15 - 8(2) = 15 - 16 = -1$ Substitute $x=2$ and $y=-1$ into $z = 4 - x - y$: $z = 4 - 2 - (-1) = 4 - 2 + 1 = 3$ Thus, $\overrightarrow c = 2\widehat i - \widehat j + 3\widehat k$.

Step 6: Calculate the projection of $\overrightarrow c$ on $\overrightarrow b$. The projection of $\overrightarrow c$ on $\overrightarrow b$ is given by $\frac{\overrightarrow c \cdot \overrightarrow b}{|\overrightarrow b|}$. First, find the magnitude of $\overrightarrow b$: $|\overrightarrow b| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$ Next, find the dot product $\overrightarrow c \cdot \overrightarrow b$: $\overrightarrow c \cdot \overrightarrow b = (2\widehat i - \widehat j + 3\widehat k) \cdot (3\widehat i - 4\widehat j + 5\widehat k)$ $\overrightarrow c \cdot \overrightarrow b = (2)(3) + (-1)(-4) + (3)(5) = 6 + 4 + 15 = 25$ Finally, calculate the projection: $\text{Proj}_{\overrightarrow b} \overrightarrow c = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$

Common Mistakes & Tips

• Arithmetic Errors: Be extremely careful with signs and basic arithmetic when calculating dot and cross products, as errors can easily propagate.
• Formula Recall: Ensure accurate recall of the formulas for cross product, dot product, magnitude, and vector projection.
• Systematic Solving: Treat the problem as setting up and solving a system of linear equations. This structured approach helps avoid missing conditions or making logical errors.

Summary

We first computed the cross product $\overrightarrow a \times \overrightarrow b$. Then, we used the three given conditions to establish a system of three linear equations in terms of the components of $\overrightarrow c$. By solving this system, we determined $\overrightarrow c$. Finally, we used the definition of scalar projection to find the projection of $\overrightarrow c$ onto $\overrightarrow b$.

The final answer is $\boxed{\frac{5}{\sqrt2}}$.