Key Concepts and Formulas

• Vector Addition: To add two vectors, add their corresponding components. If $\vec{u} = u_x \widehat i + u_y \widehat j + u_z \widehat k$ and $\vec{v} = v_x \widehat i + v_y \widehat j + v_z \widehat k$, then $\vec{u} + \vec{v} = (u_x + v_x) \widehat i + (u_y + v_y) \widehat j + (u_z + v_z) \widehat k$.
• Dot Product of Vectors: The dot product of two vectors $\vec{a} = a_x \widehat i + a_y \widehat j + a_z \widehat k$ and $\vec{b} = b_x \widehat i + b_y \widehat j + b_z \widehat k$ is given by $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$.
• Magnitude of a Vector: The magnitude of a vector $\vec{v} = v_x \widehat i + v_y \widehat j + v_z \widehat k$ is given by $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
• Scalar Projection: The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is the length of the projection of $\vec{a}$ onto $\vec{b}$ and is given by $\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

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Step-by-Step Solution

Step 1: Define the vectors and the sum vector. We are given the vector $\vec{a} = \widehat i + 2\widehat j + \widehat k$. We are also given two vectors, $\vec{v_1} = 2\widehat i + 4\widehat j - 5\widehat k$ and $\vec{v_2} = - \lambda \widehat i + 2\widehat j + 3\widehat k$. The problem states that the projection is onto the sum of these two vectors. Let's call this sum vector $\vec{b}$.

Step 2: Calculate the sum vector $\vec{b}$. We add the components of $\vec{v_1}$ and $\vec{v_2}$ to find $\vec{b}$. $\vec{b} = \vec{v_1} + \vec{v_2}$ $\vec{b} = (2\widehat i + 4\widehat j - 5\widehat k) + (-\lambda \widehat i + 2\widehat j + 3\widehat k)$ $\vec{b} = (2 - \lambda)\widehat i + (4 + 2)\widehat j + (-5 + 3)\widehat k$ $\vec{b} = (2 - \lambda)\widehat i + 6\widehat j - 2\widehat k$

Step 3: Set up the equation using the scalar projection formula. We are given that the projection of vector $\vec{a}$ onto vector $\vec{b}$ is 1. Using the scalar projection formula: $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 1$

Step 4: Calculate the dot product $\vec{a} \cdot \vec{b}$. We use the components of $\vec{a} = 1\widehat i + 2\widehat j + 1\widehat k$ and $\vec{b} = (2 - \lambda)\widehat i + 6\widehat j - 2\widehat k$. $\vec{a} \cdot \vec{b} = (1)(2 - \lambda) + (2)(6) + (1)(-2)$ $\vec{a} \cdot \vec{b} = 2 - \lambda + 12 - 2$ $\vec{a} \cdot \vec{b} = 12 - \lambda$

Step 5: Calculate the magnitude of $\vec{b}$, $|\vec{b}|$. We use the components of $\vec{b} = (2 - \lambda)\widehat i + 6\widehat j - 2\widehat k$. $|\vec{b}| = \sqrt{(2 - \lambda)^2 + (6)^2 + (-2)^2}$ $|\vec{b}| = \sqrt{(2 - \lambda)^2 + 36 + 4}$ $|\vec{b}| = \sqrt{(2 - \lambda)^2 + 40}$

Step 6: Substitute the dot product and magnitude into the projection equation and solve for $\lambda$. Now we substitute the expressions from Step 4 and Step 5 into the equation from Step 3: $\frac{12 - \lambda}{\sqrt{(2 - \lambda)^2 + 40}} = 1$ To solve for $\lambda$, we first multiply both sides by the denominator: $12 - \lambda = \sqrt{(2 - \lambda)^2 + 40}$ Before squaring, note that the left side, $12 - \lambda$, must be non-negative because it equals a square root. Thus, $12 - \lambda \ge 0$, which implies $\lambda \le 12$. Now, square both sides of the equation to eliminate the square root: $(12 - \lambda)^2 = \left(\sqrt{(2 - \lambda)^2 + 40}\right)^2$ $(12 - \lambda)^2 = (2 - \lambda)^2 + 40$ Expand both sides: $144 - 24\lambda + \lambda^2 = (4 - 4\lambda + \lambda^2) + 40$ $144 - 24\lambda + \lambda^2 = 44 - 4\lambda + \lambda^2$ The $\lambda^2$ terms cancel out: $144 - 24\lambda = 44 - 4\lambda$ Rearrange the terms to solve for $\lambda$: $144 - 44 = 24\lambda - 4\lambda$ $100 = 20\lambda$ Divide by 20: $\lambda = \frac{100}{20}$ $\lambda = 5$

Step 7: Verify the solution. We found $\lambda = 5$. We must check if this value satisfies the condition $\lambda \le 12$ derived earlier. Since $5 \le 12$, the solution is valid. Let's substitute $\lambda = 5$ back into the original projection equation: $\vec{a} \cdot \vec{b} = 12 - 5 = 7$. $|\vec{b}| = \sqrt{(2 - 5)^2 + 40} = \sqrt{(-3)^2 + 40} = \sqrt{9 + 40} = \sqrt{49} = 7$. The projection is $\frac{7}{7} = 1$, which matches the given condition.

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Common Mistakes & Tips

• Squaring Equations: When solving equations involving square roots, always square both sides carefully and remember to check for extraneous solutions by substituting the obtained values back into the equation before squaring.
• Vector Component Errors: Double-check the addition of vector components and the calculation of dot products. A single incorrect sign or number can lead to a wrong final answer.
• Distinguishing Scalar and Vector Projection: Be clear whether the question asks for the scalar projection (a number) or the vector projection (a vector). This question specifically asks for the scalar projection.

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Summary

This problem required the application of the scalar projection formula for vectors. We first found the resultant vector onto which the projection was made by summing the two given vectors. Then, we calculated the dot product of the vector to be projected and the resultant vector, as well as the magnitude of the resultant vector. By setting up and solving the scalar projection equation, we found the value of $\lambda$. Finally, we verified our solution to ensure it was valid.

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