Key Concepts and Formulas

• Orthogonality of Vectors: Two non-zero vectors are orthogonal (perpendicular) if the angle between them is $90^\circ$.
• Dot Product: The dot product of two vectors $\overrightarrow u = u_1\widehat{i} + u_2\widehat{j} + u_3\widehat{k}$ and $\overrightarrow v = v_1\widehat{i} + v_2\widehat{j} + v_3\widehat{k}$ is given by $\overrightarrow u \cdot \overrightarrow v = u_1v_1 + u_2v_2 + u_3v_3$.
• Condition for Orthogonality: Two vectors $\overrightarrow u$ and $\overrightarrow v$ are orthogonal if and only if their dot product is zero, i.e., $\overrightarrow u \cdot \overrightarrow v = 0$.
• Mutually Orthogonal Vectors: A set of vectors is mutually orthogonal if every pair of distinct vectors in the set is orthogonal.

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Step-by-Step Solution

Step 1: Understand the problem and given information. We are given three vectors: $\overrightarrow a = \widehat i - \widehat j + 2\widehat k$ $\overrightarrow b = 2\widehat i + 4\widehat j + \widehat k$ $\overrightarrow c = \lambda \widehat i + \widehat j + \mu \widehat k$

The problem states that these three vectors are mutually orthogonal. This means that the dot product of any two distinct vectors from this set must be zero: $\overrightarrow a \cdot \overrightarrow b = 0$ $\overrightarrow a \cdot \overrightarrow c = 0$ $\overrightarrow b \cdot \overrightarrow c = 0$

Our goal is to find the values of $\lambda$ and $\mu$.

Step 2: Apply the orthogonality condition $\overrightarrow a \cdot \overrightarrow c = 0$. Since $\overrightarrow a$ and $\overrightarrow c$ are orthogonal, their dot product is zero. $(\widehat i - \widehat j + 2\widehat k) \cdot (\lambda \widehat i + \widehat j + \mu \widehat k) = 0$ Using the dot product formula: $(1)(\lambda) + (-1)(1) + (2)(\mu) = 0$ $\lambda - 1 + 2\mu = 0$ $\lambda + 2\mu = 1 \quad \text{(Equation 1)}$ This equation relates $\lambda$ and $\mu$ based on the orthogonality of $\overrightarrow a$ and $\overrightarrow c$.

Step 3: Apply the orthogonality condition $\overrightarrow b \cdot \overrightarrow c = 0$. Since $\overrightarrow b$ and $\overrightarrow c$ are orthogonal, their dot product is zero. $(2\widehat i + 4\widehat j + \widehat k) \cdot (\lambda \widehat i + \widehat j + \mu \widehat k) = 0$ Using the dot product formula: $(2)(\lambda) + (4)(1) + (1)(\mu) = 0$ $2\lambda + 4 + \mu = 0$ $2\lambda + \mu = -4 \quad \text{(Equation 2)}$ This equation provides another relationship between $\lambda$ and $\mu$ based on the orthogonality of $\overrightarrow b$ and $\overrightarrow c$.

Step 4: Solve the system of linear equations for $\lambda$ and $\mu$. We now have a system of two linear equations with two unknowns:

1. $\lambda + 2\mu = 1$
2. $2\lambda + \mu = -4$

We can solve this system using the elimination method. Multiply Equation 2 by 2: $2 \times (2\lambda + \mu) = 2 \times (-4)$ $4\lambda + 2\mu = -8 \quad \text{(Equation 3)}$ Now, subtract Equation 1 from Equation 3: $(4\lambda + 2\mu) - (\lambda + 2\mu) = -8 - 1$ $4\lambda - \lambda + 2\mu - 2\mu = -9$ $3\lambda = -9$ $\lambda = \frac{-9}{3}$ $\lambda = -3$ Substitute the value of $\lambda = -3$ into Equation 1: $(-3) + 2\mu = 1$ $2\mu = 1 + 3$ $2\mu = 4$ $\mu = \frac{4}{2}$ $\mu = 2$ Thus, we find $\lambda = -3$ and $\mu = 2$.

Step 5: State the result and compare with the options. The values we found are $\lambda = -3$ and $\mu = 2$. Therefore, $(\lambda, \mu) = (-3, 2)$. Comparing this with the given options: (A) $(2, -3)$ (B) $(-2, 3)$ (C) $(3, -2)$ (D) $(-3, 2)$

Our result matches option (D).

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Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating dot products and solving systems of equations. A single sign error can lead to the wrong answer.
• Using only one orthogonality condition: The problem states "mutually orthogonal," which implies all pairs are orthogonal. You need to use at least two pairs to form two independent equations to solve for two unknowns.
• Algebraic Manipulation: Ensure accuracy when multiplying equations or performing subtraction/addition in the elimination method.

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Summary The problem requires the application of the dot product property for orthogonal vectors. By setting the dot product of $(\overrightarrow a, \overrightarrow c)$ and $(\overrightarrow b, \overrightarrow c)$ to zero, we form a system of two linear equations in $\lambda$ and $\mu$. Solving this system yields $\lambda = -3$ and $\mu = 2$.

The final answer is $\boxed{(-3, 2)}$.