Key Concepts and Formulas

• Scalar Triple Product: Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ is zero. This can be computed as the determinant of the matrix formed by their components. $[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$
• Projection of a Vector: The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by $Proj_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. The problem states the projection is $\sqrt{54}$ units, implying this value is positive.

Step-by-Step Solution

Step 1: Understand the given information and the goal. We are given three vectors $\overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k$, $\overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k$, and $\overrightarrow c = 2\widehat i + 3\widehat j + \widehat k$. We are told that these vectors are coplanar, and the projection of $\overrightarrow a$ on $\overrightarrow b$ is $\sqrt {54}$ units. Our goal is to find the sum of all possible values of $\lambda + \mu$.

Step 2: Apply the projection condition to find a relationship between $\lambda$ and $\mu$. The projection of $\overrightarrow a$ on $\overrightarrow b$ is given by $\frac{\overrightarrow a \cdot \overrightarrow b}{|\overrightarrow b|}$. First, calculate the dot product $\overrightarrow a \cdot \overrightarrow b$: $\overrightarrow a \cdot \overrightarrow b = (\lambda)(-2) + (\mu)(4) + (4)(-2) = -2\lambda + 4\mu - 8$ Next, calculate the magnitude of $\overrightarrow b$: $|\overrightarrow b| = \sqrt{(-2)^2 + (4)^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}$ Now, use the given projection value: $\frac{-2\lambda + 4\mu - 8}{\sqrt{24}} = \sqrt{54}$ Multiply both sides by $\sqrt{24}$: $-2\lambda + 4\mu - 8 = \sqrt{54 \times 24}$ Simplify the square root: $\sqrt{54 \times 24} = \sqrt{(9 \times 6) \times (4 \times 6)} = \sqrt{9 \times 4 \times 36} = 3 \times 2 \times 6 = 36$. So, we have: $-2\lambda + 4\mu - 8 = 36$ $-2\lambda + 4\mu = 44$ Divide by 2 to simplify: $-\lambda + 2\mu = 22 \quad \text{(Equation 1)}$

Step 3: Apply the coplanarity condition to find another relationship between $\lambda$ and $\mu$. Since $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar, their scalar triple product is zero. $[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 0$ Expand the determinant along the first row: $\lambda \begin{vmatrix} 4 & -2 \\ 3 & 1 \end{vmatrix} - \mu \begin{vmatrix} -2 & -2 \\ 2 & 1 \end{vmatrix} + 4 \begin{vmatrix} -2 & 4 \\ 2 & 3 \end{vmatrix} = 0$ Calculate the $2 \times 2$ determinants:

• $\begin{vmatrix} 4 & -2 \\ 3 & 1 \end{vmatrix} = (4)(1) - (-2)(3) = 4 + 6 = 10$
• $\begin{vmatrix} -2 & -2 \\ 2 & 1 \end{vmatrix} = (-2)(1) - (-2)(2) = -2 + 4 = 2$
• $\begin{vmatrix} -2 & 4 \\ 2 & 3 \end{vmatrix} = (-2)(3) - (4)(2) = -6 - 8 = -14$ Substitute these values back into the expanded determinant: $\lambda(10) - \mu(2) + 4(-14) = 0$ $10\lambda - 2\mu - 56 = 0$ $10\lambda - 2\mu = 56 \quad \text{(Equation 2)}$

Step 4: Solve the system of linear equations for $\lambda$ and $\mu$. We have the following system of equations:

1. $-\lambda + 2\mu = 22$
2. $10\lambda - 2\mu = 56$ Add Equation 1 and Equation 2 to eliminate $\mu$: $(-\lambda + 2\mu) + (10\lambda - 2\mu) = 22 + 56$ $9\lambda = 78$ $\lambda = \frac{78}{9} = \frac{26}{3}$ Substitute the value of $\lambda$ into Equation 1: $-\left(\frac{26}{3}\right) + 2\mu = 22$ $2\mu = 22 + \frac{26}{3}$ $2\mu = \frac{66 + 26}{3} = \frac{92}{3}$ $\mu = \frac{92}{6} = \frac{46}{3}$

Step 5: Calculate the sum $\lambda + \mu$. Now that we have the values of $\lambda$ and $\mu$, we can find their sum: $\lambda + \mu = \frac{26}{3} + \frac{46}{3} = \frac{26 + 46}{3} = \frac{72}{3} = 24$

Common Mistakes & Tips

• Projection Sign: When a projection is given as "$X$ units," it typically refers to the positive scalar projection. If the problem implied vector projection or allowed for negative scalar projection, the approach might differ.
• Determinant Expansion: Be careful with the signs when expanding determinants, especially when dealing with negative components.
• Solving Linear Equations: Double-check your arithmetic when solving systems of linear equations to avoid errors in finding $\lambda$ and $\mu$.

Summary We used the condition of coplanarity to establish one linear equation involving $\lambda$ and $\mu$ via the scalar triple product. We then used the information about the projection of $\overrightarrow a$ on $\overrightarrow b$ to derive a second linear equation. Solving this system of two linear equations simultaneously yielded the values of $\lambda$ and $\mu$. Finally, we summed these values to obtain the required result.

The final answer is \boxed{24}, which corresponds to option (A).