Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product $[\vec{u} \ \vec{v} \ \vec{w}] = 0$. Alternatively, if $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$, then $\vec{a} = \lambda \vec{b} + \mu \vec{c}$ for some scalars $\lambda$ and $\mu$.
• Perpendicular Vectors: Two vectors $\vec{u}$ and $\vec{v}$ are perpendicular if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.
• Magnitude of a Vector: For $\vec{v} = x\widehat i + y\widehat j + z\widehat k$, $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
• Scalar Triple Product (STP): $[\vec{u} \ \vec{v} \ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$. It is linear in each argument, so $[\vec{a} \ \vec{b} \ \vec{d}] + [\vec{a} \ \vec{c} \ \vec{d}] = \vec{a} \cdot (\vec{b} \times \vec{d}) + \vec{a} \cdot (\vec{c} \times \vec{d}) = \vec{a} \cdot ((\vec{b} + \vec{c}) \times \vec{d})$.

Step-by-Step Solution

Step 1: Simplify the Expression Using Coplanarity We are given that vector $\vec{a}$ is coplanar with vectors $\vec{b}$ and $\vec{c}$. The scalar triple product of three coplanar vectors is zero. Therefore, $[\vec{a} \ \vec{b} \ \vec{c}] = 0$. The expression to evaluate simplifies to: $[\vec{a} \ \vec{b} \ \vec{c}] + [\vec{a} \ \vec{b} \ \vec{d}] + [\vec{a} \ \vec{c} \ \vec{d}] = 0 + [\vec{a} \ \vec{b} \ \vec{d}] + [\vec{a} \ \vec{c} \ \vec{d}]$ $= [\vec{a} \ \vec{b} \ \vec{d}] + [\vec{a} \ \vec{c} \ \vec{d}]$ Using the linearity property of the scalar triple product, this can be rewritten as: $= \vec{a} \cdot (\vec{b} \times \vec{d}) + \vec{a} \cdot (\vec{c} \times \vec{d})$ $= \vec{a} \cdot ((\vec{b} \times \vec{d}) + (\vec{c} \times \vec{d}))$ Using the distributive property of the cross product, this becomes: $= \vec{a} \cdot ((\vec{b} + \vec{c}) \times \vec{d})$

Step 2: Express $\vec{a}$ in terms of $\vec{b}$ and $\vec{c}$ Since $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$, it can be written as a linear combination of them: $\vec{a} = \lambda \vec{b} + \mu \vec{c}$ Substituting the given vectors $\vec{b} = 2\widehat i + \widehat j + \widehat k$ and $\vec{c} = \widehat i - \widehat j + \widehat k$: $\vec{a} = \lambda (2\widehat i + \widehat j + \widehat k) + \mu (\widehat i - \widehat j + \widehat k)$ $\vec{a} = (2\lambda + \mu)\widehat i + (\lambda - \mu)\widehat j + (\lambda + \mu)\widehat k$

Step 3: Use the Perpendicularity Condition to find a relationship between $\lambda$ and $\mu$ We are given that $\vec{a}$ is perpendicular to $\vec{d} = 3\widehat i + 2\widehat j + 6\widehat k$. Thus, their dot product is zero: $\vec{a} \cdot \vec{d} = 0$. $((2\lambda + \mu)\widehat i + (\lambda - \mu)\widehat j + (\lambda + \mu)\widehat k) \cdot (3\widehat i + 2\widehat j + 6\widehat k) = 0$ $3(2\lambda + \mu) + 2(\lambda - \mu) + 6(\lambda + \mu) = 0$ $6\lambda + 3\mu + 2\lambda - 2\mu + 6\lambda + 6\mu = 0$ $14\lambda + 7\mu = 0$ Dividing by 7, we get: $2\lambda + \mu = 0 \implies \mu = -2\lambda$

Step 4: Substitute $\mu$ back into the expression for $\vec{a}$ and use the Magnitude Condition Substitute $\mu = -2\lambda$ into the expression for $\vec{a}$: $\vec{a} = (2\lambda + (-2\lambda))\widehat i + (\lambda - (-2\lambda))\widehat j + (\lambda + (-2\lambda))\widehat k$ $\vec{a} = (0)\widehat i + (3\lambda)\widehat j + (-\lambda)\widehat k$ $\vec{a} = 3\lambda\widehat j - \lambda\widehat k$ Now, use the magnitude condition $|\vec{a}| = \sqrt{10}$: $|\vec{a}| = |3\lambda\widehat j - \lambda\widehat k| = \sqrt{(0)^2 + (3\lambda)^2 + (-\lambda)^2}$ $|\vec{a}| = \sqrt{9\lambda^2 + \lambda^2} = \sqrt{10\lambda^2}$ Equating this to the given magnitude: $\sqrt{10\lambda^2} = \sqrt{10}$ Squaring both sides: $10\lambda^2 = 10$ $\lambda^2 = 1 \implies \lambda = \pm 1$ This gives two possible vectors for $\vec{a}$: If $\lambda = 1$, $\vec{a} = 3\widehat j - \widehat k$. If $\lambda = -1$, $\vec{a} = -3\widehat j + \widehat k$.

Step 5: Calculate the Value of the Simplified Expression We need to calculate $\vec{a} \cdot ((\vec{b} + \vec{c}) \times \vec{d})$. First, find $\vec{b} + \vec{c}$: $\vec{b} + \vec{c} = (2\widehat i + \widehat j + \widehat k) + (\widehat i - \widehat j + \widehat k)$ $\vec{b} + \vec{c} = (2+1)\widehat i + (1-1)\widehat j + (1+1)\widehat k$ $\vec{b} + \vec{c} = 3\widehat i + 0\widehat j + 2\widehat k$ Next, calculate the cross product $(\vec{b} + \vec{c}) \times \vec{d}$: $(\vec{b} + \vec{c}) \times \vec{d} = (3\widehat i + 2\widehat k) \times (3\widehat i + 2\widehat j + 6\widehat k)$ Using the determinant form: $\begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 3 & 0 & 2 \\ 3 & 2 & 6 \end{vmatrix} = \widehat i (0 \cdot 6 - 2 \cdot 2) - \widehat j (3 \cdot 6 - 2 \cdot 3) + \widehat k (3 \cdot 2 - 0 \cdot 3)$ $= \widehat i (0 - 4) - \widehat j (18 - 6) + \widehat k (6 - 0)$ $= -4\widehat i - 12\widehat j + 6\widehat k$ Finally, calculate the dot product $\vec{a} \cdot ((\vec{b} + \vec{c}) \times \vec{d})$. Let's use $\vec{a} = 3\widehat j - \widehat k$ (corresponding to $\lambda = 1$): $\vec{a} \cdot ((\vec{b} + \vec{c}) \times \vec{d}) = (3\widehat j - \widehat k) \cdot (-4\widehat i - 12\widehat j + 6\widehat k)$ $= (0)(-4) + (3)(-12) + (-1)(6)$ $= 0 - 36 - 6$ $= -42$ If we had chosen $\vec{a} = -3\widehat j + \widehat k$ (corresponding to $\lambda = -1$), the result would be $42$. Since $-42$ is an option, this is the correct value.

Common Mistakes & Tips

• Order of Operations: Always simplify the expression first using vector properties, especially coplanarity, before attempting to find the components of $\vec{a}$.
• Sign Errors: Be meticulous with signs during cross product and dot product calculations. The existence of $\pm$ for $\lambda$ means the final answer might have a sign ambiguity unless one option is present.
• Linearity of STP: Remember that $[\vec{a} \ \vec{b} \ \vec{d}] + [\vec{a} \ \vec{c} \ \vec{d}] = \vec{a} \cdot (\vec{b} \times \vec{d}) + \vec{a} \cdot (\vec{c} \times \vec{d}) = \vec{a} \cdot ((\vec{b}+\vec{c}) \times \vec{d})$. This is a powerful simplification.

Summary The problem required us to evaluate an expression involving scalar triple products. We first simplified the expression using the coplanarity of $\vec{a}$ with $\vec{b}$ and $\vec{c}$, which made the first term zero and allowed us to combine the remaining terms into a single scalar triple product. Then, we represented $\vec{a}$ as a linear combination of $\vec{b}$ and $\vec{c}$, and used the perpendicularity and magnitude conditions to determine the possible values of the scalar coefficients and thus the vector $\vec{a}$. Finally, we computed the simplified expression using the derived vector $\vec{a}$ and the sum of vectors $\vec{b}+\vec{c}$.

The final answer is $\boxed{-42}$.