Key Concepts and Formulas

1. Scalar Triple Product: For vectors $\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$, $\vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}$, and $\vec{c} = c_x \hat{i} + c_y \hat{j} + c_z \hat{k}$, the scalar triple product is given by $(\vec{a} \times \vec{b}) \cdot \vec{c}$, which can be computed as the determinant of the matrix formed by their components: $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix}$
2. Unit Vector: A unit vector $\vec{d}$ in the direction of a non-zero vector $\vec{v}$ is $\vec{d} = \frac{\vec{v}}{|\vec{v}|}$.
3. Dot Product: The dot product of two vectors $\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$ and $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ is $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$.

Step-by-Step Solution

Step 1: Understand the Goal The problem asks us to find the value of the scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$. We are given vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ (with an unknown component $x$ in $\vec{c}$). We are also given a condition involving a unit vector $\vec{d}$ that will help us determine the value of $x$.

Step 2: Identify the Given Vectors We have:

• $\vec{a}=\hat{i}+\hat{j}+\hat{k}$
• $\vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$
• $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}$

Step 3: Determine the Vector $\vec{b}+\vec{c}$

• Reasoning: The problem states that $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$. To find $\vec{d}$, we first need to compute the vector sum $\vec{b}+\vec{c}$.
• Calculation: $\vec{b}+\vec{c} = (2 \hat{i}+4 \hat{j}-5 \hat{k}) + (x \hat{i}+2 \hat{j}+3 \hat{k})$ $\vec{b}+\vec{c} = (2+x) \hat{i} + (4+2) \hat{j} + (-5+3) \hat{k}$ $\vec{b}+\vec{c} = (x+2) \hat{i} + 6 \hat{j} - 2 \hat{k}$

Step 4: Calculate the Magnitude of $\vec{b}+\vec{c}$

• Reasoning: To find the unit vector $\vec{d}$, we need to divide the vector $\vec{b}+\vec{c}$ by its magnitude.
• Calculation: The magnitude of a vector $p\hat{i}+q\hat{j}+r\hat{k}$ is $\sqrt{p^2+q^2+r^2}$. $|\vec{b}+\vec{c}| = \sqrt{(x+2)^2 + (6)^2 + (-2)^2}$ $|\vec{b}+\vec{c}| = \sqrt{(x+2)^2 + 36 + 4}$ $|\vec{b}+\vec{c}| = \sqrt{(x+2)^2 + 40}$

Step 5: Express the Unit Vector $\vec{d}$

• Reasoning: Using the definition of a unit vector, we can write $\vec{d}$ in terms of $x$.
• Calculation: $\vec{d} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|} = \frac{(x+2) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{(x+2)^2 + 40}}$

Step 6: Use the Condition $\vec{a} \cdot \vec{d}=1$ to Solve for $x$

• Reasoning: This condition provides an equation involving $x$, which we can solve.
• Calculation: $\vec{a} \cdot \vec{d} = (\hat{i}+\hat{j}+\hat{k}) \cdot \left(\frac{(x+2) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{(x+2)^2 + 40}}\right) = 1$ Performing the dot product: $\frac{1 \cdot (x+2) + 1 \cdot (6) + 1 \cdot (-2)}{\sqrt{(x+2)^2 + 40}} = 1$ Simplify the numerator: $\frac{x+2+6-2}{\sqrt{(x+2)^2 + 40}} = 1$ $\frac{x+6}{\sqrt{(x+2)^2 + 40}} = 1$ To solve for $x$, we square both sides. Important: Before squaring, we must ensure that both sides are non-negative. The right side, being a square root, is always non-negative. Therefore, the left side, $x+6$, must also be non-negative, meaning $x+6 \ge 0$, or $x \ge -6$. $(x+6)^2 = \left(\sqrt{(x+2)^2 + 40}\right)^2$ $x^2 + 12x + 36 = (x+2)^2 + 40$ $x^2 + 12x + 36 = (x^2 + 4x + 4) + 40$ $x^2 + 12x + 36 = x^2 + 4x + 44$ Subtract $x^2$ from both sides: $12x + 36 = 4x + 44$ Rearrange terms: $12x - 4x = 44 - 36$ $8x = 8$ $x = 1$
• Verification: Our solution $x=1$ satisfies the condition $x \ge -6$.

Step 7: Substitute the Value of $x$ into $\vec{c}$

• Reasoning: Now that we have found $x$, we can write the complete vector $\vec{c}$.
• Calculation: With $x=1$, vector $\vec{c}$ becomes: $\vec{c} = 1 \hat{i} + 2 \hat{j} + 3 \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

Step 8: Calculate the Scalar Triple Product $(\vec{a} \times \vec{b}) \cdot \vec{c}$

• Reasoning: The scalar triple product can be efficiently calculated using the determinant of the matrix formed by the components of the three vectors.
• Calculation: The components of the vectors are:
  • $\vec{a} = (1, 1, 1)$
  • $\vec{b} = (2, 4, -5)$
  • $\vec{c} = (1, 2, 3)$ The scalar triple product is: $(\vec{a} \times \vec{b}) \cdot \vec{c} = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{vmatrix}$ Expand the determinant along the first row: $= 1 \begin{vmatrix} 4 & -5 \\ 2 & 3 \end{vmatrix} - 1 \begin{vmatrix} 2 & -5 \\ 1 & 3 \end{vmatrix} + 1 \begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix}$ $= 1((4)(3) - (-5)(2)) - 1((2)(3) - (-5)(1)) + 1((2)(2) - (4)(1))$ $= 1(12 - (-10)) - 1(6 - (-5)) + 1(4 - 4)$ $= 1(12 + 10) - 1(6 + 5) + 1(0)$ $= 1(22) - 1(11) + 0$ $= 22 - 11$ $= 11$

Common Mistakes & Tips

• Squaring Equations: Always check the sign of both sides of an equation before squaring, especially when dealing with square roots. This prevents introducing extraneous solutions.
• Determinant Calculation: Be careful with the signs when expanding determinants, particularly with the alternating signs ($+,-,+$ for a 3x3 determinant expanded along the first row).
• Vector Component Addition: Ensure that corresponding components are added correctly when finding the sum of vectors.

Summary The problem required us to find the scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$. We first used the given information about the unit vector $\vec{d}$ and the dot product $\vec{a} \cdot \vec{d}=1$ to solve for the unknown component $x$ in vector $\vec{c}$. Once $x$ was determined to be 1, we substituted it back into $\vec{c}$ and then computed the scalar triple product using the determinant method.

The final answer is $\boxed{11}$.