To find the value of $\alpha$ such that the vectors $\mathbf{a} = (40, -19, 14)$, $\mathbf{b} = (42, 8, 22)$, and $\mathbf{c} = (\alpha, -135, -40)$ are coplanar, we need to use the condition that the scalar triple product of the vectors is zero.

The scalar triple product is defined as:

$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$

This gives us a determinant form:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 40 & -19 & 14 \\ 42 & 8 & 22 \\ \alpha & -135 & -40 \end{vmatrix} = 0$$ We will compute this determinant to find $$\alpha$$. ### Step 1: Expand the determinant $$\det \begin{vmatrix} 40 & -19 & 14 \\ 42 & 8 & 22 \\ \alpha & -135 & -40 \end{vmatrix}$$ Expanding this determinant: $$= 40 \cdot \det \begin{vmatrix} 8 & 22 \\ -135 & -40 \end{vmatrix} - (-19) \cdot \det \begin{vmatrix} 42 & 22 \\ \alpha & -40 \end{vmatrix} + 14 \cdot \det \begin{vmatrix} 42 & 8 \\ \alpha & -135 \end{vmatrix}$$ ### Step 2: Compute the 2x2 determinants Now we compute the individual 2x2 determinants. 1. $$\det \begin{vmatrix} 8 & 22 \\ -135 & -40 \end{vmatrix} = 8(-40) - 22(-135) = -320 + 2970 = 2650$$ 2. $$\det \begin{vmatrix} 42 & 22 \\ \alpha & -40 \end{vmatrix} = 42(-40) - 22(\alpha) = -1680 - 22\alpha$$ 3. $$\det \begin{vmatrix} 42 & 8 \\ \alpha & -135 \end{vmatrix} = 42(-135) - 8(\alpha) = -5670 - 8\alpha$$ ### Step 3: Substitute back into the expansion Substitute these values back into the determinant expansion: $$40(2650) - (-19)(-1680 - 22\alpha) + 14(-5670 - 8\alpha) = 0$$ Simplify: $$106000 - 19(1680 + 22\alpha) + 14(-5670 - 8\alpha) = 0$$ $$106000 - 31920 - 418\alpha - 79380 - 112\alpha = 0$$ $$-4520 - 530\alpha = 0$$ ### Step 4: Solve for $$\alpha$$ Now, solve for $$\alpha$$: $$530\alpha = -4520$$ $$\alpha = \frac{-4520}{530} = -8.53$$ So, the value of $$\alpha$$ is approximately: $$\alpha \approx -8.53$$ Would you like further details on any step, or a clearer explanation of the scalar triple product? --- Here are 5 related questions: 1. How does the scalar triple product determine coplanarity of vectors? 2. Can vectors in higher dimensions also use the scalar triple product? 3. How does changing one vector affect the coplanarity condition? 4. What other methods exist for determining if vectors are coplanar? 5. What are the geometric implications of a scalar triple product being zero? **Tip:** When using determinants to solve problems, practice expanding 3x3 determinants for accuracy and speed!