AP DSC School Assistant Math Model Test 2

Concept:Three vectors (position vectors of points) are collinear if the determinant of their components (arranged in a matrix) is zero. This condition checks that the points lie on the same straight line.Explanation:Given vectors: $\hat{i}+2\hat{j}+3\hat{k}$, $\lambda\hat{i}+4\hat{j}+7\hat{k}$, $-3\hat{i}-2\hat{j}-5\hat{k}$.List their components:First vector: (1, 2, 3)Second vector: (λ, 4, 7)Third vector: (-3, -2, -5)For collinearity, the determinant of the 3×3 matrix formed by these components must be zero:$$\begin{vmatrix}1 & 2 & 3 \\ \lambda & 4 & 7 \\ -3 & -2 & -5\end{vmatrix}=0$$Expand the determinant:$$1(4\cdot(-5)-7\cdot(-2)) - 2(\lambda\cdot(-5)-7\cdot(-3)) + 3(\lambda\cdot(-2)-4\cdot(-3)) =0$$Simplify each term:First: $1(-20+14)=1(-6)=-6$Second: $-2(-5\lambda+21)= -2(-5\lambda+21)=10\lambda-42$Third: $3(-2\lambda+12)= -6\lambda+36$Add them: $-6 + 10\lambda -42 -6\lambda +36 =0$Combine like terms: $(10\lambda-6\lambda) + (-6-42+36) = 4\lambda -12 =0$Thus $4\lambda =12$ ⇒ $\lambda =3$.Answer:λ = 3 (Option A)