Review : Matrices & Vectors — Question 6

Question 6 – Solution

Strategy. A cross product encodes both a perpendicular direction and an orientation. Translation affects a plane’s points, not its direction space.

Step 1: Find and verify the normal. Direct calculation gives $$n=u\times v=(1,-1,-1)^T.$$ Both dot products $n\cdot u$ and $n\cdot v$ vanish. The plane equation is $n\cdot(x-p)=0$, or $$\boxed{x_1-x_2-x_3=-4.}$$ The point $p$ satisfies it, and adding any combination of $u,v$ to $p$ leaves its left side unchanged.

Step 2: Compute area and volume. The parallelogram area is $\|n\|=\boxed{\sqrt 3}$. The scalar triple product is $$\boxed{n\cdot w_\tau=\tau-2,\qquad \text{unsigned volume}=|\tau-2|.}$$ For example, $\tau=0$ gives signed volume $-2$ and volume $2$.

Step 3: Prove the independence classification. If $\alpha u+\beta v+\gamma w_\tau=0$, dotting with $n$ gives $\gamma(\tau-2)=0$. For $\tau\ne 2$, this forces $\gamma=0$. The second and third coordinates of $\alpha u+\beta v=0$ then give $\alpha=\beta=0$. Thus the vectors are independent exactly for $\tau\ne 2$. At $\tau=2$, $\boxed{w_2=u+v}$ is a nontrivial dependence, and the volume collapses to zero.

Step 4: Separate orientation from translation. Reversing the order gives $v\times u=-n$. The plane equation becomes $-x_1+x_2+x_3=4$, describing the same plane. Signed volume changes sign, while unsigned volume does not. As a direction, $w_\tau$ lies in the plane’s direction space exactly when $n\cdot w_\tau=0$, or $\tau=2$. As a point with those coordinates, it lies on the translated plane exactly when $\tau-2=-4$, or $\tau=-2$. These are different questions because the plane does not pass through the origin.