Multiplying Vectors — Question 9

Question 9

Recover a vector from its dot and cross data

Let $\vec a=(2,0,0)$ be dimensionless. An unknown dimensionless vector b satisfies $\vec a\cdot\vec b=6$ and $\vec a\times\vec b=(0,-8,4)$.

1. Find all three components of b and verify both products.

2. Explain which part of b is fixed by the dot product and which by the cross product.

3. Could the stated cross product instead be $(1,-8,4)$? Explain the consistency condition.

Question 9 - Solution

Recover a vector from its dot and cross data

(a) Convert the product data into scalar equations.

$$2b_x=6,\quad-2b_z=-8,\quad2b_y=4\Rightarrow\boxed{\vec b=(3,2,4)}$$ Writing $\vec b=(b_x,b_y,b_z)$ gives $2b_x=6$, $-2b_z=-8$, and $2b_y=4$. Thus $\vec b=(3,2,4)$. Direct substitution returns dot product 6 and cross product $(0,-8,4)$.

(b) Use the two complementary kinds of information.

$$\vec b_\parallel=(3,0,0),\quad\vec b_\perp=(0,2,4),\quad\vec a\times\vec b_\parallel=\vec0$$ The dot product fixes the component parallel to a, namely $b_x=3$. The cross product fixes the two perpendicular components $b_y=2$ and $b_z=4$. The cross product alone would leave $b_x$ arbitrary. There is no scalar-like operation of dividing by a vector; solve the component equations instead.

(c) Check perpendicularity before solving.

$$\vec a\cdot(\vec a\times\vec b)=0,\quad(2,0,0)\cdot(1,-8,4)=2\ne0$$ No. Every $\vec a\times\vec b$ must be perpendicular to a, so its x component must be zero here. The candidate $(1,-8,4)$ has dot product 2 with a and violates this requirement. Incompatible product data cannot be repaired by choosing a different b.