Let a have magnitude 9.

Question

Let $\vec a$ have magnitude 9. Let $\vec b$ be such that for every $(x,y)\in\mathbb R	imes\mathbb R-\{(0,0)\}$, $x\vec a+y\vec b$ is perpendicular to $6y\vec a-18x\vec b$. Then $|\vec a	imes\vec b|$ is:

Accepted Answer

$(x\vec a+y\vec b)\cdot(6y\vec a-18x\vec b)=0$ for all $x,y$:
$$6xy|\vec a|^2-18xy|\vec b|^2+(6y^2-18x^2)(\vec a\cdot\vec b)=0$$

As an identity in $x,y$, each coefficient vanishes: coefficient of $xy$: $6|\vec a|^2=18|\vec b|^2 \Rightarrow |\vec b|^2=\dfrac{|\vec a|^2}3=27 \Rightarrow |\vec b|=3\sqrt3$; coefficient of $x^2$ (or $y^2$): $\vec a\cdot\vec b=0$.

$$|\vec a\times\vec b| = |\vec a||\vec b|\sin90^\circ = 9\times3\sqrt3 = 27\sqrt3$$

Solutions · 1