Let | a|=2, | b|=3, angle between them 4.

Question

Let $|\vec a|=2$, $|\vec b|=3$, angle between them $\dfrac\pi4$. Then $|(\vec a+2\vec b)	imes(2\vec a-3\vec b)|^2$ is:

Accepted Answer

$\vec a\cdot\vec b = 2\cdot3\cdot\cos\frac\pi4=3\sqrt2$. Let $\vec p=\vec a+2\vec b$, $\vec q=2\vec a-3\vec b$.

$$|\vec p|^2 = 4+4(9)+4(3\sqrt2)=40+12\sqrt2,\qquad |\vec q|^2 = 4(4)+9(9)-12(3\sqrt2)=97-36\sqrt2$$
$$\vec p\cdot\vec q = 2(4)-6(9)+3\sqrt2 = -46+3\sqrt2$$

$$|\vec p\times\vec q|^2 = |\vec p|^2|\vec q|^2-(\vec p\cdot\vec q)^2 = (40+12\sqrt2)(97-36\sqrt2)-(3\sqrt2-46)^2 = 882$$

Solutions · 1