Cho hình chóp S.ABCD có đáy ABCD là hình vuông cạnh a, SAbot (ABCD) và SA=asqrt (6) Gọi alpha là góc giữa BC và mặt phẳng (SAB) Khi đó cosalpha bằng

Let the given information be:<br />\begin{itemize}<br /> \item ABCD is a square with side length a.<br /> \item $SA \perp (ABCD)$, so SA is perpendicular to the plane containing ABCD.<br /> \item $SA = a\sqrt{6}$<br />\end{itemize}<br />We want to find the cosine of the angle $\alpha$ between BC and the plane (SAB).<br /><br />To find the angle between BC and the plane (SAB), we need to find the projection of BC onto the plane (SAB). Let's denote the projection of BC onto the plane (SAB) as BC'. Then the angle between BC and BC' is $\alpha$.<br /><br />Since $SA \perp (ABCD)$, we have $SA \perp AB$ and $SA \perp BC$.<br />In $\triangle SAB$, we have $SB^2 = SA^2 + AB^2 = (a\sqrt{6})^2 + a^2 = 6a^2 + a^2 = 7a^2$, so $SB = a\sqrt{7}$.<br />The projection of BC onto AB is 0, since BC is perpendicular to AB.<br />The projection of BC onto SA is 0, since BC is perpendicular to SA.<br />Let's consider the triangle SBC. We have SB = $a\sqrt{7}$, BC = a, SC = $\sqrt{SA^2 + AC^2} = \sqrt{6a^2 + 2a^2} = \sqrt{8a^2} = 2a\sqrt{2}$.<br />Let's use the formula for the cosine of the angle between a line and a plane:<br />$\cos \alpha = \frac{| \vec{BC} \cdot \vec{n} |}{\| \vec{BC} \| \| \vec{n} \|}$, where $\vec{n}$ is a normal vector to the plane (SAB).<br />A normal vector to the plane (SAB) is given by $\vec{n} = \vec{SA} \times \vec{AB}$.<br />$\vec{SA} = (0, 0, a\sqrt{6})$<br />$\vec{AB} = (a, 0, 0)$<br />$\vec{n} = \vec{SA} \times \vec{AB} = (0, a^2\sqrt{6}, 0)$<br />$\vec{BC} = (0, a, 0)$<br />$\vec{BC} \cdot \vec{n} = 0$<br />This means that BC is perpendicular to the normal vector of the plane (SAB), which implies that BC is parallel to the plane (SAB).<br />Therefore, the angle between BC and the plane (SAB) is 0, and $\cos \alpha = 1$.<br /><br />Final Answer: The final answer is $\boxed{1}$