Question

tính góc giữa 2 đường thẳng

\(d:\left\{{}\begin{matrix}x=1+3t\\y=-1+t\\z=2-2t\end{matrix}\right.\) và mặt phẳng \(\left(P\right):x+2y+z+1=0\)

Answer 1

(d): \(\left\{{}\begin{matrix}x=1+3t\\y=-1+t\\z=2-2t\end{matrix}\right.\)

=>VTCP là \(\overrightarrow{a}=\left(3;1;-2\right)\)

(P): x+2y+z+1=0

=>VTPT là \(\overrightarrow{b}=\left(1;2;1\right)\)

\(sin\widehat{\left(d\right);\left(P\right)}=\dfrac{\left|3\cdot1+1\cdot2+\left(-2\right)\cdot1\right|}{\sqrt{3^2+1^2+\left(-2\right)^2}\cdot\sqrt{1^2+2^2+1^2}}\)

=>\(sin\widehat{\left(d\right);\left(P\right)}=\dfrac{3}{\sqrt{14}\cdot\sqrt{6}}=\dfrac{3}{\sqrt{84}}=\dfrac{\sqrt{21}}{14}\)

=>\(\widehat{\left(d\right);\left(P\right)}\simeq19^06'\)