(−1,3,6),B(−6,2,6),C(−3,7,10).
1)
AB
=(−6+1,2−3,6−6)=(−5,−1,0)
AB
=−5
i
−
j
,∣
AB
∣=
25+1
=
26
AC
=(−3+1,7−3,10−6)=(−2,4,4)
AC
=−2
i
+4
j
+4
k
,∣
AC
∣=
4+16+16
=
36
=6
\begin{gathered}2)\; \; \overline {AB}\cdot \overline {AC}=10-4+0=6cos\varphi =\frac{\overline {AB}\cdot \overline {AC}}{|\overline {AB}|\cdot |\overline {AC}|} =\frac{6}{\sqrt{26}\cdot 6}=\frac{1}{\sqrt{26}}varphi =arccos\frac{1}{\sqrt{26}}\end{gathered}
2)
AB
⋅
AC
=10−4+0=6
cosφ=
∣
AB
∣⋅∣
AC
∣
AB
⋅
AC
=
26
⋅6
6
=
26
1
φ=arccos
26
1
\begin{gathered}3)\; \; A(x-x_0)+B(y-y_0)+C(z-z_0)=0-5\cdot (x+3)-1\cdot (y-7)+0\cdot (z-10)=0-5x-y-8=0pi :\; \; 5x+y+8=0\end{gathered}
3)A(x−x
0
)+B(y−y
0
)+C(z−z
0
)=0
−5⋅(x+3)−1⋅(y−7)+0⋅(z−10)=0
−5x−y−8=0
π:5x+y+8=0
ну вроде так
