Trong không gian Oxyz, cho tam giác MNP có M(2; 3; 4), N(-1; 2; 3) và P(3; 2; -3), u→=2MN→−NP→\overset{\rightarrow}{u} = 2\overset{\rightarrow}{MN} - \overset{\rightarrow}{NP}u→=2MN→−NP→ có tọa độ là
(-10; -2; 4).
(-2; -2; -4).
(10; 2; -4).
(5; 1; -2).
Đáp án: <p>(-10; -2; 4).</p>
Phương pháp giải:
+ AB→=(xB−xA;yB−yA;zB−zA)\overrightarrow {AB} = \left( {{x_B} - {x_A};{y_B} - {y_A};{z_B} - {z_A}} \right)AB=(xB−xA;yB−yA;zB−zA).
+ a→−b→=(xa−xb;ya−yb;za−zb)\overrightarrow a - \overrightarrow b = ({x_a} - {x_b};{y_a} - {y_b};{z_a} - {z_b})a−b=(xa−xb;ya−yb;za−zb).
+ ka→=(kxa;kya;kza)k\overrightarrow a = (k{x_a};k{y_a};k{z_a})ka=(kxa;kya;kza).
Lời giải chi tiết:
MN→=(−1−2;2−3;3−4)=(−3;−1;−1)⇒2MN→=(−6;−2;−2)\left. \overset{\rightarrow}{MN} = ( - 1 - 2;2 - 3;3 - 4) = ( - 3; - 1; - 1)\Rightarrow 2\overset{\rightarrow}{MN} = ( - 6; - 2; - 2) \right.MN→=(−1−2;2−3;3−4)=(−3;−1;−1)⇒2MN→=(−6;−2;−2).
NP→=(3+1;2−2;−3−3)=(4;0;−6)\overset{\rightarrow}{NP} = (3 + 1;2 - 2; - 3 - 3) = (4;0; - 6)NP→=(3+1;2−2;−3−3)=(4;0;−6).
u→=2MN→−NP→=(−6−4;−2−0;−2+6)=(−10;−2;4)\overset{\rightarrow}{u} = 2\overset{\rightarrow}{MN} - \overset{\rightarrow}{NP} = ( - 6 - 4; - 2 - 0; - 2 + 6) = ( - 10; - 2;4)u→=2MN→−NP→=(−6−4;−2−0;−2+6)=(−10;−2;4).
