题目内容
已知空间三点A(-2,0,2),B(-1,1,2),C(-3,0,4),设a=
,b=
.若向量ka+b与ka-2b互相垂直,则k的值是( )
| A.2 | B. |
| C. | D. |
D
a=(-1+2, 1-0,2-2)=(1,1,0),b=(-3+2,0-0,4-2)=(-1,0,2).
ka+b=(k,k,0)+(-1,0,2)=(k-1,k,2),
ka-2b=(k,k,0)-(-2,0,4)=(k+2,k,-4).
∵ (ka+b)⊥(ka-2b),
∴ (k-1,k,2)·(k+2, k,-4)=(k-1)(k+2)+k2-8=0,即2k2+k-10=0,
∴k=
或k=2.选D.
ka+b=(k,k,0)+(-1,0,2)=(k-1,k,2),
ka-2b=(k,k,0)-(-2,0,4)=(k+2,k,-4).
∵ (ka+b)⊥(ka-2b),
∴ (k-1,k,2)·(k+2, k,-4)=(k-1)(k+2)+k2-8=0,即2k2+k-10=0,
∴k=
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