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Mathematics examination questions for the next semester.
As shown in the figure, at △ABC, |AB? →? |=3,|AC? →? |= 1, l is the middle vertical line of BC and intersects BC at point D, e is any point on L different from D, and f is any point on line segment AD.

(1) Did you find the advertisement? →(AB? →AC? →? ) value;

(2) judging AE? →(AB? →AC? →? Whether the value of) is constant, and explain the reasons;

(3) If AC⊥BC, ask for AF? →(FB? →? +FC? →? The maximum value of).

Scalar product operation of plane vector

(1) According to the vector parallelogram rule,

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=

1

2

(

ab blood type

+

Alternating current

), so you can bring it in to solve it.

(2)

Automatic exposure device

(

ab blood type

-

Alternating current

) is a constant, just look it up.

Automatic exposure device

=

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+

Delaware

Bring it in, because DE⊥BC, so

Delaware

(

ab blood type

-

Alternating current

)=

Delaware

Civil band

=0, so we can find its value.

(3) Because

Freight bill

+

Football club

=2

disk drive

, so

very

(

Freight bill

+

Football club

)=2

very

disk drive

At this time,

very

and

disk drive

* * * line, can be used.

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Express delivery. set up

very

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, then

disk drive

=( 1-λ)

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So bring it in and you get 2λ( 1-λ).

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2. Find the length of AD according to the conditions.

( 1)AD? →(AB? →AC? →? )= 12(AB? →? +AC? →? )(AB? →AC? →? )= 12(AB? →? 2? AC? →? 2)=4= 12(AB? →? 2? AC? →? 2)=4,

(2)AE? →(AB? →AC? →? )=(AD? →? +DE? →? )? (AB? →AC? →? )=AD? →(AB? →AC? →? )+DE? →CB? →? =4,

∴AE? →(AB? →AC? →? ) is a constant.

(3)∵AC⊥BC,|AB? →? |=3,|AC? →? |= 1;

∴|BC? →? |=22√,|DC? →? |=2√;

∴|AD? →? |= 1+2? √=3√, set AF? →? =λAD? →? What about FD? →? =( 1? λ)AD? →? ;

∴AF? →(FB? →? +FC? →? )=λAD? →(2FD? →? )=λAD? →[2( 1? λ)AD? →? ]=6λ( 1? λ)=? 6(λ? 12)2+32

∴λ= 12,AF? →(FB? →? +FC? →? ) take the maximum value of 32.