In triangle ABC, vector AN=1/3 vector NC, P is a point on BN, if vector AP=m vector AB0+2/11 vector AC, what is the value of real number m?

Vector AP=AB+BP=AB+tBN (BP and BN are collinear, so BP=tBN)
=AB+t (AN-AB)
=AB+t (1/4AC-AB)
=(1-T) AB + t/4AC.
Because vector AP=m vector AB+2/11 vector AC,
The comparison coefficient is:1-t=m, t/4=2/11.
Solution m =3/11.

Let P be a point in △ABC, and AP=2 5 AB+1 5 AC, the ratio of △ABP area to △ABC area is ___.

Connect CP and extend, hand over AB to D,
Ze

AP=2
5

AB+1
5

AC=4
5

AD+1
5

AC,
I.e.

CP=4

PD,
Gu

CD =5

PD,
The ratio of △ABP area to △ABC area is 1
5.
Therefore, the answer is:1
5

Connect CP and extend, hand over AB to D,
Ze

AP=2
5

AB+1
5

AC=4
5

AD+1
5

AC,
I.e.

CP=4

PD,
Gu

CD =5

PD,
The ratio of △ABP to △ABC is 1
5.
Therefore, the answer is:1
5

In triangle ABC, AM: AB=1:3, AN: AC=1:4, BN and CM intersect at P. If AB vector=a vector, AC vector=b vector, find AP vector.PS, In triangle ABC, AM: AB=1:3, AN: AC=1:4, BN and CM intersect at P. If AB vector=a vector, AC vector=b vector, find AP vector. PS,

Let BP=xBN, CP=yCM
AC+CP=AP=AB+BP
AC+y (CA+AM)= AB+x (BA+AN)
B+y (-b+1/3a)=a+x (-a+1/4b)
1-Y=x/4
1-X=y/3
X =8/11
Y =9/11
Vector AP=AB+BP=a+8/11(-a+1/4b)=3/11 vector a+2/11 vector b

Given that point P is a point on the plane of triangle ABC, and vector AP=1/3 vector AB + t vector AC, where t is a real number, if point P falls in the triangle, find T norm Given that the point P is a point on the plane of the triangle ABC, and the vector AP=1/3 vector AB+t vector AC, where t is a real number, if the point P falls in the triangle, find the t range, including the application of the linear vector parameter equation in this problem.

Extended AP intersects BC at D, point P falls in triangle ABC,
AP=mAD,0

In triangle ABC, N is a point on AC P is a point on BN, vector AN=1/3NC vector, vector AP=m vector AB+2/11 vector AC, factor m Picture yourself.

I have A, B, P on the line. If vector AB =3 vector BP, then P is the ratio of vector AB. I don't understand the meaning of the last sentence. I have A, B, P on the line. If vector AB =3 vector BP, then P points the ratio of vector AB. I don't understand the meaning of the last sentence. I have A, B, P on the line. If vector AB =3 vector BP, then P points to the ratio of vector AB.

1 When the point P is on the line segment AB,|AB|=3|BP|, i.e. AP=2PB, then: P is the ratio of the component AB: AP/PB=22, when the point P is on the extension line of the line segment AB,|AB|=3|BP|, i.e. AP=-4PB, then: P is the ratio of the component AB: AP/PB=-4----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

In triangle ABC, vector AN=1/3 vector NC, P is a point on BN, if vector AP=m vector AB0+2/11 vector AC, what is the value of real number m?