# Answer to Question #22182 in Vector Calculus for Matthew Lind

Question #22182

A triangle has vertices at A(2,3,1), B(-1,1,2), C(1,-2,3). Find a) The length of the median drawn from B to side AC and b)The acute angle which this median makes with side BC.

Expert's answer

a) The length of the median drawn from B to side AC

Let the other point of median is D(Dx, Dy, Dz).

Dx = (Ax + Cx) / 2 = (2 + 1) / 2 = 1.5

Dy = (Ay + Cy) / 2 = (3 - 2) / 2 = 0.5

Dz = (Az + Cz) / 2 = (1 + 3) / 2 = 2.0

So, getting D(1.5, 0.5, 2.0).

BD = sqrt( (1.5 + 1.0)^2 + (0.5 - 1.0)^2 + (2.0 - 2.0)^2 ) = sqrt( 6.25 + 0.25

+ 0 ) = sqrt( 6.5 )

b)The acute angle which this median makes with side BC.

BC = sqrt( (1.0 + 1.0)^2 + (-2.0 - 1.0)^2 + (3.0 - 2.0)^2 ) = sqrt( 4.0 + 9.0 +

1.0 ) = sqrt( 14.0 )

cosDBC = (BDx * BCx + BDy * BCy + BDz * BCz) / (BD * BC) = (2.5 * 2.0 + (-0.5 *

-3.0) + 0 * 1.0) / (sqrt( 6.5 ) * sqrt( 14.0 )) = (5.0 + 1.5 + 0.0) / sqrt (

6.5 * 14.0 ) = 6.5 / sqrt ( 6.5 * 14.0 ) = sqrt ( 6.5 / 14.0 )

Angle DBC = acos( sqrt ( 6.5 / 14.0 ) ).

Let the other point of median is D(Dx, Dy, Dz).

Dx = (Ax + Cx) / 2 = (2 + 1) / 2 = 1.5

Dy = (Ay + Cy) / 2 = (3 - 2) / 2 = 0.5

Dz = (Az + Cz) / 2 = (1 + 3) / 2 = 2.0

So, getting D(1.5, 0.5, 2.0).

BD = sqrt( (1.5 + 1.0)^2 + (0.5 - 1.0)^2 + (2.0 - 2.0)^2 ) = sqrt( 6.25 + 0.25

+ 0 ) = sqrt( 6.5 )

b)The acute angle which this median makes with side BC.

BC = sqrt( (1.0 + 1.0)^2 + (-2.0 - 1.0)^2 + (3.0 - 2.0)^2 ) = sqrt( 4.0 + 9.0 +

1.0 ) = sqrt( 14.0 )

cosDBC = (BDx * BCx + BDy * BCy + BDz * BCz) / (BD * BC) = (2.5 * 2.0 + (-0.5 *

-3.0) + 0 * 1.0) / (sqrt( 6.5 ) * sqrt( 14.0 )) = (5.0 + 1.5 + 0.0) / sqrt (

6.5 * 14.0 ) = 6.5 / sqrt ( 6.5 * 14.0 ) = sqrt ( 6.5 / 14.0 )

Angle DBC = acos( sqrt ( 6.5 / 14.0 ) ).

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