ln a quadrilateral `ABCD, vec(AC)` is the bisector of the `(vec(AB) ^^ vec(AD))` which is `(2pi)/3 , 15|vec(AC)| = 2 |vec(AB)| = 5|vec(AD)|` then `cos (vec(BA) ^^ vec(CD))` is

Options

A. `cos^(-1)""sqrt14/(7sqrt2)`
B. `cos^(-1)""sqrt21/(7sqrt3)`
C. `cos^(-1)""2/(sqrt7)`
D. `cos^(-1)""(2sqrt7)/14`

Answer

c

Explanation

Let `|vec(AC)|= lambda gt 0 `
Then from ` 15 |vec(AC)| = 3|vec(AB)|= 5 |vec(AD)| |vec(AB|= 5lambda`
Let `tehta` the angle between `vec(BA) and vec(CD)`. Then
` cos theta = (vec(BA).vec(CD))/(|vec(BA)||vec(CD)|) =- (-vecb. (vecd - vecc))/(|vecb||vecd - vecc|)` (i)
Now ` - vecb . (vecd-vecc) = vecv .vecc - vecb.vecd`
`|vecb||vecc| cos pi/3 - |vecb||vecd|cos (2pi)/3`
` (5 lambda) (lambda) 1/2 + (5lambda) (3lambda) 1/2`
` (5 lambda^(2) + 15 lambda^(2))/2`
`10 lambda^(2)`
Denominator of (i) = `|vecb||vecd-vecc|`
now `|vecd-vecc|^(2) = |vecd|^(2) + |vecc|^(2) -2vecc.vecd`
`10 lambda^(2) - 3lambda^(2)`
` 7 lambda^(2)`
Denominator of (i) = `(5 lambda) (sqrt7 lambda) = 5 sqrt7 lambda^(2)`
` cos theta = (10lambda^(2))/(5 sqrt7 lambda^(2))= 2/sqrt7`