The angle between the diagonals of the parallelogram formed by the points (1, 2, 3), (-1, -2, -1), (2, 3, 2), (4, 7, 6) is

Question

The angle between the diagonals of the parallelogram formed by the points (1, 2, 3), (-1, -2, -1), (2, 3, 2), (4, 7, 6) is (A)  cos-1 (7) (B)  cos-1 ((7/√155)) (C)  cos-1 ((7/√465)) (D)  cos-1 ((7/465))

Tardigrade Admin · Accepted Answer

D.r's of AC = (1,1,-1), D.r's of BD = (5,9,7) cos θ (5+9-7/√3 . √155 ) ⇒ θ = cos-1 ((7/√465))

Q. The angle between the diagonals of the parallelogram formed by the points $(1, 2, 3), (- 1, - 2, - 1), (2, 3, 2), (4, 7, 6)$ is

$cos^{- 1} (7)$

$cos^{- 1} (\frac{7}{155})$

$cos^{- 1} (\frac{7}{465})$

$cos^{- 1} (\frac{7}{465})$

D.r's of AC = (1,1,-1), D.r's of BD = (5,9,7)
$cos θ \frac{5 + 9 - 7}{3 . 155} \Rightarrow θ = cos^{- 1} (\frac{7}{465})$

Q. The angle between the diagonals of the parallelogram formed by the points (1,2,3),(−1,−2,−1),(2,3,2),(4,7,6) is

cos−1(7)

cos−1(155​7​)

cos−1(465​7​)

cos−1(4657​)

D.r's of AC = (1,1,-1), D.r's of BD = (5,9,7) cosθ3​.155​5+9−7​⇒θ=cos−1(465​7​)

Q. The angle between the diagonals of the parallelogram formed by the points $(1, 2, 3), (- 1, - 2, - 1), (2, 3, 2), (4, 7, 6)$ is

$cos^{- 1} (7)$

$cos^{- 1} (\frac{7}{155})$

$cos^{- 1} (\frac{7}{465})$

$cos^{- 1} (\frac{7}{465})$

D.r's of AC = (1,1,-1), D.r's of BD = (5,9,7)
$cos θ \frac{5 + 9 - 7}{3 . 155} \Rightarrow θ = cos^{- 1} (\frac{7}{465})$