If the coordinates of the points A, B, C, D are (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2), then find the angle between AB and CD.

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Solution

$The given points are A (1, 2, 3), B (4, 5, 7), C (- 4, 3, - 6) and D (2, 9, 2) . We know that the direction ratios of the line joining the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) are x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1} . The direction ratios of A B are (4 - 1), (5 - 2), (7 - 3), i . e . 3, 3, 4 . The direction ratios of C D are [2 - (- 4)], (9 - 3), [2 - (- 6)], i . e . 6, 6, 8 . Let θ be the angle between A B and C D . We have a_{1} = 3, b_{1} = 3, c_{1} = 4 a_{2} = 6, b_{2} = 6, c_{2} = 8 ∴ \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} = \frac{18 + 18 + 32}{\sqrt{9 + 9 + 16} \sqrt{36 + 36 + 64}} = \frac{68}{68} = 1 \Rightarrow θ = 0 ° Thus, the angle between A B and C D measures 0 ° .$

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Q. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

If the co-ordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3, -6) and (2, 9, 2) respectively, then the angle between the lines AB and CD is

A = (1, 2, 3), B = (4, 5, 7), C = (- 4, 3, - 6), D = (2, k, 2)

are four points. If the lines

A B

and

C D

are parallel, then

k =

A = (1, 2, 3), B = (4, 5, 7), C = (- 4, 3, - 6), D = (2, k, 2)

are four points. If the lines

A B

and

C D

are parallel, then

k =

If^i +^j +^k,

2^i + 5^j,

3^i + 2^j - 3^k,

^i - 6^j -^k

respectively are the position vectors of points

A,

B,

C

and

D,

then find the angle between the straight lines

A B

and

C D .

Find whether

- - \to A B

and

- - \to C D

are collinear or not.

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