Question

find the point of intersection of the lines AB and CD where A=(7, -6, 1) B=(17, -18, -3)and C=(1, 4 -5) and D=(3, -4, 11)​

Answer 1

Given,

• A = (7, -6, 1)
• B = (17, -18, -3)
• C = (1, 4 -5)
• D = (3, -4, 11)

Let,

The point 'p' divides AB in the ratio λ:1 internally

• A = (7, -6, 1)
• B = (17, -18, -3)

$\qquad \displaystyle \sf P = \bigg( \frac{17 λ+ 7}{λ + 1} , \: \frac{ - 18λ - 6}{λ + 1} , \: \frac{ - 3λ + 1}{λ + 1} \bigg) \: - - - 1$

and the point 'p' divides CD in the ratio μ:1 internally

• C = (1, 4 -5)
• D = (3, -4, 11)

$\qquad \displaystyle \sf P^{'} = \bigg( \frac{3μ + 1}{ μ+ 1} , \: \frac{ - 4μ + 6}{μ + 1} , \: \frac{ 11μ + 5}{μ+ 1} \bigg) \: - - - 2$

Equating 1 and 2,

$\qquad \displaystyle \sf \frac{17λ + 7}{λ + 1} = \frac{3μ + 1}{μ + 1}$

$\qquad \displaystyle \sf 17 λμ+ 7μ + 17λ + 7= 3λμ + 3μ+ λ + 1$

$\qquad \displaystyle \sf 14 λμ+ 4μ + 16λ + 6= 0 \: - - - 3$

and

$\qquad \displaystyle \sf \frac{ - 18λ - 6}{λ + 1} = \frac{ - 4μ + 4}{μ + 1}$

$\qquad \displaystyle \sf (- 18λ - 6)(μ + 1)= (λ + 1)( - 4μ + 4)$

$\qquad \displaystyle \sf - 18μλ - 6μ - 18λ - 6 = - 4μλ + 4λ -4μ + 4$

$\qquad \displaystyle \sf - 14μλ -2μ- 22λ - 10 = 0$

$\qquad \displaystyle \bf - 7μλ -μ- 11λ - 5= 0$

Hence,

∴ The point of intersection is -7μλ-μ-11λ-5=0

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