Using vectors show that the point $A (- 2, 3, 5)$ , $B (7, 0, - 1)$ $C (- 3, - 2, - 5)$ and $D (3, 4, 7)$ are such that $A B$ and $C D$ intersect at the point $P (1, 2, 3)$ .

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Solution

Given points are

A(-2 , 3 , 5) B(7 , 0 , 1) C(-3 , -2 , -5) D(3 , 4 , 7)

line passing through the points A and B is

$\begin{matrix} ¯ ¯¯ ¯ A + t (¯ ¯¯ ¯ B - ¯ ¯¯ ¯ A) - 2 ˆ i + 3 ˆ j + 5 ˆ k + t (9 ˆ i - 3 ˆ j - 4 ˆ k) \Rightarrow (9 t - 2) ˆ i + (3 - 3 t) ˆ j + (5 - 4 t) ˆ k \to (1) l i n e p a s s i n t h r o u g h p o i n t C a n d D ¯ ¯¯ ¯ C + S (¯ ¯¯¯ ¯ D - ¯ ¯¯ ¯ C) - 3 ˆ i - 2 ˆ j - 5 ˆ k + S (6 ˆ i + 6 ˆ j + 12 ˆ k) (6 S - 3) ˆ i + (6 S - 2) ˆ j + (12 S - 5) ˆ k \to (2) \end{matrix}$

given that equation (1) and (2) intersect each other at a point , so by equating the term , we get

$\begin{matrix} 9 t - 2 = 6 S - 3 \to ˆ i c o e f f i c e n t 3 - 3 t = 6 S - 2 \to ˆ j c o e f f i c e n t 12 t - 5 = - 1 12 t = 4 t = \frac{1}{3} p u t t v a l u e i n ˆ k c o e f f i c e n t 5 - 4 \frac{1}{3} = 12 S - 5 \frac{26}{3} = 12 S S = \frac{13}{18} \end{matrix}$

So , equation og the line will be

$\begin{matrix} (9 \times \frac{1}{3} - 2) ˆ i + (3 - 3 \frac{1}{3}) ˆ j + (5 - 4 \times \frac{1}{3}) ˆ k ˆ i + 2 ˆ j + \frac{11}{3} ˆ k a n d a l s o \frac{4}{3} ˆ i + \frac{7}{3} ˆ j + \frac{11}{3} ˆ k \end{matrix}$

point of intersection (1 , 2 , 3)

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Similar questions

A (- 2, 3, 5)

B (7, 0, - 1)

C (- 3, - 2, - 5)

D (3, 4, 7)

P (1, 2, 3)

Chords $A B$ and $C D$ of a circle intersect each other at point $P$ such that $A P = C P$ .

Show that : $A B = C D$ .

| - - \to A B | = \frac{1}{2} | - - \to C D | a n d A B | | C D .

¯ ¯¯¯¯¯¯ ¯ A C

¯ ¯¯¯¯¯¯¯ ¯ B D

¯ ¯¯¯¯¯¯ ¯ A B = ¯ ¯¯¯¯¯¯ ¯ B C = ¯ ¯¯¯¯¯¯¯ ¯ C D = \frac{1}{3}

¯ ¯¯¯¯¯¯¯ ¯ A D

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