Let $- - \to A B = 3^i -^j +^k$ and $- - \to C D = 3^i + 2^j + 4^k$ are two vectors. The position vectors of the points $\to A$ and $\to C$ are $6^i + 7^j + 4^k$ and $- 9^j + 2^k$ , respectively. Find the position vector of a point $P$ on the line $A B$ and a point $Q$ on the line $C D$ such that $- - \to P Q$ is perpendicular to the vectors $- - \to A B$ and $- - \to C D$ .

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Solution

Equation of the line $A B$ is $\frac{x - 6}{3} = \frac{y - 7}{- 1} = \frac{z - 4}{1}$
Equation of the line $C D$ is $\frac{x - 0}{- 3} = \frac{y + 9}{2} = \frac{z - 2}{4}$
Let $\frac{x - 6}{3} = \frac{y - 7}{- 1} = \frac{z - 4}{1} = λ$ and $\frac{x - 0}{- 3} = \frac{y + 9}{2} = \frac{z - 2}{4} = k$

So, any point on the line $A B$ and $C D$ can be written as $(3 λ + 6, - λ + 7, λ + 4)$ and $(- 3 k, 2 k - 9, 4 k + 2)$ respectively.

Let the coordinates of the points $P$ and $Q$ are $(3 λ + 6, - λ + 7, λ + 4)$ and $(- 3 k, 2 k - 9, 4 k + 2)$ respectively.

$∴ - - \to P Q = (3 λ + 3 k + 6)^i - (λ + 2 k - 16)^j + (λ - 4 k + 2)^k$
$∵ - - \to P Q$ is perpendicular to both $- - \to A B$ and $- - \to C D$ .
$∴ - - \to P Q . - - \to A B = 0$ and $- - \to P Q . - - \to C D = 0$

$\Rightarrow - - \to P Q . - - \to A B = 0 \Rightarrow 9 λ + 9 k + 18 + λ + 2 k - 16 + λ - 4 k + 2 = 0 \Rightarrow 11 λ + 7 k + 4 = 0 . . . (1)$

$\Rightarrow - - \to P Q . - - \to C D = 0 \Rightarrow - 9 λ - 9 k - 18 - 2 λ - 4 k + 32 + 4 λ - 16 k + 8 = 0 \Rightarrow - 7 λ - 29 k + 22 = 0 . . . (2)$

Solving equations (1) and (2), we get
$λ = - 1$ and $k = 1$ .

So, the coordinates of $P$ and $Q$ are $(3, 8, 3)$ and $(- 3, - 7, 6)$ respectively.

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$- - \to A B = 3^i -^j +^k a n d - - \to C D = - 3^i + 2^j + 4^k$ are two vectors.The position vectors of the points A and C are $6^i + 7^j + 4^k a n d - 9^i + 2^j$ ,respectively. Find the position of vector of a point P on the line AB and a point Q on the line CD such that $- - \to P Q$ is perpendicular to $- - \to A B a n d - - \to C D$ both.