Find the vector equation of a line passing through the point $A$ whose position vector is $3 ¯ i + ¯ j - ¯ ¯ ¯ k$ and which is parallel to the vector $2 ¯ i - ¯ j + 2 ¯ ¯ ¯ k$ .If $P$ is a point of this line such that $A P = 15$ ,find the position vector of $P$ .

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Solution

Given: $A (3 \to i + \to j - \to k)$ and $P (2 \to i - \to j + 2 \to k)$

The equation of line passing through $(3 \to i + \to j - \to k)$ and parallel to $(2 \to i - \to j + 2 \to k)$

is $\to r = (3 \to i + \to j - \to k) + λ (2 \to i - \to j + 2 \to k)$

Direction ratio of $A P = (a - 3, b - 1, c + 1)$ , $A P$ is parallel to $2 \to i - \to j + 2 \to k$ then

$\frac{a - 3}{2} = \frac{b - 1}{- 1} = \frac{c + 1}{2} = t$ (constant)

$A P = 15$

$\sqrt{{(a - 3)}^{2} + {(b - 1)}^{2} + {(c + 1)}^{2}} = 15$

$\Rightarrow \sqrt{{(2 t + 3 - 3)}^{2} + {(- t + 1 - 1)}^{2} + {(2 t - 1 + 1)}^{2}} = 15$

$\Rightarrow \sqrt{4 t^{2} + t^{2} + 4 t^{2}} = 15$

$\Rightarrow \sqrt{9 t^{2}} = 15$

$\Rightarrow t = 5$

$\begin{matrix} a = 2 \times 5 + 3 = 13 b = - 5 + 1 = 4 c = 2 \times 5 - 1 = 9 \end{matrix}$

Position vector of $P = 13 \to i + 4 \to j + 9 \to k$

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