The coordinates of the image of the point P (5, 4, 2) in the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \mu(2\hat{i} + 3\hat{j} - \hat{k})$, where $\lambda$ is a parameter, is

The line is:

$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \mu(2\hat{i} + 3\hat{j} - \hat{k})$

From this:

• Point on the line, $A = (-1, 3, 1)$
• Direction ratios of the line, $\vec{d} = (2, 3, -1)$

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Any point $Q$ on the line can be written as:

$Q = (-1 + 2\mu,\ 3 + 3\mu,\ 1 - \mu)$

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$\vec{PQ} = Q - P$

$= (-6 + 2\mu,\ -1 + 3\mu,\ -1 - \mu)$

Since $\vec{PQ} \perp \vec{d}$, their dot product $= 0$:

$2(-6 + 2\mu) + 3(-1 + 3\mu) + (-1)(-1 - \mu) = 0$

$-12 + 4\mu - 3 + 9\mu + 1 + \mu = 0$

$14\mu - 14 = 0$

$\mu = 1$

The foot of perpendicular is:

$F = (-1 + 2(1),\ 3 + 3(1),\ 1 - 1)$

$= (1, 6, 0)$

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Since $F$ is the midpoint of $P$ and its image $P'(x, y, z)$:

$\dfrac{x + 5}{2} = 1 \implies x = -3$

$\dfrac{y + 4}{2} = 6 \implies y = 8$

$\dfrac{z + 2}{2} = 0 \implies z = -2$

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$P' = (-3,\ 8,\ -2)$