Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ to the plane $\vec{r} . (2 \hat{i} + \hat{j} + 3 \hat{k}) - 26 = 0$ . Also find image of P in the plane.

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Solution

Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) in the plane $\vec{r} . (2 \hat{i} + \hat{j} + 3 \hat{k}) - 26 = 0 or 2 x + y + 3 z - 26 = 0$ .
Then, PM is the normal to the plane. So, the direction ratios of PM are proportional to 2, 1, 3.
Since PM passes through P(2, 3, 4) and has direction ratios proportional to 2, 1, 3, so the equation of PM is
$\frac{x - 2}{2} = \frac{y - 3}{1} = \frac{z - 4}{3} = r (say)$
Let the coordinates of M be (2r + 2, r + 3, 3r + 4). Since M lies in the plane 2x + y + 3z − 26 = 0,so
$2 (2 r + 2) + r + 3 + 3 (3 r + 4) - 26 = 0 \Rightarrow 4 r + 4 + r + 3 + 9 r + 12 - 26 = 0 \Rightarrow 14 r - 7 = 0 \Rightarrow r = \frac{1}{2}$
Therefore, the coordinates of M are $(2 r + 2, r + 3, 3 r + 4) = (2 \times \frac{1}{2} + 2, \frac{1}{2} + 3, 3 \times \frac{1}{2} + 4) = (3, \frac{7}{2}, \frac{11}{2})$
Thus, the position vector of the foot of perpendicular are $3 \hat{i} + \frac{7}{2} \hat{j} + \frac{11}{2} \hat{k}$ .
Now,
Length of the perpendicular from P on to the given plane
$= |\frac{2 \times 2 + 1 \times 3 + 3 \times 4 - 26}{\sqrt{4 + 1 + 9}}| = \frac{7}{\sqrt{14}} = \sqrt{\frac{7}{2}} units$
Let $Q (x_{1}, y_{1}, z_{1})$ be the image of point P in the given plane.
Then, the coordinates of M are $(\frac{x_{1} + 2}{2}, \frac{y_{1} + 3}{2}, \frac{z_{1} + 4}{2})$ .
But, the coordinates of M are $(3, \frac{7}{2}, \frac{11}{2})$ .
$∴ (\frac{x_{1} + 2}{2}, \frac{y_{1} + 3}{2}, \frac{z_{1} + 4}{2}) = (3, \frac{7}{2}, \frac{11}{2}) \Rightarrow \frac{x_{1} + 2}{2} = 3, \frac{y_{1} + 3}{2} = \frac{7}{2}, \frac{z_{1} + 4}{2} = \frac{11}{2} \Rightarrow x_{1} = 4, y_{1} = 4, z_{1} = 7$
Thus, the coordinates of the image of the point P in the given plane are (4, 4, 7).

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