The distance between the line r-2i-2j-3k-i-j-4k- and the plane r- i-5j-k-5 is

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Perpendicular Distance of a Point from a Plane

The distance ...

Question

The distance between the line $\to r = 2 \to i - 2 \to j + 3 \to k + λ (\to i - \to j + 4 \to k)$ and the plane $\to r \cdot (\to i + 5 \to j + \to k) = 5$ is

\frac{10}{3 \sqrt{3}}

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\frac{10}{9}

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\frac{10}{3}

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Solution

The correct option is A $\frac{10}{3 \sqrt{3}}$
Clearly given line and plane are parallel
Equation of line perpendicular to both is given by,
$\frac{x - 2}{1} = \frac{y + 2}{5} = \frac{z - 3}{1} = k$ (say) ... $(1)$
So any point on this line is given by, $P (k + 2, 5 k - 2, k + 3)$
Distance between given line and plane will be along line $(1)$
To know intersection of $(1)$ and plane we should have,
$k + 2 + 5 (5 k - 2) + k + 3 = 5 \Rightarrow k = \frac{10}{27}$
Thus $P = (\frac{10}{27} - 2, \frac{50}{27} - 2, \frac{10}{27} + 3)$
Hence shortest distance is $= \sqrt{(\frac{10}{27})^{2} + (\frac{50}{27})^{2} + (\frac{10}{27})^{2}} = \sqrt{\frac{2700}{27^{2}}} = \frac{10}{3 \sqrt{3}}$
Hence, option 'A' is correct.

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The distance between the line →r=2→i−2→j+3→k+λ(→i−→j+4→k) and the plane →r⋅(→i+5→j+→k)=5 is

The distance between the line $\to r = 2 \to i - 2 \to j + 3 \to k + λ (\to i - \to j + 4 \to k)$ and the plane $\to r \cdot (\to i + 5 \to j + \to k) = 5$ is