Find the distance of the point $(2, 12, 5)$ from the point of intersection of the line $\to r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)$ and the plane $\to r (^i - 2^j +^k) = 0$

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Solution

consider the equation of the line
$\to r = 2^i + 4^j + 2^k + λ (3^i + 4^j + 2 k)$ ---- $(i)$
Let, $\to r = x^i + y^j + z^k$
Therefore,
$x^i + y^j + z^k = (2 + 3 λ)^i + (4 + 4 λ)^j + (2 + 2 λ)^k$
Where, $x = 2 + 3 λ, y = 4 + 4 λ, z = 2 + 2 λ$ ---- $(i i i)$
And
$\to r (^i - 2^j +^k) = 0$ --- $(i i)$
Hence, $(x^i + y^j + z^k) (^i - 2^j +^k) = 0$
$x - 2 y + z = 0$ --- $(i v)$
and the line $(i)$ and line $(i i)$ intersect,
Therefore, from $(i i i)$ and $(i v)$
+ $2 + 3 λ - 2 (4 + 4 λ) + 2 + 2 λ = 0$
$2 + 3 λ - 8 - 8 λ + 2 + 2 λ = 0$
$- 3 λ = 4$
$λ = - \frac{4}{3}$
put in $(i i i)$
$x = 2 + 3 (- \frac{4}{3}), y = 4 + 4 (- \frac{4}{3}), z = 2 + 2 (- \frac{4}{3})$
Therefore, $x = - 2, y = - \frac{4}{3}, z = - \frac{2}{3}$
Distance of point $(- 2, - \frac{4}{3}, - \frac{2}{3})$ from $(2, 12, 5)$
$= \sqrt{{(2 + 2)}^{2} + {(12 + \frac{4}{3})}^{2} + {(5 + \frac{2}{3})}^{2}}$
$= \sqrt{4^{2} + {(\frac{40}{3})}^{2} + {(\frac{17}{3})}^{2}}$
$= \sqrt{16 + \frac{1600}{9} + \frac{289}{9}}$
$= \sqrt{\frac{144 + 1600 + 289}{9}}$
$= \sqrt{\frac{2033}{9}}$

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