Find the length of the perpendicular from the point $(5, 4, - 1)$ to the line $\to r =^i + λ (2^i + 9^j + 5^k)$ .

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Solution

$\begin{matrix} \to r =^i + λ (2^i + 9^j + 5^k) \frac{x - 1}{2} = \frac{y - 0}{9} = \frac{z - 0}{5} L e t p o i n t B b e : (2 r + 1, 9 r, 5 r) D i r e c t i o n r a t i o o f A B \equiv (2 r - 4, 9 r - 4, 5 r + 1) 2 (2 r - 4) + 9 (9 r - 4) + 5 (5 r + 1) = 0 110 r = 39 r = \frac{39}{110} L e n g t h o f p e r p e n d i c u l a r = A B = \sqrt{{(2 r - 4)}^{2} + {(9 r - 4)}^{2} + {(5 r + 1)}^{2}} = \sqrt{110 r^{2} + 33 - 78 r} = \sqrt{\frac{{(39)}^{2}}{110} + 33 - \frac{78 \times 39}{110}} = \sqrt{\frac{33 (110) - {(39)}^{2}}{110}} = \frac{\sqrt{33 (110) - {(39)}^{2}}}{\sqrt{110}} = \frac{\sqrt{2109}}{\sqrt{110}} \end{matrix}$

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Similar questions

The length of perpendicular drawn from the point (5,4, -1) to the line $\to r =^i + λ (2^i + 9^j + 5^k)$ is

\to r =^i + 2^j + 3^k + λ (2^i +^j + 4^k)

$\to r = 2^i + 4^j + 5^k + μ (3^i + 4^j + 5^k)$

\to r =^i + 2^j + 3^k + λ (2^i + 3^j + 4^k)

\to r = 2^i + 4^j + 5^k + μ (4^i + 6^j + 8^k)

l_{1}

l_{2}

\to r =^i + 2^j - 4^k + λ (2^i + 3^j + 6^k)

\to r = 3^i + 3^j - 5^k + μ (2^i + 3^j + 6^k)

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