Let Q be the orgin, and M and N be the points on the lines and respectively such that MN is the shortest distance between the given lines. Then is equal to __________. [2024]

Question

Let O be the orgin, and M and N be the points on the lines $\frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3}$ and $\frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9}$ respectively such that MN is the shortest distance between the given lines. Then $\vec{O M} \cdot \vec{O N}$ is equal to __________. [2024]

Smart Achievers · Accepted Answer

(9)

Let, $L_{1} : \frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3} = λ \in R$

$L_{2} : \frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9} = μ \in R$

M is a point on $L_{1}$ .

So, $M = (4 λ + 5, λ + 4, 3 λ + 5)$

and N is a point on $L_{2}$ .

So, $N = (12 μ - 8, 5 μ - 2, 9 μ - 11)$

The direction ratios of MN is

$(12 μ - 4 λ - 13, 5 μ - λ - 6, 9 μ - 3 λ - 16)$

$∵ M N ⊥ L_{1} and M N ⊥ L_{2}$

$∴ 4 (12 μ - 4 λ - 13) + 5 μ - λ - 6 + 3 (9 μ - 3 λ - 16) = 0$

$\Rightarrow 48 μ - 16 λ - 52 + 5 μ - λ - 6 + 27 μ - 9 λ - 48 = 0$

$\Rightarrow 80 μ - 26 λ - 106 = 0 \Rightarrow 40 μ - 13 λ - 53 = 0$ ... (i)

Also,

$12 (12 μ - 4 λ - 13) + 5 (5 μ - λ - 6) + 9 (9 μ - 3 λ - 16) = 0$

$144 μ - 48 λ - 156 + 25 μ - 5 λ - 30 + 81 μ - 27 λ - 144 = 0$

$\Rightarrow 250 μ - 80 λ - 330 = 0$ $\Rightarrow 25 μ - 8 λ - 33 = 0$ ... (ii)

On solving equation (i) and (ii), we get $μ = 1$ and $λ = - 1$

M = (1, 3, 2) and N = (4, 3, –2)

$∴ \vec{O M} = \hat{i} + 3 \hat{j} + 2 \hat{k} and \vec{O N} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$

$\vec{O M} \cdot \vec{O N}$ = 4 + 9 – 4 = 9.

Solution

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Solution

Q. Let O be the orgin, and M and N be the points on the lines x–54=y–41=z–53 and x+812=y+25=z+119 respectively such that MN is the shortest distance between the given lines. Then OM→·ON→ is equal to __________. [2024]

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