The distance of the point (7, –2, 11) from the line along the line is : [2024]

Question

The distance of the point (7, –2, 11) from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ is : [2024]

Smart Achievers · Accepted Answer

(4)

Let A be the point (7, –2, 11).

Equation of line AB is $\frac{x - 7}{2} = \frac{y + 2}{- 3} = \frac{z - 11}{6} = λ$

$\Rightarrow x = 2 λ + 7, y = - 3 λ - 2 and z = 6 λ + 11$

Point B is of the form $(2 λ + 7, - 3 λ - 2, 6 λ + 11)$ .

Now, point B lies on the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$

$\Rightarrow \frac{2 λ + 7 - 6}{1} = \frac{- 3 λ - 2 - 4}{0} = \frac{6 λ + 11 - 8}{3}$

$\Rightarrow - 3 λ - 6 = 0 \Rightarrow λ = - 2$

Thus point B is (3, 4, –1).

$∴$ Required distance = $A B = \sqrt{4^{2} + {(- 6)}^{2} + 12^{2}}$

. $= \sqrt{16 + 36 + 144} = \sqrt{196} = 14 units$ .

Solution

Q.
The distance of the point (7, –2, 11) from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ is : [2024]

1 18

2 12

3 21

4 14

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